1.8.2: Enthalpy
There is considerable merit in identifying an extensive property of a closed system called the enthalpy, \(\mathrm{H}\). The enthalpy of a closed system is a state variable and defined by equation (a).
\[\mathrm{H}=\mathrm{U}+\mathrm{p} \, \mathrm{V} \nonumber \]
We identify a given state by the symbol I having enthalpy \(\mathrm{H}[\mathrm{I}]\), energy \(\mathrm{U}[\mathrm{I}]\) and volume \(\mathrm{V}[\mathrm{I}]\) at pressure \(\mathrm{p}\).
\[\mathrm{H}[\mathrm{I}]=\mathrm{U}[\mathrm{I}]+\mathrm{p} \, \mathrm{V}[\mathrm{I}] \nonumber \]
This system is displaced to a neighbouring state such that the differential change in enthalpy is \(\mathrm{dH}\). Using equation (a),
\[\mathrm{dH}=\mathrm{dU}+\mathrm{p} \, \mathrm{dV}+\mathrm{V} \, \mathrm{dp} \nonumber \]
But according to the first law of thermodynamics, the differential change in thermodynamic energy \(\mathrm{dU}\) is given by ‘\(q-p \, d V\)’ where \(\mathrm{q}\) is the heat accompanying the change. Then,
\[\mathrm{dH}=\mathrm{q}-\mathrm{p} \, \mathrm{dV}+\mathrm{p} \, \mathrm{dV}+\mathrm{V} \, \mathrm{dp} \nonumber \]
or,
\[\mathrm{dH}=\mathrm{q}+\mathrm{V} \, \mathrm{dp} \nonumber \]
At constant pressure,
\[\mathrm{dH}=\mathrm{q} \nonumber \]
For a change from state I to state II the change in enthalpy is given by equation (g).
\[\Delta \mathrm{H}=\int_{\mathrm{I}}^{\mathrm{II}} \mathrm{dH}=\mathrm{H}(\mathrm{II})-\mathrm{H}(\mathrm{I})=\mathrm{q} \nonumber \]
In equation (g) we replace the integral of dH by the difference \(\mathrm{H}(\mathrm{II}) - \mathrm{H}(\mathrm{I})\) because enthalpy is a state variable and so \(\Delta \mathrm{H}\) is independent of the path between the two states and hence so is \(\mathrm{q}\). In liquid solutions, the recorded heat is also independent of the rate of change in chemical composition between state I and state II.


8.6: Dual Status / Dependency vs Delinquency
It is important to understand the concept of “ dual status ” in juvenile law and detention. It may be shocking to learn that a majority of juvenile “delinquents” entered the juvenile justice system through dependency. As a result of parental abuse, neglect or incarceration, a child can enter the system at any point in their life as a dependent . This means their parent can no longer care for them. A majority of dependents are placed with another relative, but not all. Those children often end up in foster homes or group homes.
When a juvenile enters the juvenile justice system as a result of parental neglect or abuse it is called dependency . The area of law that governs how we treat juveniles in California falls under the Welfare and Institutions Code section 300:
ARTICLE 6. Dependent Children—Jurisdiction [300 - 304.7]
(Article 6 added by Stats. 1976, Ch. 1068.)
300.
A child who comes within any of the following descriptions is within the jurisdiction of the juvenile court which may adjudge that person to be a dependent child of the court.
When we look at juvenile delinquency, these are juveniles that have committed a crime. In juvenile courts, this is called wardship and is also guided by the Welfare and Institutions Codes but falls under section 602 in California.
ARTICLE 14. Wards—Jurisdiction [601 - 608]
(Heading of Article 14 renumbered from Article 5 by Stats. 1976, Ch. 1068.)
602.
Except as provided in Section 707, any person who is under 18 years of age when he or she violates any law of this state or of the United States or any ordinance of any city or county of this state defining crime other than an ordinance establishing a curfew based solely on age, is within the jurisdiction of the juvenile court, which may adjudge such person to be a ward of the court.
(Amended November 8, 2016, by initiative Proposition 57, Sec. 4.1.)
The concerning part of this distinction is many “juvenile delinquents” who commit crimes began in the juvenile dependency court as a result of abuse or neglect. We refer to these juveniles who fall under both dependency as an abused or neglected child and as a ward after committing a crime as “Dual Status” because they fall under both conditions in the Welfare and Institutions Code.
Pin It! Dual Status
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20.2: Physical Barreirs - Skin and Mucous Membranes
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- Ying Liu
- City College of San Francisco
Learning Objectives
- Summarize cellular junctions in physical barriers
- Describe how skin acts as a barrier to protect us from infections from the outside world
- Evaluate the roles of mucous membranes in protecting against infections from inside of the body
- Identify the importance of endothelia in CNS immune defense
Physical Barriers
Physical barriers play an important role in preventing microbes from reaching tissues that are susceptible to infection. At the cellular level, barriers consist of cells that are tightly joined to prevent invaders from crossing through to deeper tissue. For example, the endothelial cells that line blood vessels have very tight cell-to-cell junctions, blocking microbes from gaining access to the bloodstream.
Cell junctions are generally composed of cell membrane proteins that may connect with the extracellular matrix or with complementary proteins from neighboring cells. Tissues in various parts of the body have different types of cell junctions. These include tight junctions, desmosomes, and gap junctions , as illustrated in Figure \(\PageIndex{1}\). Invading microorganisms may attempt to break down these substances chemically, using enzymes such as proteases that can cause structural damage to create a point of entry for pathogens.
Query \(\PageIndex{1}\)
The Skin Barrier
One of the body’s most important physical barriers is the skin barrier , which is composed of three layers of closely packed cells. The thin upper layer is called the epidermis . A second, thicker layer, called the dermis , contains hair follicles, sweat glands, nerves, and blood vessels. A layer of fatty tissue called the hypodermis lies beneath the dermis and contains blood and lymph vessels (Figure \(\PageIndex{2}\)).
The topmost layer of skin, the epidermis, consists of cells that are packed with keratin . These dead cells remain as a tightly connected, dense layer of protein-filled cell husks on the surface of the skin. The keratin makes the skin’s surface mechanically tough and resistant to degradation by bacterial enzymes. Fatty acids on the skin’s surface create a dry, salty, and acidic environment that inhibits the growth of some microbes and is highly resistant to breakdown by bacterial enzymes. In addition, the dead cells of the epidermis are frequently shed, along with any microbes that may be clinging to them. Shed skin cells are continually replaced with new cells from below, providing a new barrier that will soon be shed in the same way.
Infections can occur when the skin barrier is compromised or broken. A wound can serve as a point of entry for opportunistic pathogens, which can infect the skin tissue surrounding the wound and possibly spread to deeper tissues.
Query \(\PageIndex{1}\)
Mucous Membranes
The mucous membranes lining the nose, mouth, lungs, and urinary and digestive tracts provide another nonspecific barrier against potential pathogens. Mucous membranes consist of a layer of epithelial cells bound by tight junctions. The epithelial cells secrete a moist, sticky substance called mucus , which covers and protects the more fragile cell layers beneath it and traps debris and particulate matter, including microbes. Mucus secretions also contain antimicrobial peptides .
In many regions of the body, mechanical actions serve to flush mucus (along with trapped or dead microbes) out of the body or away from potential sites of infection. For example, in the respiratory system, inhalation can bring microbes, dust, mold spores, and other small airborne debris into the body. This debris becomes trapped in the mucus lining the respiratory tract, a layer known as the mucociliary blanket. The epithelial cells lining the upper parts of the respiratory tract are called ciliated epithelial cells because they have hair-like appendages known as cilia. Movement of the cilia propels debris-laden mucus out and away from the lungs. The expelled mucus is then swallowed and destroyed in the stomach, or coughed up, or sneezed out (Figure \(\PageIndex{4}\)). This system of removal is often called the mucociliary escalator .
The mucociliary escalator is such an effective barrier to microbes that the lungs, the lowermost (and most sensitive) portion of the respiratory tract, were long considered to be a sterile environment in healthy individuals. Only recently has research suggested that healthy lungs may have a small normal microbiota. Disruption of the mucociliary escalator by the damaging effects of smoking or diseases such as cystic fibrosis can lead to increased colonization of bacteria in the lower respiratory tract and frequent infections, which highlights the importance of this physical barrier to host defenses.
Like the respiratory tract, the digestive tract is a portal of entry through which microbes enter the body, and the mucous membranes lining the digestive tract provide a nonspecific physical barrier against ingested microbes. The intestinal tract is lined with epithelial cells, interspersed with mucus-secreting goblet cells (Figure \(\PageIndex{5}\)). This mucus mixes with material received from the stomach, trapping foodborne microbes and debris. The mechanical action of peristalsis , a series of muscular contractions in the digestive tract, moves the sloughed mucus and other material through the intestines, rectum, and anus, excreting the material in feces.
Query \(\PageIndex{1}\)
Query \(\PageIndex{1}\)
Endothelia
The epithelial cells lining the urogenital tract, blood vessels, lymphatic vessels, and certain other tissues are known as endothelia. These tightly packed cells provide a particularly effective frontline barrier against invaders. The endothelia of the blood-brain barrier , for example, protect the central nervous system (CNS), which consists of the brain and the spinal cord. The CNS is one of the most sensitive and important areas of the body, as microbial infection of the CNS can quickly lead to serious and often fatal inflammation. The cell junctions in the blood vessels traveling through the CNS are some of the tightest and toughest in the body, preventing any transient microbes in the bloodstream from entering the CNS. This keeps the cerebrospinal fluid that surrounds and bathes the brain and spinal cord sterile under normal conditions.
Query \(\PageIndex{1}\)
Key Concepts and Summary
- The physical defenses of innate immunity include physical barriers, mechanical actions that remove microbes and debris, and the microbiome, which competes with and inhibits the growth of pathogens.
- The skin, mucous membranes, and endothelia throughout the body serve as physical barriers that prevent microbes from reaching potential sites of infection. Tight cell junctions in these tissues prevent microbes from passing through.
- Microbes trapped in dead skin cells or mucus are removed from the body by mechanical actions such as shedding of skin cells, mucociliary sweeping, coughing, peristalsis , and flushing of bodily fluids (e.g., urination, tears)


8: Groups, Teams, and Teamwork Last updated Save as PDF Page ID 47567 Lumen Learning Lumen Learning 8.1: Introduction to Conflict Within Teams 8.2: Conflict Within Teams 8.3: Putting It Together- Groups, Teams, and Teamwork 8.4: Assignment- Team Plan for Sun City Boards 8.5: Assignment- Leadership Competition 8.6: Discussion- Team Dynamics at Sun City Boards 8.7: Discussion- What Makes Teams Successful? 8.8: Assignment- Midterm Muddiest Point 8.9: Why It Matters- Groups, Teams, and Teamwork 8.10: Introduction to Common Group Behaviors 8.11: Common Group Behaviors 8.12: Introduction to Types of Teams 8.13: Types of Teams 8.14: Introduction to Building Effective Teams 8.15: Building Effective Teams 8.16: Introduction to the Five Stages of Team Development 8.17: The Five Stages of Team Development


4.10: Spanning, Linear Independence and Basis in Rⁿ
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Outcomes
- Determine the span of a set of vectors, and determine if a vector is contained in a specified span.
- Determine if a set of vectors is linearly independent.
- Understand the concepts of subspace, basis, and dimension.
- Find the row space, column space, and null space of a matrix.
By generating all linear combinations of a set of vectors one can obtain various subsets of \(\mathbb{R}^{n}\) which we call subspaces. For example what set of vectors in \(\mathbb{R}^{3}\) generate the \(XY\)-plane? What is the smallest such set of vectors can you find? The tools of spanning, linear independence and basis are exactly what is needed to answer these and similar questions and are the focus of this section. The following definition is essential.
Spanning Set of Vectors
We begin this section with a definition.
Consider the following example.
Example \(\PageIndex{1}\): Span of Vectors
Describe the span of the vectors \(\vec{u}=\left[ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right]^T\) and \(\vec{v}=\left[ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right]^T \in \mathbb{R}^{3}\).
Solution
You can see that any linear combination of the vectors \(\vec{u}\) and \(\vec{v}\) yields a vector of the form \(\left[ \begin{array}{rrr} x & y & 0 \end{array} \right]^T\) in the \(XY\)-plane.
Moreover every vector in the \(XY\)-plane is in fact such a linear combination of the vectors \(\vec{u}\) and \(\vec{v}\). That’s because \[\left[ \begin{array}{r} x \\ y \\ 0 \end{array} \right] = (-2x+3y) \left[ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right] + (x-y)\left[ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right]\nonumber \]
Thus \(\mathrm{span}\{\vec{u},\vec{v}\}\) is precisely the \(XY\)-plane.
You can convince yourself that no single vector can span the \(XY\)-plane. In fact, take a moment to consider what is meant by the span of a single vector.
