<s>[INST] Determine the matrix relative to a given basis Question: (a) Let $f: V \rightarrow W$ with $ V,W \simeq \mathbb{R}^{3}$ given by: $$f(x_1, x_2, x_3) = (x_1 - x_3, 2x_1 -5x_2 -x_3, x_2 + x_3).$$
Determine the matrix of $f$ relative to the basis $\{(0,2,1),(-1,1,1),(2,-1,1)\}$ of $V$ and $\{(-1,-1,0),(1,-1,2),(0,2,0)\}$ of $W$.
(b) Let $n \in \mathbb{N}$ and $U_n$ the vector space of real polynomials of degree $\leq n$. The linear map $f: U_n \rightarrow U_n$ is given by $f(p) = p'$. Determine the matrix of $f$ relative to the basis $\{1,t,t^{2},...,t^{n}\}$ of $U_n$.
My attempt so far: 
(a): First relative to the bases of $W$ I found the coordinates of an arbitrary vector: $\left( \begin{array}{r} a \\ b \\ c \end{array} \right) = x \left( \begin{array}{r} -1 \\ -1 \\ 0 \end{array} \right) + y \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) + z \left( \begin{array}{c} 0 \\ 2 \\ 0 \end{array} \right)$
$\begin{array}{l} a = -x + y \\ b = - x - y + 2z \\ c = 2y \end{array}$ or $\begin{array}{l} x = -a + \frac{1}{2}c \\ z = -\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c \\ y = \frac{1}{2}c \end{array}$
At this point I believe I have the linear combinations of the given basis in $W$ for an arbitrary vector, so next I take the vectors from $V$ and send them to $W$ using the given function:
$\begin{array}{l} f(v_1) = f(0,2,1) = (-1,-11,3) = (1 + \frac{3}{2})w_1 + \frac{3}{2}w_2 + (\frac{1}{2} - \frac{11}{2} + \frac{3}{2})w_3 \\ f(v_2) = f(-1,1,1) = (-2,-8,2) = (2+1)w_1 + w_2 + (1 - 4 +1)w_3 \\ f(v_3) = f(2,-1,1) = (1,8,0) = w_1 + (-\frac{1}{2} + 4)w_3 \end{array}$
or $\left( \begin{array}{rrc} \frac{5}{2} & 3 & 1 \\ \frac{3}{2} & 1 & 0 \\ -\frac{7}{2} & -2 & \frac{7}{2}\end{array} \right)$
Was I taking the correct steps? I didn't really do anything differently based on the fact that $V,W$ were isometric... Is there a particular significance or interpretation for the resulting matrix?
(b): Not really sure here... 
$f(p) = p'$
would it make sense to write something like:
$f(1,t,t^{2},\dots, t^{n}) = (0,1,2t, \dots, nt^{n-1})$?
and if a basis for $(1,t,t^{2},\dots, t^{n})$ would be $A = \left( \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & \cdots & \cdots & 0 & 1 \end{array} \right)$
could i write: 
$A' =  \left( \begin{array}{ccccc} 0 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0 & 0 & 1 & 0 \end{array} \right)$?
 [/INST] Here's a suggestion for seeing the bigger picture in your first question:


*

*The key idea below is that the column vectors of a matrix $M$ are the images of the standard ordered basis when the matrix $M$ is viewed as a transformation
$M: \mathbb{R}^n \to \mathbb{R}^n$.

*It is easy to find a matrix of f with respect to the standard ordered basis of $\mathbb{R}^3$: it is given by a matrix with columns $\{1, 2, 0\}$, $\{0, -5, 1\}$, and $\{-1, -1, 1\}$.  Call this matrix $C$.

*The matrix which transforms the given ordered basis of $V$ to the standard basis of $\mathbb{R}^3$ is given by writing the basis vectors as columns of this matrix. Similarly, this can be done for the basis of $W$.  Let's call these matrices $A$ and $B$.

*You desire a matrix which represents $f$.  Interpreting the matrices above as transformations, consider the following diagram:
$$\begin{array}{ccc}
V & \to & \mathbb{R}^3\\
\downarrow & \qquad & \downarrow\\
W & \to & \mathbb{R}^3
\end{array}$$
where the horizontal maps are $A$ and $B$ and the right vertical map is $C$.
Your desired matrix is the left vertical arrow.  It is now clear that it is given by the matrix $B^{-1} C A$.
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