# Finding a transformation matrix with respect to a coordinate mapping

Determine the transformation matrix of $T$ with respect to the coordinate mapping:

$\xi(p) = (p(-1),p(0),p(1))$, and we define $T:\mathcal{P}_2 \to > \mathcal{P}_2$, with $T(p(t))=p(t+1)$ where $\mathcal{P}_2$ is the vectorspace of polynomials of maximum degree 2 over a field $F$.

I'm not sure how to do this. The answer should be:

$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{bmatrix}$

I know how to find the "standard" transformation matrix for $T$, by letting $T$ act on the standard basis $(1,t,t^2)$ of $\mathcal{P}_2$, then I can obtain a matrix which describes the linear transformation.

I know that a coordinate mapping/evaluation is isomorphism a from a vectorspace $V$, with dimension $n$ over a field $F$ to $F^n$, like this $\xi: V\to F^n$, then $\xi_j =\pi_j\xi$, where $\pi_j:F^n \to F, (x_1,x_2,\dots,x_n) \mapsto x_j$. Since this is a linear transformation I can also find a matrix for this as above with $\xi(p) = (p(-1),p(0),p(1))$.

I'm thinking that I should somehow should multiply the matrix for the transformation of $T$ and the transformation of coordinate mapping, but I can't get it to work so that I end up with the desired answer. It feels like I'm missing something (or rather a lot), I'm not sure how these should interact properly. Can someone help me out?

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Fix a basis, in this case you have $b_1(t)=1,b_2(t)=t,b_3(t)=t^2$.

Then you have $T(b_1) = t \mapsto 1 = b_1$, $T(b_2) = t \mapsto t+1 = b_1+b_2$ and $T(b_3) = t \mapsto t^2+2t+1 = b_3+2b_2+b_1$.

Then you have $\xi (b_1) = (1,1,1)^T$, $\xi (b_2) = (-1,0,1)^T$ and $\xi (b_3) = (1,0,1)^T$.

You have $\xi (T(b_1)) = \xi (b_1)$, $\xi (T(b_2)) = \xi(b_1)+\xi(b_2)$ and $\xi (T(b_3)) = \xi(b_3)+2\xi(b_2)+\xi(b_1)$.

Now compute the matrix that maps the vectors $\xi(b_k)$ to the vectors $\xi(T(b_k))$. In fact, you can read off the matrix from the above (using the basis $\xi(b_k)$) as:

$T_\xi=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$

Addendum: The matrix in the question is the same operator in a different basis, but it is not clear to me why this basis would be 'natural' in some sense (it is the controller canonical form, but deriving this is a minor pain).

A basis that 'works' is $\beta_1(t) = 1 -{3 \over 2} t + {1 \over 2} t^2$, $\beta_2(t) = 2t -t^2$, $\beta_3(t) = -{1 \over 2}t + {1 \over 2} t^2$. If $p \in {\cal P}_2$ and $x,y$ represent the coordinates in the bases $b_k, \beta_k$ respectively, then we have $x=S y$, where $S = \begin{bmatrix} 1 & 0 & 0 \\ -{3 \over 2} & 2 & -{1 \over 2} \\ {1 \over 2} & -1 & {1 \over 2} \end{bmatrix}$. The bases $\xi(b_k)$ and $\xi(\beta_k)$ are related in the same matter, and we have the matrix that maps the vectors $\xi(\beta_k)$ to the vectors $\xi(T(\beta_k))$ given by $S^{-1} T_\xi S = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{bmatrix}$.

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Thank you for your great answer, I really appreciate it. I got exactly the same answer as you when fooling around with this problem for a bit which confuses me. Should I conclude that the solution to the problem (the matrix in my post) is wrong? Or can it be obtained with another choice of basis? I'm absolutely not doubting your answer but I have a hard time wrapping my head around this. Thanks again! ( I should perhaps add that I found the problem + solution in a pdf by an old lecturer of mine btw) – John Smith Feb 21 '14 at 22:01
Well, with a different basis you can get the matrix above, but it is not immediately clear to me how that basis would be natural (this is similar to obtaining a controller canonical form in a linear control system). The required basis is $\beta_1(t) = 1 -{3 \over 2} t + {1 \over 2} t^2$, $\beta_2(t) = 2t -t^2$ and $\beta_3(t) = -{1 \over 2}t + {1 \over 2} t^2$. – copper.hat Feb 21 '14 at 22:07
@JohnSmith: I added some more detail. – copper.hat Feb 21 '14 at 23:02