However you can make the set larger if you wish. For example consider the larger set of vectors \(\{ \vec{u}, \vec{v}, \vec{w}\}\) where \(\vec{w}=\left[ \begin{array}{rrr} 4 & 5 & 0 \end{array} \right]^T\). Since the first two vectors already span the entire \(XY\)-plane, the span is once again precisely the \(XY\)-plane and nothing has been gained. Of course if you add a new vector such as \(\vec{w}=\left[ \begin{array}{rrr} 0 & 0 & 1 \end{array} \right]^T\) then it does span a different space. What is the span of \(\vec{u}, \vec{v}, \vec{w}\) in this case?
The distinction between the sets \(\{ \vec{u}, \vec{v}\}\) and \(\{ \vec{u}, \vec{v}, \vec{w}\}\) will be made using the concept of linear independence.
Consider the vectors \(\vec{u}, \vec{v}\), and \(\vec{w}\) discussed above. In the next example, we will show how to formally demonstrate that \(\vec{w}\) is in the span of \(\vec{u}\) and \(\vec{v}\).
Example \(\PageIndex{2}\): Vector in a Span
Let \(\vec{u}=\left[ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right]^T\) and \(\vec{v}=\left[ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right]^T \in \mathbb{R}^{3}\). Show that \(\vec{w} = \left[ \begin{array}{rrr} 4 & 5 & 0 \end{array} \right]^{T}\) is in \(\mathrm{span} \left\{ \vec{u}, \vec{v} \right\}\).
Solution
For a vector to be in \(\mathrm{span} \left\{ \vec{u}, \vec{v} \right\}\), it must be a linear combination of these vectors. If \(\vec{w} \in \mathrm{span} \left\{ \vec{u}, \vec{v} \right\}\), we must be able to find scalars \(a,b\) such that\[\vec{w} = a \vec{u} +b \vec{v}\nonumber \]
We proceed as follows. \[\left[ \begin{array}{r} 4 \\ 5 \\ 0 \end{array} \right] = a \left[ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right] + b \left[ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right]\nonumber \] This is equivalent to the following system of equations \[\begin{aligned} a + 3b &= 4 \\ a + 2b &= 5\end{aligned}\]
We solving this system the usual way, constructing the augmented matrix and row reducing to find the reduced row-echelon form. \[\left[ \begin{array}{rr|r} 1 & 3 & 4 \\ 1 & 2 & 5 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rr|r} 1 & 0 & 7 \\ 0 & 1 & -1 \end{array} \right]\nonumber \] The solution is \(a=7, b=-1\). This means that \[\vec{w} = 7 \vec{u} - \vec{v}\nonumber \] Therefore we can say that \(\vec{w}\) is in \(\mathrm{span} \left\{ \vec{u}, \vec{v} \right\}\).
Linearly Independent Set of Vectors
We now turn our attention to the following question: what linear combinations of a given set of vectors \(\{ \vec{u}_1, \cdots ,\vec{u}_k\}\) in \(\mathbb{R}^{n}\) yields the zero vector? Clearly \(0\vec{u}_1 + 0\vec{u}_2+ \cdots + 0 \vec{u}_k = \vec{0}\), but is it possible to have \(\sum_{i=1}^{k}a_{i}\vec{u}_{i}=\vec{0}\) without all coefficients being zero?
You can create examples where this easily happens. For example if \(\vec{u}_1=\vec{u}_2\), then \(1\vec{u}_1 - \vec{u}_2+ 0 \vec{u}_3 + \cdots + 0 \vec{u}_k = \vec{0}\), no matter the vectors \(\{ \vec{u}_3, \cdots ,\vec{u}_k\}\). 0But sometimes it can be more subtle.
Example \(\PageIndex{3}\): Linearly Dependent Set of Vectors
Consider the vectors \[\vec{u}_1=\left[ \begin{array}{rrr} 0 & 1 & -2 \end{array} \right]^T, \vec{u}_2=\left[ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right]^T, \vec{u}_3=\left[ \begin{array}{rrr} -2 & 3 & 2 \end{array} \right]^T, \mbox{ and } \vec{u}_4=\left[ \begin{array}{rrr} 1 & -2 & 0 \end{array} \right]^T\nonumber \] in \(\mathbb{R}^{3}\).
Then verify that \[1\vec{u}_1 +0 \vec{u}_2+ - \vec{u}_3 -2 \vec{u}_4 = \vec{0}\nonumber \]
You can see that the linear combination does yield the zero vector but has some non-zero coefficients. Thus we define a set of vectors to be linearly dependent if this happens.
Note that if \(\sum_{i=1}^{k}a_{i}\vec{u}_{i}=\vec{0}\) and some coefficient is non-zero, say \(a_1 \neq 0\), then \[\vec{u}_1 = \frac{-1}{a_1} \sum_{i=2}^{k}a_{i}\vec{u}_{i}\nonumber \] and thus \(\vec{u}_1\) is in the span of the other vectors. And the converse clearly works as well, so we get that a set of vectors is linearly dependent precisely when one of its vector is in the span of the other vectors of that set.
In particular, you can show that the vector \(\vec{u}_1\) in the above example is in the span of the vectors \(\{ \vec{u}_2, \vec{u}_3, \vec{u}_4 \}\).
If a set of vectors is NOT linearly dependent, then it must be that any linear combination of these vectors which yields the zero vector must use all zero coefficients. This is a very important notion, and we give it its own name of linear independence .
Note also that we require all vectors to be non-zero to form a linearly independent set.
To view this in a more familiar setting, form the \(n \times k\) matrix \(A\) having these vectors as columns. Then all we are saying is that the set \(\{ \vec{u}_1, \cdots ,\vec{u}_k\}\) is linearly independent precisely when \(AX=0\) has only the trivial solution.
Here is an example.
Example \(\PageIndex{4}\): Linearly Independent Vectors
Consider the vectors \(\vec{u}=\left[ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right]^T\), \(\vec{v}=\left[ \begin{array}{rrr} 1 & 0 & 1 \end{array} \right]^T\), and \(\vec{w}=\left[ \begin{array}{rrr} 0 & 1 & 1 \end{array} \right]^T\) in \(\mathbb{R}^{3}\). Verify whether the set \(\{\vec{u}, \vec{v}, \vec{w}\}\) is linearly independent.
Solution
So suppose that we have a linear combinations \(a\vec{u} + b \vec{v} + c\vec{w} = \vec{0}\). Then you can see that this can only happen with \(a=b=c=0\).
As mentioned above, you can equivalently form the \(3 \times 3\) matrix \(A = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{array} \right]\), and show that \(AX=0\) has only the trivial solution.
Thus this means the set \(\left\{ \vec{u}, \vec{v}, \vec{w} \right\}\) is linearly independent.
In terms of spanning, a set of vectors is linearly independent if it does not contain unnecessary vectors, that is not vector is in the span of the others.
Thus we put all this together in the following important theorem.
Theorem \(\PageIndex{1}\): Linear Independence as a Linear Combination
Let \(\left\{\vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) be a collection of vectors in \(\mathbb{R}^{n}\). Then the following are equivalent:
- It is linearly independent, that is whenever \[\sum_{i=1}^{k}a_{i}\vec{u}_{i}=\vec{0}\nonumber \] it follows that each coefficient \(a_{i}=0\).
- No vector is in the span of the others.
- The system of linear equations \(AX=0\) has only the trivial solution, where \(A\) is the \(n \times k\) matrix having these vectors as columns.
The last sentence of this theorem is useful as it allows us to use the reduced row-echelon form of a matrix to determine if a set of vectors is linearly independent. Let the vectors be columns of a matrix \(A\). Find the reduced row-echelon form of \(A\). If each column has a leading one, then it follows that the vectors are linearly independent.
Sometimes we refer to the condition regarding sums as follows: The set of vectors, \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) is linearly independent if and only if there is no nontrivial linear combination which equals the zero vector. A nontrivial linear combination is one in which not all the scalars equal zero. Similarly, a trivial linear combination is one in which all scalars equal zero.
Here is a detailed example in \(\mathbb{R}^{4}\).
Example \(\PageIndex{5}\): Linear Independence
Determine whether the set of vectors given by \[\left\{ \left[ \begin{array}{r} 1 \\ 2 \\ 3 \\ 0 \end{array} \right], \; \left[ \begin{array}{r} 2 \\ 1 \\ 0 \\ 1 \end{array} \right] , \; \left[ \begin{array}{r} 0 \\ 1 \\ 1 \\ 2 \end{array} \right] , \; \left[ \begin{array}{r} 3 \\ 2 \\ 2 \\ 0 \end{array} \right] \right\}\nonumber \] is linearly independent. If it is linearly dependent, express one of the vectors as a linear combination of the others.
Solution
In this case the matrix of the corresponding homogeneous system of linear equations is \[\left[ \begin{array}{rrrr|r} 1 & 2 & 0 & 3 & 0\\ 2 & 1 & 1 & 2 & 0 \\ 3 & 0 & 1 & 2 & 0 \\ 0 & 1 & 2 & 0 & 0 \end{array} \right]\nonumber \]
The reduced row-echelon form is \[\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right]\nonumber \]
and so every column is a pivot column and the corresponding system \(AX=0\) only has the trivial solution. Therefore, these vectors are linearly independent and there is no way to obtain one of the vectors as a linear combination of the others.
Consider another example.
Example \(\PageIndex{6}\): Linear Independence
Determine whether the set of vectors given by \[\left\{ \left[ \begin{array}{r} 1 \\ 2 \\ 3 \\ 0 \end{array} \right], \; \left[ \begin{array}{r} 2 \\ 1 \\ 0 \\ 1 \end{array} \right], \; \left[ \begin{array}{r} 0 \\ 1 \\ 1 \\ 2 \end{array} \right], \; \left[ \begin{array}{r} 3 \\ 2 \\ 2 \\ -1 \end{array} \right] \right\}\nonumber \] is linearly independent. If it is linearly dependent, express one of the vectors as a linear combination of the others.
Solution
Form the \(4 \times 4\) matrix \(A\) having these vectors as columns: \[A= \left[ \begin{array}{rrrr} 1 & 2 & 0 & 3 \\ 2 & 1 & 1 & 2 \\ 3 & 0 & 1 & 2 \\ 0 & 1 & 2 & -1 \end{array} \right]\nonumber \] Then by Theorem \(\PageIndex{1}\) , the given set of vectors is linearly independent exactly if the system \(AX=0\) has only the trivial solution.
The augmented matrix for this system and corresponding reduced row-echelon form are given by \[\left[ \begin{array}{rrrr|r} 1 & 2 & 0 & 3 & 0 \\ 2 & 1 & 1 & 2 & 0 \\ 3 & 0 & 1 & 2 & 0 \\ 0 & 1 & 2 & -1 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] Not all the columns of the coefficient matrix are pivot columns and so the vectors are not linearly independent. In this case, we say the vectors are linearly dependent.
It follows that there are infinitely many solutions to \(AX=0\), one of which is \[\left[ \begin{array}{r} 1 \\ 1 \\ -1 \\ -1 \end{array} \right]\nonumber \] Therefore we can write \[1\left[ \begin{array}{r} 1 \\ 2 \\ 3 \\ 0 \end{array} \right] +1\left[ \begin{array}{r} 2 \\ 1 \\ 0 \\ 1 \end{array} \right] -1 \left[ \begin{array}{r} 0 \\ 1 \\ 1 \\ 2 \end{array} \right] -1 \left[ \begin{array}{r} 3 \\ 2 \\ 2 \\ -1 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array} \right]\nonumber \]
This can be rearranged as follows \[1\left[ \begin{array}{r} 1 \\ 2 \\ 3 \\ 0 \end{array} \right] +1\left[ \begin{array}{r} 2 \\ 1 \\ 0 \\ 1 \end{array} \right] -1 \left[ \begin{array}{r} 0 \\ 1 \\ 1 \\ 2 \end{array} \right] =\left[ \begin{array}{r} 3 \\ 2 \\ 2 \\ -1 \end{array} \right]\nonumber \] This gives the last vector as a linear combination of the first three vectors.
Notice that we could rearrange this equation to write any of the four vectors as a linear combination of the other three.
When given a linearly independent set of vectors, we can determine if related sets are linearly independent.
Example \(\PageIndex{7}\): Related Sets of Vectors
Let \(\{ \vec{u},\vec{v},\vec{w}\}\) be an independent set of \(\mathbb{R}^n\). Is \(\{\vec{u}+\vec{v}, 2\vec{u}+\vec{w}, \vec{v}-5\vec{w}\}\) linearly independent?
Solution
Suppose \(a(\vec{u}+\vec{v}) + b(2\vec{u}+\vec{w}) + c(\vec{v}-5\vec{w})=\vec{0}_n\) for some \(a,b,c\in\mathbb{R}\). Then \[(a+2b)\vec{u} + (a+c)\vec{v} + (b-5c)\vec{w}=\vec{0}_n.\nonumber \]
Since \(\{\vec{u},\vec{v},\vec{w}\}\) is independent, \[\begin{aligned} a + 2b & = 0 \\ a + c & = 0 \\ b - 5c & = 0 \end{aligned}\]
This system of three equations in three variables has the unique solution \(a=b=c=0\). Therefore, \(\{\vec{u}+\vec{v}, 2\vec{u}+\vec{w}, \vec{v}-5\vec{w}\}\) is independent.
