Here is the given question:
Consider the linear transformation: $T:\mathbb{R}^2 \to \mathbb{R}^2$ by: $T(x; y) = (x-y; -x + y)$: Write the matrix representation of $T$ with respect to the basis ${(1; 2);(-1; 1)}$ (use the same basis for domain and co-domain).
Now in the solution they did the following:
- Picked the transition matrix
\begin{equation*} P= \begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} \end{equation*}
2.Picked the matrix $A$ which is the matrix representation of the linear transformation with respect to the standard basis.
\begin{equation*} A= \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \end{equation*}
The found the matrix representation $B$ relative to the basis ${(1; 2);(-1; 1)}$ by solving $B=P^{-1}AP$
My question is the following: Why didn't we find $B$ by doing $B=PAP^{-1}$ since we are going from the standard basis to another one, the set-up is something I would do if I am going from ${(1; 2);(-1; 1)}$ to the standard basis.
$\newcommand{\B}{\mathcal{B}}\newcommand{\S}{\mathcal{S}}\newcommand{\x}{\mathbf{x}}$Let $\B$ be the basis in question. Note that $P$ is the transition matrix from $\B$ to the standard basis $\S$. That is, $[\x]_{\S} = P[\x]_{\B}$, or equivalently, $[\x]_{\B} = P^{-1}[\x]_{\S}$ for all $\x\in \Bbb{R}^2$.
The matrix of $T$ with respect to $\B$ in domain and codomain (which you are calling $B$) is defined as satisfying $[T(\x)]_{\B} = \color{blue}{B}[\x]_{\B}$ for all $\x\in \Bbb{R}^2$. But
$$\begin{align*}[T(\x)]_{\B} &= P^{-1}[T(\x)]_{\S} \\ &= P^{-1}A [\x]_{\S} \quad (\text{since }[T(\x)]_{\S} = A [\x]_{\S} \text{ by definition of }A)\\ &= \color{blue}{P^{-1} AP}[\x]_{\B}.\end{align*}$$
So the matrix you want is $B =P^{-1} AP$.