<s>[INST] What Is This Diagonalizable Matrix Doing to Its EigenVector? Let A = $\begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix}$.  It is diagonalizable with:
$$\begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \ \ \ \ \ \begin{bmatrix} 2 & 1 \\ 2 & 3\end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 8 \end{bmatrix}$$
We may say that $A$ doesn't do anything on $(1, -1)^t$ but elongates $(1, 2)^t$ by $4$.  Now stack the eigenvectors into a matrix called $X$ and we'll have $AX = XD$:
$$\begin{align} \begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix} \ \ & = \ \ \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} \\
\begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix} \ \ & = \ \ \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} 2/3 & -1/3 \\ 1/3 & 1/3 \end{bmatrix} 
\end{align}$$
Now, I'm going to feed $(1, -1)^t$ into the last equation.  On the left-hand side, as stated before, $A$ will do nothing to it.  But what is happening on the right-hand side?  For example, in the first multiplication, we'll have:
$$\begin{align} \begin{bmatrix} 2/3 & -1/3 \\ 1/3 & 1/3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \ \ & = \ \ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \end{align}$$
I'm aware that $X^{-1} = \begin{bmatrix} 2/3 & -1/3 \\ 1/3 & 1/3\end{bmatrix}$ is a change of coordinate matrix.  So it takes $(1, -1)^t$, the coordinates corresponding to the standard basis into $(1, 0)^t$ the coordinates corresponding to the eigenbasis.  But what is happening in the second step?
$$\begin{align} \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \ \ = \ \ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \end{align}$$
 [/INST] The second step $\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ scales $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ by the eigenvalue $1$, resulting a scaled standard vector to "select" one eigenvector column from $X$.
Trying with the other eigenvector $XDX^{-1} \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ may be clearer.
</s>