The following corollary follows from the fact that if the augmented matrix of a homogeneous system of linear equations has more columns than rows, the system has infinitely many solutions.
Corollary \(\PageIndex{1}\): Linear Dependence in \(\mathbb{R}''\)
Let \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) be a set of vectors in \(\mathbb{R}^{n}\). If \(k>n\), then the set is linearly dependent (i.e. NOT linearly independent).
- Proof
-
Form the \(n \times k\) matrix \(A\) having the vectors \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) as its columns and suppose \(k > n\). Then \(A\) has rank \(r \leq n <k\), so the system \(AX=0\) has a nontrivial solution and thus not linearly independent by Theorem \(\PageIndex{1}\) .
Example \(\PageIndex{8}\): Linear Dependence
Consider the vectors \[\left\{ \left[ \begin{array}{r} 1 \\ 4 \end{array} \right], \left[ \begin{array}{r} 2 \\ 3 \end{array} \right], \left[ \begin{array}{r} 3 \\ 2 \end{array} \right] \right\}\nonumber \] Are these vectors linearly independent?
Solution
This set contains three vectors in \(\mathbb{R}^2\). By Corollary \(\PageIndex{1}\) these vectors are linearly dependent. In fact, we can write \[(-1) \left[ \begin{array}{r} 1 \\ 4 \end{array} \right] + (2) \left[ \begin{array}{r} 2 \\ 3 \end{array} \right] = \left[ \begin{array}{r} 3 \\ 2 \end{array} \right]\nonumber \] showing that this set is linearly dependent.
The third vector in the previous example is in the span of the first two vectors. We could find a way to write this vector as a linear combination of the other two vectors. It turns out that the linear combination which we found is the only one, provided that the set is linearly independent.
Theorem \(\PageIndex{2}\): Unique Linear Combination
Let \(U \subseteq\mathbb{R}^n\) be an independent set. Then any vector \(\vec{x}\in\mathrm{span}(U)\) can be written uniquely as a linear combination of vectors of \(U\).
- Proof
-
To prove this theorem, we will show that two linear combinations of vectors in \(U\) that equal \(\vec{x}\) must be the same. Let \(U =\{ \vec{u}_1, \vec{u}_2, \ldots, \vec{u}_k\}\). Suppose that there is a vector \(\vec{x}\in \mathrm{span}(U)\) such that \[\begin{aligned} \vec{x} & = s_1\vec{u}_1 + s_2\vec{u}_2 + \cdots + s_k\vec{u}_k, \mbox{ for some } s_1, s_2, \ldots, s_k\in\mathbb{R}, \mbox{ and} \\ \vec{x} & = t_1\vec{u}_1 + t_2\vec{u}_2 + \cdots + t_k\vec{u}_k, \mbox{ for some } t_1, t_2, \ldots, t_k\in\mathbb{R}.\end{aligned}\] Then \(\vec{0}_n=\vec{x}-\vec{x} = (s_1-t_1)\vec{u}_1 + (s_2-t_2)\vec{u}_2 + \cdots + (s_k-t_k)\vec{u}_k\).
Since \(U\) is independent, the only linear combination that vanishes is the trivial one, so \(s_i-t_i=0\) for all \(i\), \(1\leq i\leq k\).
Therefore, \(s_i=t_i\) for all \(i\), \(1\leq i\leq k\), and the representation is unique.Let \(U \subseteq\mathbb{R}^n\) be an independent set. Then any vector \(\vec{x}\in\mathrm{span}(U)\) can be written uniquely as a linear combination of vectors of \(U\).
Suppose that \(\vec{u},\vec{v}\) and \(\vec{w}\) are nonzero vectors in \(\mathbb{R}^3\), and that \(\{ \vec{v},\vec{w}\}\) is independent. Consider the set \(\{ \vec{u},\vec{v},\vec{w}\}\). When can we know that this set is independent? It turns out that this follows exactly when \(\vec{u}\not\in\mathrm{span}\{\vec{v},\vec{w}\}\).
Example \(\PageIndex{9}\)
Suppose that \(\vec{u},\vec{v}\) and \(\vec{w}\) are nonzero vectors in \(\mathbb{R}^3\), and that \(\{ \vec{v},\vec{w}\}\) is independent. Prove that \(\{ \vec{u},\vec{v},\vec{w}\}\) is independent if and only if \(\vec{u}\not\in\mathrm{span}\{\vec{v},\vec{w}\}\).
Solution
If \(\vec{u}\in\mathrm{span}\{\vec{v},\vec{w}\}\), then there exist \(a,b\in\mathbb{R}\) so that \(\vec{u}=a\vec{v} + b\vec{w}\). This implies that \(\vec{u}-a\vec{v} - b\vec{w}=\vec{0}_3\), so \(\vec{u}-a\vec{v} - b\vec{w}\) is a nontrivial linear combination of \(\{ \vec{u},\vec{v},\vec{w}\}\) that vanishes, and thus \(\{ \vec{u},\vec{v},\vec{w}\}\) is dependent.
Now suppose that \(\vec{u}\not\in\mathrm{span}\{\vec{v},\vec{w}\}\), and suppose that there exist \(a,b,c\in\mathbb{R}\) such that \(a\vec{u}+b\vec{v}+c\vec{w}=\vec{0}_3\). If \(a\neq 0\), then \(\vec{u}=-\frac{b}{a}\vec{v}-\frac{c}{a}\vec{w}\), and \(\vec{u}\in\mathrm{span}\{\vec{v},\vec{w}\}\), a contradiction. Therefore, \(a=0\), implying that \(b\vec{v}+c\vec{w}=\vec{0}_3\). Since \(\{ \vec{v},\vec{w}\}\) is independent, \(b=c=0\), and thus \(a=b=c=0\), i.e., the only linear combination of \(\vec{u},\vec{v}\) and \(\vec{w}\) that vanishes is the trivial one.
Therefore, \(\{ \vec{u},\vec{v},\vec{w}\}\) is independent.
Consider the following useful theorem.
This theorem also allows us to determine if a matrix is invertible. If an \(n \times n\) matrix \(A\) has columns which are independent, or span \(\mathbb{R}^n\), then it follows that \(A\) is invertible. If it has rows that are independent, or span the set of all \(1 \times n\) vectors, then \(A\) is invertible.
A Short Application to Chemistry
The following section applies the concepts of spanning and linear independence to the subject of chemistry.
When working with chemical reactions, there are sometimes a large number of reactions and some are in a sense redundant. Suppose you have the following chemical reactions. \[\begin{array}{c} CO+\frac{1}{2}O_{2}\rightarrow CO_{2} \\ H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O \\ CH_{4}+\frac{3}{2}O_{2}\rightarrow CO+2H_{2}O \\ CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O \end{array}\nonumber \] There are four chemical reactions here but they are not independent reactions. There is some redundancy. What are the independent reactions? Is there a way to consider a shorter list of reactions? To analyze this situation, we can write the reactions in a matrix as follows \[\left[ \begin{array}{cccccc} CO & O_{2} & CO_{2} & H_{2} & H_{2}O & CH_{4} \\ 1 & 1/2 & -1 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 1 & -1 & 0 \\ -1 & 3/2 & 0 & 0 & -2 & 1 \\ 0 & 2 & -1 & 0 & -2 & 1 \end{array} \right]\nonumber \]
Each row contains the coefficients of the respective elements in each reaction. For example, the top row of numbers comes from \(CO+\frac{1}{2}O_{2}-CO_{2}=0\) which represents the first of the chemical reactions.
We can write these coefficients in the following matrix \[\left[ \begin{array}{rrrrrr} 1 & 1/2 & -1 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 1 & -1 & 0 \\ -1 & 3/2 & 0 & 0 & -2 & 1 \\ 0 & 2 & -1 & 0 & -2 & 1 \end{array} \right]\nonumber \] Rather than listing all of the reactions as above, it would be more efficient to only list those which are independent by throwing out that which is redundant. We can use the concepts of the previous section to accomplish this.
First, take the reduced row-echelon form of the above matrix. \[\left[ \begin{array}{rrrrrr} 1 & 0 & 0 & 3 & -1 & -1 \\ 0 & 1 & 0 & 2 & -2 & 0 \\ 0 & 0 & 1 & 4 & -2 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] The top three rows represent “independent" reactions which come from the original four reactions. One can obtain each of the original four rows of the matrix given above by taking a suitable linear combination of rows of this reduced row-echelon matrix.
With the redundant reaction removed, we can consider the simplified reactions as the following equations \[\begin{array}{c} CO+3H_{2}-1H_{2}O-1CH_{4}=0 \\ O_{2}+2H_{2}-2H_{2}O=0 \\ CO_{2}+4H_{2}-2H_{2}O-1CH_{4}=0 \end{array}\nonumber \] In terms of the original notation, these are the reactions \[\begin{array}{c} CO+3H_{2}\rightarrow H_{2}O+CH_{4} \\ O_{2}+2H_{2}\rightarrow 2H_{2}O \\ CO_{2}+4H_{2}\rightarrow 2H_{2}O+CH_{4} \end{array}\nonumber \]
These three reactions provide an equivalent system to the original four equations. The idea is that, in terms of what happens chemically, you obtain the same information with the shorter list of reactions. Such a simplification is especially useful when dealing with very large lists of reactions which may result from experimental evidence.
Subspaces and Basis
The goal of this section is to develop an understanding of a subspace of \(\mathbb{R}^n\). Before a precise definition is considered, we first examine the subspace test given below.
Theorem \(\PageIndex{4}\): Subspace Test
A subset \(V\) of \(\mathbb{R}^n\) is a subspace of \(\mathbb{R}^n\) if
- the zero vector of \(\mathbb{R}^n\), \(\vec{0}_n\), is in \(V\);
- \(V\) is closed under addition, i.e., for all \(\vec{u},\vec{w}\in V\), \(\vec{u}+\vec{w}\in V\);
- \(V\) is closed under scalar multiplication, i.e., for all \(\vec{u}\in V\) and \(k\in\mathbb{R}\), \(k\vec{u}\in V\).
This test allows us to determine if a given set is a subspace of \(\mathbb{R}^n\). Notice that the subset \(V = \left\{ \vec{0} \right\}\) is a subspace of \(\mathbb{R}^n\) (called the zero subspace ), as is \(\mathbb{R}^n\) itself. A subspace which is not the zero subspace of \(\mathbb{R}^n\) is referred to as a proper subspace.
A subspace is simply a set of vectors with the property that linear combinations of these vectors remain in the set. Geometrically in \(\mathbb{R}^{3}\), it turns out that a subspace can be represented by either the origin as a single point, lines and planes which contain the origin, or the entire space \(\mathbb{R}^{3}\).
Consider the following example of a line in \(\mathbb{R}^3\).
Example \(\PageIndex{10}\): Subspace of \(\mathbb{R}^3\)
In \(\mathbb{R}^3\), the line \(L\) through the origin that is parallel to the vector \({\vec{d}}= \left[ \begin{array}{r} -5 \\ 1 \\ -4 \end{array}\right]\) has (vector) equation \(\left[ \begin{array}{r} x \\ y \\ z \end{array}\right] =t\left[ \begin{array}{r} -5 \\ 1 \\ -4 \end{array}\right], t\in\mathbb{R}\), so \[L=\left\{ t{\vec{d}} ~|~ t\in\mathbb{R}\right\}.\nonumber \] Then \(L\) is a subspace of \(\mathbb{R}^3\).
Solution
Using the subspace test given above we can verify that \(L\) is a subspace of \(\mathbb{R}^3\).
- First: \(\vec{0}_3\in L\) since \(0\vec{d}=\vec{0}_3\).
- Suppose \(\vec{u},\vec{v}\in L\). Then by definition, \(\vec{u}=s\vec{d}\) and \(\vec{v}=t\vec{d}\), for some \(s,t\in\mathbb{R}\). Thus \[\vec{u}+\vec{v} = s\vec{d}+t\vec{d} = (s+t)\vec{d}.\nonumber \] Since \(s+t\in\mathbb{R}\), \(\vec{u}+\vec{v}\in L\); i.e., \(L\) is closed under addition.
- Suppose \(\vec{u}\in L\) and \(k\in\mathbb{R}\) (\(k\) is a scalar). Then \(\vec{u}=t\vec{d}\), for some \(t\in\mathbb{R}\), so \[k\vec{u}=k(t\vec{d})=(kt)\vec{d}.\nonumber \] Since \(kt\in\mathbb{R}\), \(k\vec{u}\in L\); i.e., \(L\) is closed under scalar multiplication.
