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:
$A' = \left( \begin{array}{ccccc} 0 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0 & 0 & 1 & 0 \end{array} \right)$?
Best Answer
Yes, your solution to (a) is correct assuming your computations for how to express the elements is correct. Though you don't need to find how to write an arbitrary vector in terms of the basis of $W$, you only need to find out how to write $f(0,2,1)$, $f(-1,1,1)$, and $f(2,-1,1)$ (the images of the vectors in the given basis for $V$) in terms of the given basis of $W$.
No, you never need to use the fact that $V$ and $W$ are isomorphic (we don't consider a metric here, so "isometric" is not appropriate here).
Your answer for (b), on the other hand, is not well done (or correct).
The basis for your vector space is $\{1,t,\ldots,t^n\}$. To find the matrix representation of $f$ relative to this basis, you need to find the image of each basis vector in the domain, and express it in terms of the basis vectors of the range. But remember: the vectors in $U_n$ are not tuples, they are polynomials. So $(1,t,\ldots,t^n)$ is not an element of $U_n$.
So, you would have to find each of $f(1)$, $f(t)$, $f(t^2),\ldots, f(t^n)$, and then express them in terms of the basis of the range, which happens to again be $\{1,t,t^2,\ldots,t^n\}$. So for instance, $f(t^2) = 2t$, so $$f(t^2) = 0\cdot 1 + 2\cdot t + 0\cdot t^2 + \cdots + 0\cdot t^n.$$ That means that the third column of the matrix (the one corresponding to the third basis vector, which is $t^2$) will be the transpose of $(0,2,0,\ldots,0)$, that is $$\left(\begin{array}{c} 0\\2\\0\\ \vdots \\ 0\end{array}\right).$$ Etc.