how to understand this new way to solve matrix under basis change? I just found this exercise using a new shortcut to solve matrix under new basis given the old matrix, please enlighten me why this shortcut works:
The Exercise: relative to the standard unit vectors $e_1$, $e_2$, a linear map $A:\mathbb{R}^2 \to \mathbb{R}^2$ is described by the matrix $ A=\begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix}$. Find the representing matrix $A'$ for this linear map relative to the basis $v'_1 = \begin{pmatrix} \frac{1}{\sqrt2} \\ -\frac{1}{\sqrt2} \\ \end{pmatrix}$, $v'_2 = \begin{pmatrix} \frac{1}{\sqrt2} \\ \frac{1}{\sqrt2} \\ \end{pmatrix}$
Instead of traditional $A'=PAP^{-1}$ way, the answer give a new shortcut:
compute the images of the basis vectors and express them in $v'_i$ then the coefficients give the columns of $A'$.
$$Av'_1 = 0v'_1 + 1v'_2 , \ \ Av'_1 = 1v'_1 + 0v'_2$$
so $ A' = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}$
note1: I feel this shortcut only work when one of the basis is standard unit vectors.
note2: the book is using $[A]^e_e$ multiply $[v'_i]_e$, the result is $[Av'_i]_e$ in space 2 of the digram below, my confusion is why expressing them using $[v'_i]_e$ will give the matrix $[A']^{v'}_{v'}$ which should be get by expressing $[A'v'_i]_{v'}$ using $[v'_i]_{v'}$ that relating space 3, 4 of the digram.
$\require{AMScd}$
\begin{CD}
1\  \mathbb{R}^2 \text{ using basis } e @>A>>2\ \mathbb{R}^2 \text{ using basis } e\\
@V P=[I]^e_{v'} VV  @VV P V\\
3\ \mathbb{R}^2 \text{ using basis } v' @>>A'>4\ \mathbb{R}^2 \text{ using basis } v'
\end{CD}

Maybe reading 《linear algebra done right》chapter3 is helpful. In this problem, linear transformation is known. Matrix is a way to show it. To finding its matrix under bases $v_{1}^{\prime}, v_{2}^{\prime}$,the only thing we need is how this transformation affect on bases $v_{1}^{\prime}, v_{2}^{\prime}$.Solve $Av_{1}^{\prime}, Av_{2}^{\prime}$ and rewrite them as a linear combination of $v_{1}^{\prime}, v_{2}^{\prime}$,then the column of matrix solved because of matrix definition.