Since \(L\) satisfies all conditions of the subspace test, it follows that \(L\) is a subspace.
Note that there is nothing special about the vector \(\vec{d}\) used in this example; the same proof works for any nonzero vector \(\vec{d}\in\mathbb{R}^3\), so any line through the origin is a subspace of \(\mathbb{R}^3\).
We are now prepared to examine the precise definition of a subspace as follows.
More generally this means that a subspace contains the span of any finite collection vectors in that subspace. It turns out that in \(\mathbb{R}^{n}\), a subspace is exactly the span of finitely many of its vectors.
Theorem \(\PageIndex{5}\): Subspaces are Spans
Let \(V\) be a nonempty collection of vectors in \(\mathbb{R}^{n}.\) Then \(V\) is a subspace of \(\mathbb{R}^{n}\) if and only if there exist vectors \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) in \(V\) such that \[V= \mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\nonumber \] Furthermore, let \(W\) be another subspace of \(\mathbb{R}^n\) and suppose \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\} \in W\). Then it follows that \(V\) is a subset of \(W\).
Note that since \(W\) is arbitrary, the statement that \(V \subseteq W\) means that any other subspace of \(\mathbb{R}^n\) that contains these vectors will also contain \(V\).
- Proof
-
We first show that if \(V\) is a subspace, then it can be written as \(V= \mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\). Pick a vector \(\vec{u}_{1}\) in \(V\). If \(V= \mathrm{span}\left\{ \vec{u}_{1}\right\} ,\) then you have found your list of vectors and are done. If \(V\neq \mathrm{span}\left\{ \vec{u}_{1}\right\} ,\) then there exists \(\vec{u}_{2}\) a vector of \(V\) which is not in \(\mathrm{span}\left\{ \vec{u}_{1}\right\} .\) Consider \(\mathrm{span}\left\{ \vec{u}_{1},\vec{u}_{2}\right\}.\) If \(V=\mathrm{span}\left\{ \vec{u}_{1},\vec{u}_{2}\right\}\), we are done. Otherwise, pick \(\vec{u}_{3}\) not in \(\mathrm{span}\left\{ \vec{u}_{1},\vec{u}_{2}\right\} .\) Continue this way. Note that since \(V\) is a subspace, these spans are each contained in \(V\). The process must stop with \(\vec{u}_{k}\) for some \(k\leq n\) by Corollary \(\PageIndex{1}\) , and thus \(V=\mathrm{span}\left\{ \vec{u}_{1},\cdots , \vec{u}_{k}\right\}\).
Now suppose \(V=\mathrm{span}\left\{ \vec{u}_{1},\cdots , \vec{u}_{k}\right\}\), we must show this is a subspace. So let \(\sum_{i=1}^{k}c_{i}\vec{u}_{i}\) and \(\sum_{i=1}^{k}d_{i}\vec{u}_{i}\) be two vectors in \(V\), and let \(a\) and \(b\) be two scalars. Then \[a \sum_{i=1}^{k}c_{i}\vec{u}_{i}+ b \sum_{i=1}^{k}d_{i}\vec{u}_{i}= \sum_{i=1}^{k}\left( a c_{i}+b d_{i}\right) \vec{u}_{i}\nonumber \] which is one of the vectors in \(\mathrm{span}\left\{ \vec{u}_{1},\cdots , \vec{u}_{k}\right\}\) and is therefore contained in \(V\). This shows that \(\mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) has the properties of a subspace.
To prove that \(V \subseteq W\), we prove that if \(\vec{u}_i\in V\), then \(\vec{u}_i \in W\).
Suppose \(\vec{u}\in V\). Then \(\vec{u}=a_1\vec{u}_1 + a_2\vec{u}_2 + \cdots + a_k\vec{u}_k\) for some \(a_i\in\mathbb{R}\), \(1\leq i\leq k\). Since \(W\) contain each \(\vec{u}_i\) and \(W\) is a vector space, it follows that \(a_1\vec{u}_1 + a_2\vec{u}_2 + \cdots + a_k\vec{u}_k \in W\).
Since the vectors \(\vec{u}_i\) we constructed in the proof above are not in the span of the previous vectors (by definition), they must be linearly independent and thus we obtain the following corollary.
Corollary \(\PageIndex{2}\): Subspaces are Spans of Independent Vectors
If \(V\) is a subspace of \(\mathbb{R}^{n},\) then there exist linearly independent vectors \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) in \(V\) such that \(V=\mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\).
In summary, subspaces of \(\mathbb{R}^{n}\) consist of spans of finite, linearly independent collections of vectors of \(\mathbb{R}^{n}\). Such a collection of vectors is called a basis.
Definition \(\PageIndex{6}\): Basis of a Subspace
Let \(V\) be a subspace of \(\mathbb{R}^{n}\). Then \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) is a basis for \(V\) if the following two conditions hold.
- \(\mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\} =V\)
- \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) is linearly independent
Note the plural of basis is bases .
The following is a simple but very useful example of a basis, called the standard basis.
Definition \(\PageIndex{7}\): Standard Basis of \(\mathbb{R}^n\)
Let \(\vec{e}_i\) be the vector in \(\mathbb{R}^n\) which has a \(1\) in the \(i^{th}\) entry and zeros elsewhere, that is the \(i^{th}\) column of the identity matrix. Then the collection \(\left\{\vec{e}_1, \vec{e}_2, \cdots, \vec{e}_n \right\}\) is a basis for \(\mathbb{R}^n\) and is called the standard basis of \(\mathbb{R}^n\).
The main theorem about bases is not only they exist, but that they must be of the same size. To show this, we will need the the following fundamental result, called the Exchange Theorem.
Theorem \(\PageIndex{6}\): Exchange Theorem
Suppose \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{r}\right\}\) is a linearly independent set of vectors in \(\mathbb{R}^n\), and each \(\vec{u}_{k}\) is contained in \(\mathrm{span}\left\{ \vec{v}_{1},\cdots ,\vec{v}_{s}\right\}\) Then \(s\geq r.\)
In words, spanning sets have at least as many vectors as linearly independent sets.
- Proof
-
Since each \(\vec{u}_j\) is in \(\mathrm{span}\left\{ \vec{v}_{1},\cdots ,\vec{v}_{s}\right\}\), there exist scalars \(a_{ij}\) such that \[\vec{u}_{j}=\sum_{i=1}^{s}a_{ij}\vec{v}_{i}\nonumber \] Suppose for a contradiction that \(s<r\). Then the matrix \(A = \left[ a_{ij} \right]\) has fewer rows, \(s\) than columns, \(r\). Then the system \(AX=0\) has a non trivial solution \(\vec{d}\), that is there is a \(\vec{d}\neq \vec{0}\) such that \(A\vec{d}=\vec{0}\). In other words, \[\sum_{j=1}^{r}a_{ij}d_{j}=0,\;i=1,2,\cdots ,s\nonumber \] Therefore, \[\begin{aligned} \sum_{j=1}^{r}d_{j}\vec{u}_{j} &=\sum_{j=1}^{r}d_{j}\sum_{i=1}^{s}a_{ij} \vec{v}_{i} \\ &=\sum_{i=1}^{s}\left( \sum_{j=1}^{r}a_{ij}d_{j}\right) \vec{v} _{i}=\sum_{i=1}^{s}0\vec{v}_{i}=0\end{aligned}\] which contradicts the assumption that \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{r}\right\}\) is linearly independent, because not all the \(d_{j}\) are zero. Thus this contradiction indicates that \(s\geq r\).
We are now ready to show that any two bases are of the same size.
Theorem \(\PageIndex{7}\): Bases of \(\mathbb{R}^{n}\) are of the Same Size
Let \(V\) be a subspace of \(\mathbb{R}^{n}\) with two bases \(B_1\) and \(B_2\). Suppose \(B_1\) contains \(s\) vectors and \(B_2\) contains \(r\) vectors. Then \(s=r.\)
- Proof
-
This follows right away from Theorem 9.4.4 . Indeed observe that \(B_1 = \left\{ \vec{u}_{1},\cdots ,\vec{u}_{s}\right\}\) is a spanning set for \(V\) while \(B_2 = \left\{ \vec{v}_{1},\cdots ,\vec{v}_{r}\right\}\) is linearly independent, so \(s \geq r.\) Similarly \(B_2 = \left\{ \vec{v}_{1},\cdots ,\vec{v} _{r}\right\}\) is a spanning set for \(V\) while \(B_1 = \left\{ \vec{u}_{1},\cdots , \vec{u}_{s}\right\}\) is linearly independent, so \(r\geq s\).
The following definition can now be stated.
The next result follows.
Consider the following example.
Example \(\PageIndex{11}\): Basis of Subspace
Let \[V=\left\{ \left[\begin{array}{c} a\\ b\\ c\\ d\end{array}\right]\in\mathbb{R}^4 ~:~ a-b=d-c \right\}.\nonumber \] Show that \(V\) is a subspace of \(\mathbb{R}^4\), find a basis of \(V\), and find \(\dim(V)\).
Solution
The condition \(a-b=d-c\) is equivalent to the condition \(a=b-c+d\), so we may write
\[V =\left\{ \left[\begin{array}{c} b-c+d\\ b\\ c\\ d\end{array}\right] ~:~b,c,d \in\mathbb{R} \right\} = \left\{ b\left[\begin{array}{c} 1\\ 1\\ 0\\ 0\end{array}\right] +c\left[\begin{array}{c} -1\\ 0\\ 1\\ 0\end{array}\right] +d\left[\begin{array}{c} 1\\ 0\\ 0\\ 1\end{array}\right] ~:~ b,c,d\in\mathbb{R} \right\}\nonumber \]
This shows that \(V\) is a subspace of \(\mathbb{R}^4\), since \(V=\mathrm{span}\{ \vec{u}_1, \vec{u}_2, \vec{u}_3 \}\) where
\[\vec{u}_1 = \left[\begin{array}{r} 1 \\ 1 \\ 0 \\ 0 \end{array}\right], \vec{u}_2 = \left[\begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \end{array}\right], \vec{u}_3 = \left[\begin{array}{r} 1 \\ 0 \\ 0 \\ 1 \end{array}\right]\nonumber \]
Furthermore,
\[\left\{ \left[\begin{array}{c} 1\\ 1\\ 0\\ 0\end{array}\right], \left[\begin{array}{c} -1\\ 0\\ 1\\ 0\end{array}\right], \left[\begin{array}{c} 1\\ 0\\ 0\\ 1\end{array}\right] \right\}\nonumber \] is linearly independent, as can be seen by taking the reduced row-echelon form of the matrix whose columns are \(\vec{u}_1, \vec{u}_2\) and \(\vec{u}_3\).
\[\left[\begin{array}{rrr} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \rightarrow \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right]\nonumber \]
Since every column of the reduced row-echelon form matrix has a leading one, the columns are linearly independent.
Therefore \(\{ \vec{u}_1, \vec{u}_2, \vec{u}_3 \}\) is linearly independent and spans \(V\), so is a basis of \(V\). Hence \(V\) has dimension three.
We continue by stating further properties of a set of vectors in \(\mathbb{R}^{n}\).
Corollary \(\PageIndex{4}\): Linearly Independent and Spanning Sets in \(\mathbb{R}^{n}\)
The following properties hold in \(\mathbb{R}^{n}\):
- Suppose \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{n}\right\}\) is linearly independent. Then \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{n}\right\}\) is a basis for \(\mathbb{R}^{n}\).
- Suppose \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{m}\right\}\) spans \(\mathbb{R}^{n}.\) Then \(m\geq n.\)
- If \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{n}\right\}\) spans \(\mathbb{R}^{n},\) then \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{n}\right\}\) is linearly independent.
- Proof
-
Assume first that \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{n}\right\}\) is linearly independent, and we need to show that this set spans \(\mathbb{R}^{n}\). To do so, let \(\vec{v}\) be a vector of \(\mathbb{R}^{n}\), and we need to write \(\vec{v}\) as a linear combination of \(\vec{u}_i\)’s. Consider the matrix \(A\) having the vectors \(\vec{u}_i\) as columns: \[A = \left[ \begin{array}{rrr} \vec{u}_{1} & \cdots & \vec{u}_{n} \end{array} \right]\nonumber \]
By linear independence of the \(\vec{u}_i\)’s, the reduced row-echelon form of \(A\) is the identity matrix. Therefore the system \(A\vec{x}= \vec{v}\) has a (unique) solution, so \(\vec{v}\) is a linear combination of the \(\vec{u}_i\)’s.
To establish the second claim, suppose that \(m<n.\) Then letting \(\vec{u}_{i_{1}},\cdots ,\vec{u}_{i_{k}}\) be the pivot columns of the matrix \[\left[ \begin{array}{ccc} \vec{u}_{1} & \cdots & \vec{u}_{m} \end{array} \right]\nonumber \] it follows \(k\leq m<n\) and these \(k\) pivot columns would be a basis for \(\mathbb{R}^{n}\) having fewer than \(n\) vectors, contrary to Corollary \(\PageIndex{3}\) .
