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Info provided:

$T:P_2→P_2$ given by $T(p(x)) = p(kx)$ where $k>0$ Find matrix $[T]_{B,B}$ representing $T$ in the basis $B$

Attempted Solution:

Using the standard basis for $P_2$, {$1,x,x^2$}, would $[T]_{B,B}$ simply be represented as $([T(1)]_B,[T(x)]_B,[T(x^2)]_B)$ ?

Is it that in the general case $[T]_{W,V}$ is from $V$ to $W$, therefore as in this case $[T]_{B,B}$ is simply from $B$ to $B$?

So that $[T]_{B,B}$ = $$ \left( \begin{matrix} 1 & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k^2 \\ \end{matrix} \right) $$

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Here is the detail of the case when $B = \{1,x,x^2\}$. You can imitate it to solve for a more general $B$.

We have \begin{equation*} \begin{cases} T(1) = 1+0x+0x^2\\ T(x) = 0+kx+0x^2\\ T(x^2) = 0+0x+k^2x^2\\ \end{cases} \end{equation*} and so the matrix $[T]_{B,B}$ with respect to $B = \{1,x,x^2\}$ is of your form.

The general case is not $[T]_{V,W}$ but $[T]_{B_1,B_2}$, where $B_1 = \{p_1,p_2,p_3\}$ and $B_2 = \{q_1,q_2,q_3\}$ are bases of $P_2$. Just calculate the coefficients in \begin{equation*} \begin{cases} T(p_1) = a_{11}q_1+a_{21}q_2+a_{31}q_3\\ T(p_2) = a_{12}q_1+a_{22}q_2+a_{32}q_3\\ T(p_3) = a_{13}q_1+a_{23}q_2+a_{33}q_3 \end{cases} \end{equation*} and the matrix $[T]_{B_1,B_2}$ is \begin{equation*} \begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{pmatrix}. \end{equation*}

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