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I am going through some practice problems for my test this week. This one is similar to a HW problem, but different enough that I am having trouble.

The linear Transformation $A: P_2 \to P_2$ is given by $Ap(t) = p(t+1)$. Find its matrix in the basis $1,t,t^2$.

So I understand that the columns 1,2,3 of the matrix will be made up of the linear mapping of $1,t,t^2$ respectively. What I don't see is how to get the components of those column vectors.

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  • $\begingroup$ What is $P_2$ ? And what is $p$ ? $\endgroup$ Commented Oct 26, 2014 at 16:46
  • $\begingroup$ $P_2$ is just a vector space. $\endgroup$ Commented Oct 26, 2014 at 16:47
  • $\begingroup$ I am assuming that $p$ is a polynomial. The question is just that one line that you see above. $\endgroup$ Commented Oct 26, 2014 at 16:49
  • $\begingroup$ I'm pretty sure Snagglewhen means that $P_2$ is the space of polynomials (in $t$) of degree $\le 2$. And $p \in P_2$. $\endgroup$ Commented Oct 26, 2014 at 16:49
  • $\begingroup$ @Bye_World: I am also pretty sure, but should not be the OP the person to be sure ? $\endgroup$ Commented Oct 26, 2014 at 16:50

2 Answers 2

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Let $A:P_2\to P_2$ be the linear map defined by $$ A(p)(t)=p(t+1) $$ For $0\leq k\leq 2$ let $e_k\in P_3$ be $e_k(t)=t^k$ and note that $\beta=\{e_0,e_1,e_2\}$ is a basis of $P_2$.

To compute the entries of $[A]_\beta$, note that \begin{array}{lclclcl} A(e_0) & = & \color{red}{1}\ e_0 & + & \color{red}{0}\ e_1 & + & \color{red}{0}\ e_2 \\ A(e_1) & = & \color{blue}{1}\ e_0 & + & \color{blue}{1}\ e_1 & + & \color{blue}{0}\ e_2 \\ A(e_2) & = & \color{green}{1}\ e_0 & + & \color{green}{2}\ e_1 & + & \color{green}{1}\ e_2 \end{array} (Check these computations yourself!) This implies $$ [A]_\beta = \begin{bmatrix} \color{red}{1} & \color{blue}{1} & \color{green}{1} \\ \color{red}{0} & \color{blue}{1} & \color{green}{2} \\ \color{red}{0} & \color{blue}{0} & \color{green}{1} \end{bmatrix} $$

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  • $\begingroup$ How did you come up with the entries of T? I understand that $\beta=\{e_0,e_1,e_2\} = \{1,t,t^2\}$ $\endgroup$ Commented Oct 26, 2014 at 18:37
  • $\begingroup$ For example $A(e_2)(t)=e_2(t+1)=(t+1)^2=1+2t+t^2$ so $A(e_2)=1 e_0+2e_1+1e_2$. $\endgroup$ Commented Oct 26, 2014 at 18:42
  • $\begingroup$ Your matrix is not good, you have written its transposition. What you're really looking for is writing the image of each element of the basis, and express it in this same basis, which means that you should color the ligns with the same color, not the colums. The first column of A contains the coefficients of A$e_o$ in the basis $(e_o,e_1,e_2)$, not the coefficient of each $(e_o,e_1,e_2)$ relatively to $e_o$ which is what you wrote. $\endgroup$ Commented Oct 27, 2014 at 13:08
  • $\begingroup$ $[A]_\beta = (Ae_o,Ae_1,Ae_2)_\beta $ $a_{2,1}$ is the coefficient of $Ae_o$ relatively to $e_1$ for instance, which is 0 here, not 1 like you wrote it $\endgroup$ Commented Oct 27, 2014 at 13:14
  • $\begingroup$ @mvggz Yes of course. Edited. $\endgroup$ Commented Oct 27, 2014 at 21:22
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I'll treat the case $P_n$ since it's the same mechanism:

For: 0 $\leq$ j $\leq$ n :

$ At^j = (t+1)^j = \sum_{i=0}^j C_i^j*t^i*1^{n-i} = \sum_{i=0}^j C_i^j*t^i $

So if you let $a_{i,j}$ be the coefficient of your matrix A, you have :

$a_{i,j} = 0$ when i > j

$ a_{i,i} = 1 $

$a_{i,j} = C_i^j $ when i < j

So your matrix is triangular, with zeros below the diagonal and ones on it.

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