Finally consider the third claim. If \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{n}\right\}\) is not linearly independent, then replace this list with \(\left\{ \vec{u}_{i_{1}},\cdots ,\vec{u}_{i_{k}}\right\}\) where these are the pivot columns of the matrix \[\left[ \begin{array}{ccc} \vec{u}_{1} & \cdots & \vec{u}_{n} \end{array} \right]\nonumber \] Then \(\left\{ \vec{u}_{i_{1}},\cdots ,\vec{u}_{i_{k}}\right\}\) spans \(\mathbb{R}^{n}\) and is linearly independent, so it is a basis having less than \(n\) vectors again contrary to Corollary \(\PageIndex{3}\) .
The next theorem follows from the above claim.
Consider Corollary \(\PageIndex{4}\) together with Theorem \(\PageIndex{8}\) . Let \(\dim(V) = r\). Suppose there exists an independent set of vectors in \(V\). If this set contains \(r\) vectors, then it is a basis for \(V\). If it contains less than \(r\) vectors, then vectors can be added to the set to create a basis of \(V\). Similarly, any spanning set of \(V\) which contains more than \(r\) vectors can have vectors removed to create a basis of \(V\).
We illustrate this concept in the next example.
Example \(\PageIndex{12}\): Extending an Independent Set
Consider the set \(U\) given by \[U=\left\{ \left.\left[\begin{array}{c} a\\ b\\ c\\ d\end{array}\right] \in\mathbb{R}^4 ~\right|~ a-b=d-c \right\}\nonumber \] Then \(U\) is a subspace of \(\mathbb{R}^4\) and \(\dim(U)=3\).
Then \[S=\left\{ \left[\begin{array}{c} 1\\ 1\\ 1\\ 1\end{array}\right], \left[\begin{array}{c} 2\\ 3\\ 3\\ 2\end{array}\right] \right\},\nonumber \] is an independent subset of \(U\). Therefore \(S\) can be extended to a basis of \(U\).
Solution
To extend \(S\) to a basis of \(U\), find a vector in \(U\) that is not in \(\mathrm{span}(S)\). \[\left[\begin{array}{rrr} 1 & 2 & ? \\ 1 & 3 & ? \\ 1 & 3 & ? \\ 1 & 2 & ? \end{array}\right]\nonumber \]
\[\left[\begin{array}{rrr} 1 & 2 & 1 \\ 1 & 3 & 0 \\ 1 & 3 & -1 \\ 1 & 2 & 0 \end{array}\right] \rightarrow \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right]\nonumber \]
Therefore, \(S\) can be extended to the following basis of \(U\): \[\left\{ \left[\begin{array}{r} 1\\ 1\\ 1\\ 1\end{array}\right], \left[\begin{array}{r} 2\\ 3\\ 3\\ 2\end{array}\right], \left[\begin{array}{r} 1\\ 0\\ -1\\ 0\end{array}\right] \right\},\nonumber \]
Next we consider the case of removing vectors from a spanning set to result in a basis.
Theorem \(\PageIndex{9}\): Finding a Basis from a Span
Let \(W\) be a subspace. Also suppose that \(W=span\left\{ \vec{w} _{1},\cdots ,\vec{w}_{m}\right\}\). Then there exists a subset of \(\left\{ \vec{w}_{1},\cdots ,\vec{w}_{m}\right\}\) which is a basis for \(W\).
- Proof
-
Let \(S\) denote the set of positive integers such that for \(k\in S,\) there exists a subset of \(\left\{ \vec{w}_{1},\cdots ,\vec{w}_{m}\right\}\) consisting of exactly \(k\) vectors which is a spanning set for \(W\). Thus \(m\in S\). Pick the smallest positive integer in \(S\). Call it \(k\). Then there exists \(\left\{ \vec{u}_{1},\cdots , \vec{u}_{k}\right\} \subseteq \left\{ \vec{w}_{1},\cdots ,\vec{w} _{m}\right\}\) such that \(\text{span}\left\{ \vec{u}_{1},\cdots ,\vec{u} _{k}\right\} =W.\) If \[\sum_{i=1}^{k}c_{i}\vec{w}_{i}=\vec{0}\nonumber \] and not all of the \(c_{i}=0,\) then you could pick \(c_{j}\neq 0\), divide by it and solve for \(\vec{u}_{j}\) in terms of the others, \[\vec{w}_{j}=\sum_{i\neq j}\left( -\frac{c_{i}}{c_{j}}\right) \vec{w}_{i}\nonumber \] Then you could delete \(\vec{w}_{j}\) from the list and have the same span. Any linear combination involving \(\vec{w}_{j}\) would equal one in which \(\vec{w}_{j}\) is replaced with the above sum, showing that it could have been obtained as a linear combination of \(\vec{w}_{i}\) for \(i\neq j\). Thus \(k-1\in S\) contrary to the choice of \(k\). Hence each \(c_{i}=0\) and so \(\left\{ \vec{u}_{1},\cdots ,\vec{u} _{k}\right\}\) is a basis for \(W\) consisting of vectors of \(\left\{ \vec{w} _{1},\cdots ,\vec{w}_{m}\right\}\).
The following example illustrates how to carry out this shrinking process which will obtain a subset of a span of vectors which is linearly independent.
Example \(\PageIndex{13}\): Subset of a Span
Let \(W\) be the subspace \[span\left\{ \left[ \begin{array}{r} 1 \\ 2 \\ -1 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} 1 \\ 3 \\ -1 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} 8 \\ 19 \\ -8 \\ 8 \end{array} \right] ,\left[ \begin{array}{r} -6 \\ -15 \\ 6 \\ -6 \end{array} \right] ,\left[ \begin{array}{r} 1 \\ 3 \\ 0 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} 1 \\ 5 \\ 0 \\ 1 \end{array} \right] \right\}\nonumber \] Find a basis for \(W\) which consists of a subset of the given vectors.
Solution
You can use the reduced row-echelon form to accomplish this reduction. Form the matrix which has the given vectors as columns. \[\left[ \begin{array}{rrrrrr} 1 & 1 & 8 & -6 & 1 & 1 \\ 2 & 3 & 19 & -15 & 3 & 5 \\ -1 & -1 & -8 & 6 & 0 & 0 \\ 1 & 1 & 8 & -6 & 1 & 1 \end{array} \right]\nonumber \] Then take the reduced row-echelon form
\[\left[ \begin{array}{rrrrrr} 1 & 0 & 5 & -3 & 0 & -2 \\ 0 & 1 & 3 & -3 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] It follows that a basis for \(W\) is
\[\left\{ \left[ \begin{array}{r} 1 \\ 2 \\ -1 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} 1 \\ 3 \\ -1 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 1 \\ 3 \\ 0 \\ 1 \end{array} \right] \right\}\nonumber \] Since the first, second, and fifth columns are obviously a basis for the column space of the , the same is true for the matrix having the given vectors as columns.
Consider the following theorems regarding a subspace contained in another subspace.
The proof is left as an exercise but proceeds as follows. Begin with a basis for \(W,\left\{ \vec{w}_{1},\cdots ,\vec{w}_{s}\right\}\) and add in vectors from \(V\) until you obtain a basis for \(V\). Not that the process will stop because the dimension of \(V\) is no more than \(n\).
Consider the following example.
Example \(\PageIndex{14}\): Extending a Basis
Let \(V=\mathbb{R}^{4}\) and let \[W=\mathrm{span}\left\{ \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right] \right\}\nonumber \] Extend this basis of \(W\) to a basis of \(\mathbb{R}^{n}\).
Solution
An easy way to do this is to take the reduced row-echelon form of the matrix
\[\left[ \begin{array}{cccccc} 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \end{array} \right] \label{basiseq1}\]
Note how the given vectors were placed as the first two columns and then the matrix was extended in such a way that it is clear that the span of the columns of this matrix yield all of \(\mathbb{R}^{4}\). Now determine the pivot columns. The reduced row-echelon form is
\[\left[ \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & -1 \end{array} \right] \label{basiseq2}\]
Therefore the pivot columns are \[\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right]\nonumber \]
and now this is an extension of the given basis for \(W\) to a basis for \(\mathbb{R}^{4}\).
Why does this work? The columns of \(\eqref{basiseq1}\) obviously span \(\mathbb{R }^{4}\). In fact the span of the first four is the same as the span of all six.
Consider another example.
Example \(\PageIndex{15}\): Extending a Basis
Let \(W\) be the span of \(\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right]\) in \(\mathbb{R}^{4}\). Let \(V\) consist of the span of the vectors \[\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} 7 \\ -6 \\ 1 \\ -6 \end{array} \right] ,\left[ \begin{array}{r} -5 \\ 7 \\ 2 \\ 7 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right]\nonumber \] Find a basis for \(V\) which extends the basis for \(W\).
Solution
Note that the above vectors are not linearly independent, but their span, denoted as \(V\) is a subspace which does include the subspace \(W\).
Using the process outlined in the previous example, form the following matrix
\[\left[ \begin{array}{rrrrr} 1 & 0 & 7 & -5 & 0 \\ 0 & 1 & -6 & 7 & 0 \\ 1 & 1 & 1 & 2 & 0 \\ 0 & 1 & -6 & 7 & 1 \end{array} \right]\nonumber \]
Next find its reduced row-echelon form \[\left[ \begin{array}{rrrrr} 1 & 0 & 7 & -5 & 0 \\ 0 & 1 & -6 & 7 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]\nonumber \]
It follows that a basis for \(V\) consists of the first two vectors and the last. \[\left\{ \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right] \right\}\nonumber \] Thus \(V\) is of dimension 3 and it has a basis which extends the basis for \(W\).
Row Space, Column Space, and Null Space of a Matrix
We begin this section with a new definition.
Using the reduced row-echelon form, we can obtain an efficient description of the row and column space of a matrix. Consider the following lemma.
Lemma \(\PageIndex{1}\): E ffect of Row Operations on Row Space
Let \(A\) and \(B\) be \(m\times n\) matrices such that \(A\) can be carried to \(B\) by elementary row \(\left[ \mbox{column} \right]\) operations. Then \(\mathrm{row}(A)=\mathrm{row}(B)\) \(\left[\mathrm{col}(A)=\mathrm{col}(B) \right]\).
- Proof
-
We will prove that the above is true for row operations, which can be easily applied to column operations.
Let \(\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_m\) denote the rows of \(A\).
- If \(B\) is obtained from \(A\) by a interchanging two rows of \(A\), then \(A\) and \(B\) have exactly the same rows, so \(\mathrm{row}(B)=\mathrm{row}(A)\).
- Suppose \(p\neq 0\), and suppose that for some \(j\), \(1\leq j\leq m\), \(B\) is obtained from \(A\) by multiplying row \(j\) by \(p\). Then \[\mathrm{row}(B)=\mathrm{span}\{ \vec{r}_1, \ldots, p\vec{r}_{j}, \ldots, \vec{r}_m\}.\nonumber \] Since \[\{ \vec{r}_1, \ldots, p\vec{r}_{j}, \ldots, \vec{r}_m\} \subseteq\mathrm{row}(A),\nonumber \] it follows that \(\mathrm{row}(B)\subseteq\mathrm{row}(A)\). Conversely, since \[\{ \vec{r}_1, \ldots, \vec{r}_m\}\subseteq\mathrm{row}(B),\nonumber \] it follows that \(\mathrm{row}(A)\subseteq\mathrm{row}(B)\). Therefore, \(\mathrm{row}(B)=\mathrm{row}(A)\).
Suppose \(p\neq 0\), and suppose that for some \(i\) and \(j\), \(1\leq i,j\leq m\), \(B\) is obtained from \(A\) by adding \(p\) time row \(j\) to row \(i\). Without loss of generality, we may assume \(i<j\).
Then \[\mathrm{row}(B)=\mathrm{span}\{ \vec{r}_1, \ldots, \vec{r}_{i-1}, \vec{r}_i+p\vec{r}_j, \ldots,\vec{r}_j,\ldots, \vec{r}_m\}.\nonumber \]
Since \[\{ \vec{r}_1, \ldots, \vec{r}_{i-1}, \vec{r}_i+p\vec{r}_{j}, \ldots, \vec{r}_m\} \subseteq\mathrm{row}(A),\nonumber \] it follows that \(\mathrm{row}(B)\subseteq\mathrm{row}(A)\).
Conversely, since \[\{ \vec{r}_1, \ldots, \vec{r}_m\}\subseteq\mathrm{row}(B),\nonumber \] it follows that \(\mathrm{row}(A)\subseteq\mathrm{row}(B)\). Therefore, \(\mathrm{row}(B)=\mathrm{row}(A)\).
Consider the following lemma.
