<s>[INST] Checking an introductory Linear Algebra Exercise regarding linear applications. My final is few days away and I'm doing the list of exercises recommended by the teacher. Not all exercises have solutions however, and the teacher isn't answering my questions, so I just want to check whether I did the following exercise correctly:

Let $f:\mathbb{R}^3\rightarrow \mathbb{R}^3 : (2x_1-x_2+x_3,x_2+x_3,x_1+x_2+2x_3)$.

*

*Prove $f$ is a linear application.

*Find the matrix $A$ that corresponds with the linear applicaiton $f$.

*Find a basis for $\text{Im}(f)$ and $\text{Ker}(f)$.



*

*Proving that $f(c\textbf{v})=cf(\textbf{v})$ and that $f(\textbf{v}+\textbf{u})=f(\textbf{v})+f(\textbf{u})$ is easy enough.


*$$A=\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}$$


*This is the bit I'm more concerned about. Vectors of $\text{Im}(f)$ are precisely those of the form
$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\x_2 \\ x_3\end{pmatrix}$$
we can apply elementary column operations since these don't affect the column space ($\text{Im}(f)$),
$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} \sim \begin{pmatrix} 0 & -1 & 0 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \end{pmatrix} \sim \begin{pmatrix} 0 & -1 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 0 \end{pmatrix}$$
Since the first two columns are clearly linearly independent, and vectors of $\text{Im}(f)$ are preciesly those of the form
$$x_1\begin{pmatrix} 0 \\2 \\ 3\end{pmatrix}+x_2\begin{pmatrix} -1 \\1 \\ 1\end{pmatrix}$$
for $x_1,x_2\in \mathbb{R}$, we must have that $\{ (0, 2, 3),(-1, 1, 1)\}$ is a basis of $\text{Im}(f)$. Solutions of the equation $$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}\textbf{v}=\textbf{0}$$
are not changed by multiplying on the left both sides by elementary matrices.
$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 2 & -1 & 1 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & -3 & -3 \end{pmatrix}\sim\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$
Thus $\text{Ker}(f)$ is precisely the set of vectors $v$ such that $x_1=x_2=-x_3$, or $$v=t\begin{pmatrix} 1 \\ 1\\ -1\end{pmatrix}$$ for $t\in \mathbb{R}$. Therefore $\{ (1,1,-1)\}$ is a basis of $\text{Ker}(f)$.
 [/INST] That all looks fine to me (except that for item 1, you'd get no credit unless you actually wrote out the proof, of course).
An alternative to your solution for 1 is to do item 2 first; then you have
$$
f(v) = Av
$$
for all $v \in \Bbb R^3$. So
$$
f(v+w) = A(v + w) = Av + Aw = f(v)+ f(w)
$$
by the distributive law for matrix multiplication, and a similar thing works for scalar multiples.
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