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}
 A: 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.
A: *

*A linear map is completely determined by its action on a basis.

*$M e_i$ is the $i$th column of the matrix $M$.

*$P=[I]^e_{v'}$ by definition is the matrix such that $e_i=Pv_i‘$

*$Av'_i$ can be written in the basis as $\alpha_{i1}v'_1+\alpha_{i2}v'_2$

*from $A'=PAP^{-1}$ we get  \begin{align}A'e_i &\overset5= PAP^{-1} e_i \\&\overset3= PAv'_i\\&\overset4= P(\alpha_{i1}v'_1+\alpha_{i2}v'_2)\\&\overset{\llap{\text{linearity}}}=\alpha_{i1}Pv'_1+\alpha_{i2}Pv'_2\\ &\overset3=\alpha_{i1}e_1+\alpha_{i2}e_2 = \binom{\alpha_{i1}}{\alpha_{i2}}.\end{align}
In this way we see that $A'=\begin{pmatrix}\alpha_{11}& \alpha_{12} \\ \alpha_{11}& \alpha_{12} \end{pmatrix}$, the matrix of coefficients.