This lemma suggests that we can examine the reduced row-echelon form of a matrix in order to obtain the row space. Consider now the column space. The column space can be obtained by simply saying that it equals the span of all the columns. However, you can often get the column space as the span of fewer columns than this. A variation of the previous lemma provides a solution. Suppose \(A\) is row reduced to its reduced row-echelon form \(R\). Identify the pivot columns of \(R\) (columns which have leading ones), and take the corresponding columns of \(A\). It turns out that this forms a basis of \(\mathrm{col}(A)\).
Before proceeding to an example of this concept, we revisit the definition of rank.
Definition \(\PageIndex{10}\): Rank of a Matrix
Previously, we defined \(\mathrm{rank}(A)\) to be the number of leading entries in the row-echelon form of \(A\). Using an understanding of dimension and row space, we can now define rank as follows: \[\mbox{rank}(A) = \dim(\mathrm{row}(A))\nonumber \]
Consider the following example.
Example \(\PageIndex{16}\): Rank, Column and Row Space
Find the rank of the following matrix and describe the column and row spaces. \[A = \left[ \begin{array}{rrrrr} 1 & 2 & 1 & 3 & 2 \\ 1 & 3 & 6 & 0 & 2 \\ 3 & 7 & 8 & 6 & 6 \end{array} \right]\nonumber \]
Solution
The reduced row-echelon form of \(A\) is \[\left[ \begin{array}{rrrrr} 1 & 0 & -9 & 9 & 2 \\ 0 & 1 & 5 & -3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] Therefore, the rank is \(2\).
Notice that the first two columns of \(R\) are pivot columns. By the discussion following Lemma \(\PageIndex{2}\) , we find the corresponding columns of \(A\), in this case the first two columns. Therefore a basis for \(\mathrm{col}(A)\) is given by \[\left\{\left[ \begin{array}{r} 1 \\ 1 \\ 3 \end{array} \right] , \left[ \begin{array}{r} 2 \\ 3 \\ 7 \end{array} \right] \right\}\nonumber \]
For example, consider the third column of the original matrix. It can be written as a linear combination of the first two columns of the original matrix as follows. \[\left[ \begin{array}{r} 1 \\ 6 \\ 8 \end{array} \right] =-9\left[ \begin{array}{r} 1 \\ 1 \\ 3 \end{array} \right] +5\left[ \begin{array}{r} 2 \\ 3 \\ 7 \end{array} \right]\nonumber \]
What about an efficient description of the row space? By Lemma \(\PageIndex{2}\) we know that the nonzero rows of \(R\) create a basis of \(\mathrm{row}(A)\). For the above matrix, the row space equals \[\mathrm{row}(A) = \mathrm{span} \left\{ \left[ \begin{array}{rrrrr} 1 & 0 & -9 & 9 & 2 \end{array} \right], \left[ \begin{array}{rrrrr} 0 & 1 & 5 & -3 & 0 \end{array} \right] \right\}\nonumber \]
Notice that the column space of \(A\) is given as the span of columns of the original matrix, while the row space of \(A\) is the span of rows of the reduced row-echelon form of \(A\).
Consider another example.
Example \(\PageIndex{17}\): Rank, Column and Row Space
Find the rank of the following matrix and describe the column and row spaces. \[\left[ \begin{array}{rrrrrr} 1 & 2 & 1 & 3 & 2 \\ 1 & 3 & 6 & 0 & 2 \\ 1 & 2 & 1 & 3 & 2 \\ 1 & 3 & 2 & 4 & 0 \end{array} \right]\nonumber \]
Solution
The reduced row-echelon form is \[\left[ \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & \frac{13}{2} \\ 0 & 1 & 0 & 2 & -\frac{5}{2} \\ 0 & 0 & 1 & -1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] and so the rank is \(3\). The row space is given by \[\mathrm{row}(A) = \mathrm{span} \left\{ \left[ \begin{array}{ccccc} 1 & 0 & 0 & 0 & \frac{13}{2} \end{array} \right], \left[ \begin{array}{rrrrr} 0 & 1 & 0 & 2 & -\frac{5}{2} \end{array} \right] , \left[ \begin{array}{rrrrr} 0 & 0 & 1 & -1 & \frac{1}{2} \end{array} \right] \right\}\nonumber \]
Notice that the first three columns of the reduced row-echelon form are pivot columns. The column space is the span of the first three columns in the original matrix , \[\mathrm{col}(A) = \mathrm{span} \left\{ \left[ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array} \right], \; \left[ \begin{array}{r} 2 \\ 3 \\ 2 \\ 3 \end{array} \right] , \; \left[ \begin{array}{r} 1 \\ 6 \\ 1 \\ 2 \end{array} \right] \right\}\nonumber \]
Consider the solution given above for Example \(\PageIndex{17}\) , where the rank of \(A\) equals \(3\). Notice that the row space and the column space each had dimension equal to \(3\). It turns out that this is not a coincidence, and this essential result is referred to as the Rank Theorem and is given now. Recall that we defined \(\mathrm{rank}(A) = \mathrm{dim}(\mathrm{row}(A))\).
The following statements all follow from the Rank Theorem.
Corollary \(\PageIndex{5}\): Results of the Rank Theorem
Let \(A\) be a matrix. Then the following are true:
- \(\mathrm{rank}(A) = \mathrm{rank}(A^T)\).
- For \(A\) of size \(m \times n\), \(\mathrm{rank}(A) \leq m\) and \(\mathrm{rank}(A) \leq n\).
- For \(A\) of size \(n \times n\), \(A\) is invertible if and only if \(\mathrm{rank}(A) = n\).
- For invertible matrices \(B\) and \(C\) of appropriate size, \(\mathrm{rank}(A) = \mathrm{rank}(BA) = \mathrm{rank}(AC)\).
Consider the following example.
Example \(\PageIndex{18}\): Rank of the Transpose
Let \[A = \left[ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right]\nonumber \] Find \(\mathrm{rank}(A)\) and \(\mathrm{rank}(A^T)\).
Solution
To find \(\mathrm{rank}(A)\) we first row reduce to find the reduced row-echelon form. \[A = \left[ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]\nonumber \]
Therefore the rank of \(A\) is \(2\). Now consider \(A^T\) given by \[A^T = \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right]\nonumber \] Again we row reduce to find the reduced row-echelon form.
\[\left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right]\nonumber \]
You can see that \(\mathrm{rank}(A^T) = 2\), the same as \(\mathrm{rank}(A)\).
We now define what is meant by the null space of a general \(m\times n\) matrix.
It can also be referred to using the notation \(\ker \left( A\right)\). Similarly, we can discuss the image of \(A\), denoted by \(\mathrm{im}\left( A\right)\). The image of \(A\) consists of the vectors of \(\mathbb{R}^{m}\) which “get hit” by \(A\). The formal definition is as follows.
Consider \(A\) as a mapping from \(\mathbb{R}^{n}\) to \(\mathbb{R}^{m}\) whose action is given by multiplication. The following diagram displays this scenario. \[\overset{\mathrm{null} \left( A\right) }{\mathbb{R}^{n}}\ \overset{A}{\rightarrow }\ \overset{ \mathrm{im}\left( A\right) }{\mathbb{R}^{m}}\nonumber \] As indicated, \(\mathrm{im}\left( A\right)\) is a subset of \(\mathbb{R}^{m}\) while \(\mathrm{null} \left( A\right)\) is a subset of \(\mathbb{R}^{n}\).
It turns out that the null space and image of \(A\) are both subspaces. Consider the following example.
Example \(\PageIndex{19}\): Null Space
Let \(A\) be an \(m\times n\) matrix. Then the null space of \(A\), \(\mathrm{null}(A)\) is a subspace of \(\mathbb{R}^n\).
Solution
- Since \(A\vec{0}_n=\vec{0}_m\), \(\vec{0}_n\in\mathrm{null}(A)\).
- Let \(\vec{x},\vec{y}\in\mathrm{null}(A)\). Then \(A\vec{x}=\vec{0}_m\) and \(A\vec{y}=\vec{0}_m\), so \[A(\vec{x}+\vec{y})=A\vec{x}+A\vec{y} = \vec{0}_m+\vec{0}_m=\vec{0}_m,\nonumber \] and thus \(\vec{x}+\vec{y}\in\mathrm{null}(A)\).
- Let \(\vec{x}\in\mathrm{null}(A)\) and \(k\in\mathbb{R}\). Then \(A\vec{x}=\vec{0}_m\), so \[A(k\vec{x}) = k(A\vec{x})=k\vec{0}_m=\vec{0}_m,\nonumber \] and thus \(k\vec{x}\in\mathrm{null}(A)\).
Therefore by the subspace test, \(\mathrm{null}(A)\) is a subspace of \(\mathbb{R}^n\).
The proof that \(\mathrm{im}(A)\) is a subspace of \(\mathbb{R}^m\) is similar and is left as an exercise to the reader.
We now wish to find a way to describe \(\mathrm{null}(A)\) for a matrix \(A\). However, finding \(\mathrm{null} \left( A\right)\) is not new! There is just some new terminology being used, as \(\mathrm{null} \left( A\right)\) is simply the solution to the system \(A\vec{x}=\vec{0}\).
Consider the following example.
Example \(\PageIndex{20}\): Null Space of \(A\)
Let \[A=\left[ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & -1 & 1 \\ 2 & 3 & 3 \end{array} \right]\nonumber \] Find \(\mathrm{null} \left( A\right)\) and \(\mathrm{im}\left( A\right)\).
Solution
In order to find \(\mathrm{null} \left( A\right)\), we simply need to solve the equation \(A\vec{x}=\vec{0}\). This is the usual procedure of writing the augmented matrix, finding the reduced row-echelon form and then the solution. The augmented matrix and corresponding reduced row-echelon form are \[\left[ \begin{array}{rrr|r} 1 & 2 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 2 & 3 & 3 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rrr|r} 1 & 0 & 3 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]\nonumber \]
The third column is not a pivot column, and therefore the solution will contain a parameter. The solution to the system \(A\vec{x}=\vec{0}\) is given by \[\left[ \begin{array}{r} -3t \\ t \\ t \end{array} \right] :t\in \mathbb{R}\nonumber \] which can be written as \[t \left[ \begin{array}{r} -3 \\ 1 \\ 1 \end{array} \right] :t\in \mathbb{R}\nonumber \]
Therefore, the null space of \(A\) is all multiples of this vector, which we can write as \[\mathrm{null} (A) = \mathrm{span} \left\{ \left[ \begin{array}{r} -3 \\ 1 \\ 1 \end{array} \right] \right\}\nonumber \]
Finally \(\mathrm{im}\left( A\right)\) is just \(\left\{ A\vec{x} : \vec{x} \in \mathbb{R}^n \right\}\) and hence consists of the span of all columns of \(A\), that is \(\mathrm{im}\left( A\right) = \mathrm{col} (A)\).
Notice from the above calculation that that the first two columns of the reduced row-echelon form are pivot columns. Thus the column space is the span of the first two columns in the original matrix , and we get \[\mathrm{im}\left( A\right) = \mathrm{col}(A) = \mathrm{span} \left\{ \left[ \begin{array}{r} 1 \\ 0 \\ 2 \end{array} \right], \; \left[ \begin{array}{r} 2 \\ -1 \\ 3 \end{array} \right] \right\}\nonumber \]
Here is a larger example, but the method is entirely similar.
Example \(\PageIndex{21}\): Null Space of \(A\)
Let \[A=\left[ \begin{array}{rrrrr} 1 & 2 & 1 & 0 & 1 \\ 2 & -1 & 1 & 3 & 0 \\ 3 & 1 & 2 & 3 & 1 \\ 4 & -2 & 2 & 6 & 0 \end{array} \right]\nonumber \] Find the null space of \(A\).
Solution
To find the null space, we need to solve the equation \(AX=0\). The augmented matrix and corresponding reduced row-echelon form are given by
\[\left[ \begin{array}{rrrrr|r} 1 & 2 & 1 & 0 & 1 & 0 \\ 2 & -1 & 1 & 3 & 0 & 0 \\ 3 & 1 & 2 & 3 & 1 & 0 \\ 4 & -2 & 2 & 6 & 0 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rrrrr|r} 1 & 0 & \frac{3}{5} & \frac{6}{5} & \frac{1}{5} & 0 \\ 0 & 1 & \frac{1}{5} & -\frac{3}{5} & \frac{2}{5} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] It follows that the first two columns are pivot columns, and the next three correspond to parameters. Therefore, \(\mathrm{null} \left( A\right)\) is given by \[\left[ \begin{array}{c} \left( -\frac{3}{5}\right) s +\left( -\frac{6}{5}\right) t+\left( \frac{1}{5}\right) r \\ \left( -\frac{1}{5}\right) s +\left( \frac{3}{5}\right) t +\left( - \frac{2}{5}\right) r \\ s \\ t \\ r \end{array} \right] :s ,t ,r\in \mathbb{R}\text{.}\nonumber \] We write this in the form \[s \left[ \begin{array}{r} -\frac{3}{5} \\ -\frac{1}{5} \\ 1 \\ 0 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} -\frac{6}{5} \\ \frac{3}{5} \\ 0 \\ 1 \\ 0 \end{array} \right] + r \left[ \begin{array}{r} \frac{1}{5} \\ -\frac{2}{5} \\ 0 \\ 0 \\ 1 \end{array} \right] :s , t , r\in \mathbb{R}\text{.}\nonumber \] In other words, the null space of this matrix equals the span of the three vectors above. Thus \[\mathrm{null} \left( A\right) =\mathrm{span}\left\{ \left[ \begin{array}{r} -\frac{3}{5} \\ -\frac{1}{5} \\ 1 \\ 0 \\ 0 \end{array} \right] ,\left[ \begin{array}{r} -\frac{6}{5} \\ \frac{3}{5} \\ 0 \\ 1 \\ 0 \end{array} \right] ,\left[ \begin{array}{r} \frac{1}{5} \\ -\frac{2}{5} \\ 0 \\ 0 \\ 1 \end{array} \right] \right\}\nonumber \]
Notice also that the three vectors above are linearly independent and so the dimension of \(\mathrm{null} \left( A\right)\) is 3. The following is true in general, the number of parameters in the solution of \(AX=0\) equals the dimension of the null space. Recall also that the number of leading ones in the reduced row-echelon form equals the number of pivot columns, which is the rank of the matrix, which is the same as the dimension of either the column or row space.
Before we proceed to an important theorem, we first define what is meant by the nullity of a matrix.
From our observation above we can now state an important theorem.
Example \(\PageIndex{22}\): Rank and Nullity
Let \[A=\left[ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & -1 & 1 \\ 2 & 3 & 3 \end{array} \right]\nonumber \]
Find \(\mathrm{rank}\left( A\right)\) and \(\dim( \mathrm{null}\left(A\right))\).
Solution
In the above Example \(\PageIndex{20}\) we determined that the reduced row-echelon form of \(A\) is given by \[\left[ \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} \right]\nonumber \]
Therefore the rank of \(A\) is \(2\). We also determined that the null space of \(A\) is given by \[\mathrm{null} (A) = \mathrm{span} \left\{ \left[ \begin{array}{r} -3 \\ 1 \\ 1 \end{array} \right] \right\}\nonumber \]
Therefore the nullity of \(A\) is \(1\). It follows from Theorem \(\PageIndex{14}\) that \(\mathrm{rank}\left( A\right) + \dim( \mathrm{null}\left(A\right)) = 2 + 1 = 3\), which is the number of columns of \(A\).
We conclude this section with two similar, and important, theorems.
Theorem \(\PageIndex{15}\)
Let \(A\) be an \(m\times n\) matrix. The following are equivalent.
- \(\mathrm{rank}(A)=n\).
- \(\mathrm{row}(A)=\mathbb{R}^n\), i.e., the rows of \(A\) span \(\mathbb{R}^n\).
- The columns of \(A\) are independent in \(\mathbb{R}^m\).
- The \(n\times n\) matrix \(A^TA\) is invertible.
- There exists an \(n\times m\) matrix \(C\) so that \(CA=I_n\).
- If \(A\vec{x}=\vec{0}_m\) for some \(\vec{x}\in\mathbb{R}^n\), then \(\vec{x}=\vec{0}_n\).
Theorem \(\PageIndex{16}\)
Let \(A\) be an \(m\times n\) matrix. The following are equivalent.
- \(\mathrm{rank}(A)=m\).
- \(\mathrm{col}(A)=\mathbb{R}^m\), i.e., the columns of \(A\) span \(\mathbb{R}^m\).
- The rows of \(A\) are independent in \(\mathbb{R}^n\).
- The \(m\times m\) matrix \(AA^T\) is invertible.
- There exists an \(n\times m\) matrix \(C\) so that \(AC=I_m\).
- The system \(A\vec{x}=\vec{b}\) is consistent for every \(\vec{b}\in\mathbb{R}^m\).


3.2: Power Rule
“I’ve got the power!” –Snap (German band)
If you apply the definition of the derivative to several functions, you’ll see:
\(\begin{array}{cc} \frac{d}{dx} x^0 & 0 \\ \frac{d}{dx} x^1 & 1\\ \frac{d}{dx} x^2 & 2x \\ \frac{d}{dx} x^3 & 3x^2 \\ \frac{d}{dx} x^4 & 4x^3 \end{array}\)
See the pattern?
\(\boxed{\frac{d}{dx} x^n = n x^{n-1}}\)
As we’ll see in the next section, this even works for non-integers. The key is to multiply by the exponent, then decrease the exponent by one .
The next rules say that constant multiples and addition work nicely.
\(\boxed{\frac{d}{dx} c \cdot f(x) = c f'(x)}\)
\(\boxed{\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)}\)
The key is you can just worry about the derivative of each piece of a sum separately. Constant multiples “come along for the ride”. With only these three rules, you can now take the derivative of any polynomial. Check it out.
Find the following derivatives.
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\(\frac{d}{dx} 3x^2\)
To do this one, we use the power rule on the \(x^2\) part, and get \(2x\). However, we are also multiplying by \(3\), so the answer is multiplied by \(3\) as well. Hence the answer is \(3(2x) = \boxed{6x}\).
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\(\frac{d}{dx} x^3 + x\)
By the power rule, we find \(\frac{d}{dx} x^3 = 3x^2\), and \(\frac{d}{dx} x\) is \(\frac{d}{dx} x^1\) which becomes \(1x^0\) by the power rule, which is \(1\). By the addition rule, we have \(\frac{d}{dx} x^3 + x = \boxed{3x^2 + 1}\).
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\(\frac{d}{dx} 2x^3 + 5\)
You take the derivative of \(x^3\) and you have \(3x^2\). Times by \(2\), that leaves \(6x^2\). Okay, about the five? It is tempting to leave the five put, but actually \(\frac{d}{dx} 5 = 0\). Why? Well, it’s a constant, so it does not affect the slope. Hence we get \(\boxed{6x^2}\).
More power rule examples
Note that the power rule works with fractional and negative exponents as well! Here are some examples.
To apply the power rule in this case, we need to first multiply by the exponent ( ), then subtract one from the exponent . Then we have
In this problem, we just worry about the \(x^{-2}\) to start with. We multiply by the exponent \(({\color{red} -2})\), and then subtract one from that to get \({\color{blue} -2 - 1 = -3}\). Then we have
\(\frac{d}{dx} 7x^{-2} = 7 ({\color{red} -2} x^
Callstack:
at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/03:_Rules_for_Derivatives/3.02:_Power_Rule), /content/body/div[2]/div[2]/div/p[3]/span, line 1, column 1
Sometimes there are hidden fractional or negative exponents.  Just remember these rules:
\(\boxed{\frac{1}{x^n} = x^{-n}}\)
Let’s see some examples.
Find
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\(\frac{d}{dx} \sqrt{x}\)
This problem is much easier if we can rewrite the \(\sqrt{x}\). This is the same thing as , and hence we have
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\(\frac{d}{dx} 5\sqrt[3]{x}\)
To solve this, it really helps to rewrite \(\sqrt[3]{x}\) as . Once you do that, this prolem is just the power rule and constant multiple rule:
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\(\frac{d}{dx} \sqrt[5]{x^3} + x^2 + 7\)
Focus on the easy parts first: we know \(\frac{d}{dx} x^2 = 2x\), and we know that \(\frac{d}{dx} 7 = 0\). So we just need to figure out the \(\frac{d}{dx} \sqrt[5]{x^3}\). What is this? Well, we can rewrite this as . So now it is just the power rule, and we multiply by and subtract to get . Hence . Putting it all together:
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\(\frac{d}{dx} \frac{1}{x}\)
We can rewrite \(\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} \frac{1}{x^1}\) as \(\frac{d}{dx} x^{-1}\). Now we apply the power rule:
\(\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} x^{-1} = \boxed{-1 x^{-2}}.\)
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\(\frac{d}{dx} \frac{4}{x^2}\)
We can think of \(4\) as just a constant out front, and hence we want \(\frac{d}{dx} 4 \left( \frac{1}{x^2} \right )\). We can then rewrite \(\frac{1}{x^2}\) as \(x^{-2}\). And we want \(\frac{d}{dx} 4(x^{-2})\). Using the power rule, we multiply by \(-2\) and subtract one, and we have
\(\frac{d}{dx} \frac{4}{x^2} = \frac{d}{dx} 4x^{-2} = \boxed{-8 x^{-3}}.\)
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\(\frac{d}{dx} \frac{1}{\sqrt{x}}\)
This combines the fractional and denominator stuff. We first rewrite \(\sqrt{x}\) as :
We then rewrite as a negative fractional exponent.
Finally, we use the power rule.


1.3.8: Substitutes for and Neutralizers of Leadership
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Learning Objectives
- What does the concept “substitute for leadership” mean?
Several factors have been discovered that can substitute for or neutralize the effects of leader behavior (see Table 13.1). 89 Substitutes for leadership behavior can clarify role expectations, motivate organizational members, or satisfy members (making it unnecessary for the leader to attempt to do so). In some cases, these substitutes supplement the behavior of a leader. Sometimes it is a group member’s characteristics that make leadership less necessary, as when a master craftsperson or highly skilled worker performs up to his or her own high standards without needing outside prompting. Sometimes the task’s characteristics take over, as when the work itself—solving an interesting problem or working on a familiar job—is intrinsically satisfying. Sometimes the characteristics of the organization make leadership less necessary, as when work rules are so clear and specific that workers know exactly what they must do without help from the leader (seeAn Inside Lookat flat management structure and the orchestra with no leader).
| Substitutes for and Neutralizers of Leader Behavior |
| Supportive or Neutralizer | Substitute Leadership | Instrumental Leadership |
| A. Subordinate Characteristics: | ||
| 1. Experience, ability, training | Substitute | |
| 2. "Professional" orientation | Substitute | Substitute |
| 3. Indifference toward rewards offered by organization | Neutralizer | Neutralizer |
| B. Task Characteristics: | ||
| 1. Structured, routine, unambiguous task | Substitute | |
| 2. Feedback provided by task | Substitute | |
| 3. Intrinsically satisfying task | Substitute | |
| C. Organization Characteristics: | ||
| 1. Cohesive work group | Substitute | Substitute |
| 2. Low position power (leader lacks control over organizational rewards) | Neutralizer | Neutralizer |
| 3. Formalization (explicit plans, goals, areas of responsibility) | Substitute | |
| 4. Inflexibility (rigid, unyielding rules and procedures) | Neutralizer | |
| 5. Leader located apart from subordinates with only limited communication possible | Neutralizer | Neutralizer |
| Source: Adapted fromLeadership in organizationsby G. A. Yukl. |
Table 13.1
Neutralizers of leadership, on the other hand, are not helpful; they prevent leaders from acting as they wish. A computer-paced assembly line, for example, prevents a leader from using initiating structure behavior to pace the line. A union contract that specifies that workers be paid according to seniority prevents a leader from dispensing merit-based pay. Sometimes, of course, neutralizers can be beneficial. Union contracts, for example, clarify disciplinary proceedings and identify the responsibilities of both management and labor. Leaders must be aware of the presence of neutralizers and their effects so that they can eliminate troublesome neutralizers or take advantage of any potential benefits that accompany them (such as the clarity of responsibilities provided by a union contract). If a leader’s effectiveness is being neutralized by a poor communication system, for example, the leader might try to remove the neutralizer by developing (or convincing the organization to develop) a more effective system.
Followers differ considerably in their focus of attention while at work, thereby affecting the effectiveness of the act of leadership. Focus of attention is an employee’s cognitive orientation while at work. It reflects what and how strongly an individual thinks about various objects, events, or phenomena while physically present at work. Focus of attention reflects an individual difference in that not all individuals have the same cognitive orientation while at work—some think a great deal about their job, their coworkers, their leader, or off-the-job factors, while others daydream. 90 An employee’s focus of attention has both “trait” and “state” qualities. For example, there is a significant amount of minute-by-minute variation in an employee’s focus of attention (the “state” component), and there is reasonable consistency in the categories of events that employees think about while they are at work (the “trait” component).
Research suggests that the more followers focus on off-job (nonleader) factors, the less they will react to the leader’s behaviors. Thus, a strong focus on one’s life “away from work” (for example, time with family and friends) tends to neutralize the motivational, attitudinal, and/or behavioral effects associated with any particular leader behavior. It has also been observed, however, that a strong focus on the leader, either positive or negative, enhances the impact that the leader’s behaviors have on followers. 91
Managerial Leadership
You Are Now the Leader
Leading and managing are two very different things. Being a manager means something more than gaining authority or charge over former colleagues. With the title does come the power to affect company outcomes, but it also comes with something more: the power to shape the careers and personal growth of subordinates.
According to Steve Keating, a senior manager at the Toro Company, it is important not to assume that being made a manager automatically makes you a leader. Rather, being a manager means having the opportunity to lead. Enterprises need managers to guide processes, but the employees—the people—need a leader. Keating believes that leaders need a mindset that emphasizes people, and the leader’s job is to help the people in the organization to be successful. According to Keating, “If you don’t care for people, you can’t lead them” (Hakim 2017 n.p.).
For someone who has been promoted over his peers, ground rules are essential. "Promotion doesn’t mean the end of friendship but it does change it," according to Keating. If apeerhas been promoted, rather than grouse and give in to envy, it is important to step back and look at the new manager; take a hard look at why the peer was promoted and what skill or characteristic made you a less appealing fit for the position (Hakim 2017).
Carol Walker, president of Prepared to Lead, a management consulting firm, advises new managers to develop a job philosophy. She urges new managers to develop a core philosophy that provides a guide to the day-to-day job of leading. She urges managers to build up the people they are leading and work as a "servant leader." The manager’s perspective should be on employee growth and success. Leaders must bear in mind that employees don’t work for the manager; they work for the organization—and for themselves. Managers coordinate this relationship; they are not the center of it. Work should not be assigned haphazardly, but with the employee’s skills and growth in mind. “An employee who understands why she has been asked to do something is far more likely to assume true ownership for the assignment," Walker says (Yakowicz 2015 n.p.). A leader’s agenda should be on employee success, not personal glory. Employees are more receptive when they recognize that their leader is working not for their own success, but for the employee’s success.
A survey from HighGround revealed one important item that most new managers and even many seasoned managers overlook: asking for feedback. Everyone has room for growth, even managers. Traditional management dictates a top-down style in which managers review subordinates. But many companies have found it beneficial to turn things around and ask employees, “How can I be a better manager?” Of course, this upward review only works if employees believe that their opinion will be heard. Managers need to carefully cultivate a rapport where employees don't fear reprisals for negative feedback. Listening to criticism from those you are leading builds trust and helps ensure that as a manager, you are providing the sort of leadership that employees need to be successful (Kauflin 2017). Showing respect and caring for employees by asking this simple question is inspiring —an important aspect of leadership itself. Whether asking for feedback or focusing on an employee’s fit with a particular job description, a leader helps guide employees through the day-to-day, builds a positive culture, and helps employees improve their skills.
Sources
Hakim, Amy C. 2017. “When a Manager Becomes a Leader.” Psychology Today. https://www.psychologytoday.com/blog...becomes-leader
Yakowicz, Will. 2015. “How to Help a New Manager Become a Great Leader.” Inc .
www.inc.com/will-yakowicz/ho...t-leaders.html Kauflin, Jeff. 2017. “Every Manager Can Become A Better Leader By Asking This One Question.” Forbes . www.forbes.com/sites/jeffkau.../#3ca1eaff4ac1
Questions
- What do you think are the most important qualities in a leader? In a manager? Are your two lists mutually exclusive? Why?
- How do you think a leader can use feedback to model the growth process for employees?
Concept Check
- Identify and describe substitutes of leadership.
References
89. P.M. Podsakoff, B.P. Niehoff, S.B. MacKenzie, & M.L. Williams. 1993. Do substitutes for leadership really substitute for leadership: An empirical examination of Kerr and Jermier’s situational leadership model. Organizational Behavior and Human Decision Processes 54:1–44; S. Kerr. 1977. Substitutes for leadership: Some implications for organizational design. Organization and Administrative Sciences 8:135–146; S. Kerr & J.M. Jermier. 1978. Substitutes for leadership: Their meaning and measurement. Organizational Behavior and Human Performance 22:375– 403; J. P. Howell & P. W. Dorfman. 1981. Substitutes for leadership: Test of a construct. Academy of Management Journal 24:714– 728; J.L. Pierce, R.B. Dunham, & L.L. Cummings. 1984. Sources of environmental structuring and participant responses. Organizational Behavior and Human Performance 33:214–242.
90. D.G. Gardner, R.B. Dunham, L.L. Cummings, & J.L. Pierce. 1989. Focus of attention at work: Construct definition and empirical validation. Journal of Occupational Psychology 62:61–77.
91. D.G. Gardner, R.B. Dunham, L.L. Cummings, & J.L. Pierce. 1987. Focus of attention at work and leader-follower relationships. Journal of Occupational Behaviour 8:277–294.


15.16: Gap 5- Actual Performance vs. Perceived Service Gap
- Outline how a retailer can close the actual performance and perceived service gap
Gap between expected service and experienced service: This gap arises when the consumer misinterprets the service quality. The physician may keep visiting the patient to show and ensure care, but the patient may interpret this as an indication that something is really wrong.
The fifth or Customer gap represents the overall difference between the customer’s expectations for and perception of the level of service received. This gap can occur due to service issues from gaps 1–4 or can reflect an error in the customer’s judgement of the service received. The customer’s perception is influenced by a range of factors including personal experience and needs, advertising and word of mouth. For example, a customer may have experienced excellent service during a meal but incorrect charges on the bill or boisterous diners at the next table may taint the experience and leave the customer dissatisfied overall. The key point to recognize is that the fifth gap can’t be closed directly; in order to close the customer gap, a retailer needs to close gaps 1–4. Obtaining customer feedback is the key to identifying the root cause(s)—that is, the underlying service gap(s)—and, ultimately, closing the Customer gap.
Contributors and Attributions
- Gap 5: Actual Performance vs. Perceived Service Gap. Authored by : Nina Burokas. Provided by : Lumen Learning. License : CC BY: Attribution
- Introduction to Gap 5. Provided by : Boundless. Located at : https://courses.lumenlearning.com/boundless-marketing/chapter/service-quality/ . License : CC BY-SA: Attribution-ShareAlike


22.7C: Histology of the Liver
Hepatocytes are the main tissue cells of the liver. The gallbladder contains the mucosa, muscularis, perimuscular, and serosa layers.
- Describe the histology of the liver
Key Points
- A hepatocyte is the main tissue cell of the liver and makes up 70–80% of the liver’s cytoplasmic mass.
- Hepatocytes contain large amounts of rough endoplasmic reticulum and free ribosomes.
- Hepatocytes are involved in: protein synthesis; protein storage; the transformation of carbohydrates,; the synthesis of cholesterol, bile salts, and phospholipids; and detoxification, modification, and excretion of exogenous and endogenous substances.
- Hepatocytes are unique in that they are one of the few types of cells in the human body that are capable of regeneration.
- There are several different layers of the gallbladder: the mucosa (epithelium and lamina propria), the muscularis, the perimuscular, and the serosa.
Key Terms
- hepatocyte : Any of the cells in the liver responsible for the metabolism of proteins, carbohydrates, and lipids, and for detoxification.
The Liver
Hepatocytes : A cross-section of a human liver that shows hepatocytes.
A hepatocyte is the main tissue cell of the liver and makes up 70–80% of the liver’s cytoplasmic mass. Hepatocytes contain large amounts of rough endoplasmic reticulum and free ribosomes. Hepatocytes are involved in:
- Protein synthesis.
- Protein storage.
- The transformation of carbohydrates.
- The synthesis of cholesterol, bile salts, and phospholipids.
- The detoxification, modification, and excretion of exogenous and endogenous substances.
Hepatocytes also initiate the formation and secretion of bile. Hepatocytes are organized into plates separated by vascular channels (sinusoids) for blood vessels. The hepatocyte plates are one cell thick in mammals.
Hepatocytes are unique in that they are one of the few types of cell in the human body that are capable of regeneration. Hepatocytes are derived from hepatoblasts, the precursor stem cell of the liver that divides to produce new hepatocytes. The liver is capable of complete regeneration from as little as 25% of the original organ.


6.1: Molecular Ion
Molecular Ion
The molecular ion provides the molecular mass of the analyte and is the first clue used to interpret a mass spectrum. The mass to charge ratio of the molecular ion is based up on the mass of the most abundant isotope for each element in the molecule. This is not the relative atomic mass from the periodic table. Since many mass spectrometers have unit mass resolution, the isotope mass is normally rounded to the nearest whole number, this is called the nominal mass. For example the molecular ion for CHBr 3 is observed at 250 m/z ;(12 + 1 + 3 \(\times\) 79)=250) , not at the relative molecular weight of 253. The mass of the molecular ion is based upon the mass of the isotope with the highest natural abundance. The most common bromine isotope is 79 Br . Do not use the weighted average atomic weight for Br (79.9) which is based upon the natural abundance of 79 Br and 81 Br . The mass spectrum of CHBr 3 includes ions for all the naturally occurring isotopes and the intensity of each peak depends upon the probability for that combination of isotopes. These patterns are discussed in detail in the section on isotope abundance.
In many mass spectra, the molecular ion is easily identified as the ion with the highest mass to charge ratio. However, this assignment should be made with caution because the highest mass to charge ion may be an impurity, an isotope of the molecular ion, or a fragment. Many compounds fragment easily and no molecular ion is observed in the 70 eV EI spectrum. It is important to clarify that the molecular ion IS NOT necessarily the ion with the greatest abundance, the ion with the greatest abundance is called the base peak. The base peak is the peak with the greatest abundance. The mass spectrum is usually normalized so that this peak has an intensity of 100.
A list of molecular ion characteristics are in Table \(\PageIndex{1}\) to help you identify them in a mass spectrum. Low energy EI, where the ionization energy is reduced, often increases in intensity of the molecular ion. Chemical Ionization, CI, is also useful for identifying the molecular ion since the the adduct ion is often more stable than the radical cation produced by electron ionization. The adduct ion is often formed by protonating the analyte to form \((\mathrm{M}+\mathrm{H})\) and is observed at a mass to charge ratio of M+1 .
| The mass to charge ratio must correspond to a reasonable molecular formula with the proper isotope abundance. |
| Most compounds have an even molecular mass. The one common exception to this is the "Nitrogen Rule" discussed below. |
| The Nitrogen Rule: Any compound with an odd number of nitrogen atoms will have an odd molecular mass. Any compound with an even number of nitrogen atoms (including zero) will have an even molecular mass. This is because nitrogen is the only common atom where the most common isotope has an odd valence and an even mass. For example: the molecular ion for CH 4 is observed at 16 m/z , the molecular ion for NH 3 is observed at 17 m/z , and the molecular ion for N 2 H 4 is observed at 32 m/z . |
| If a peak is the molecular ion, the next highest mass fragment must correspond to the loss of a possible neutral fragment. For example, a peak that corresponds to loss of 5 u from the molecular ion is highly unlikely |
Figure \(\PageIndex{1}\) shows the mass spectrum of acetone (CH 3 COCH 3 ). The molecular ion is clearly shown at 58 m/z (12 x 3 + 6 x 1 + 16 = 58). The base peak is at 43 m/z and corresponds to loss of 15 m/z from the intact molecule, this is caused by breaking a C-C bond for loss of a CH 3 • radical to give CH 3 CO + at 43 m/z (12 x 2 + 3 x 1 + 16 = 43). The mass spectrum also includes several other minor peaks - the peak at 59 m/z is caused by the small abundance of C-13 that gives a small fraction of the acetone molecules a mass of 59; the peak at 15 m/z is caused by the CH 3 fragment retaining the charge when the C-C bond breaks. These fragmentation and isotope patterns are discussed in more detail in the following sections.


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3.5: Lecture Outline for Part II
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Suggested Lecture Material for Part II—Race and Ethnicity
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Review Master Status.
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Identify and discuss the history of race and race relations in the US since the 16
th
century.
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Definitions, discussion, and examples of race.
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Socially defined.
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A social construct.
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Definitions, discussion, and examples of ethnicity.
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Identify the various racial and ethnic groups in the US.
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Alaska Native
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American Indian
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Asian American
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Black or African American
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Latino(a)/Hispanic American
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Pacific Islander
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White or Anglo American
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How are these divisions determined?
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Why do we make these divisions?
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Discuss the various designations of racial and ethnic groups.
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What do people want to be called?
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How do racial/ethnic designations occur and change?
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Definitions, discussion, and examples of dominant group and dominant group status.
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Discuss and give examples of white privilege.
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Definitions, discussion, and examples of minority groups and minority group status.
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Discuss and give examples of bigotry against minority groups.
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Use the Internet to display population data about racial and ethnic groups in the US.
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Immigration
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Define immigration.
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Define emigration.
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Define and discuss push factors.
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Define and discuss pull factors.
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Assimilation.
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Define and discuss cultural assimilation.
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Define and discuss structural assimilation.
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Identify and discuss models of assimilation.
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Standard Model.
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Pluralistic Model.
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Use the Internet to display statistical information concerning immigration into the US.
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Historical data.
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Current data.
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Future data.
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What is the projected population of each racial and ethnic group?
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Use the Internet to display statistical information concerning race and ethnicity including such statistical referents as housing, health care, home ownership, business ownership, educational attainment, and labor force participation of each of the racial and ethnic groups in the US.
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Compare and contrast the data found.
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Use the Internet to display statistical information concerning the relationship between race and ethnicity and the criminal justice system in the United States.