When a matrix is diagonalized as $A = P^{-1}DP$, must $P$ consists of eigenvectors of $A$?

Question

A matrix $A$ is diagonalizable when it can be expressed as P^{-1}DP. I know that $D$ must consists of eigenvalues of $A$. However, I am not sure if $P$ must consists of eigenvectors of $A$. Is it a must for $P$ to consists of eigenvectors of $A$? It seems to be that it is not a must. However, I cannot prove it.

So the question is when a matrix is diagonalized as $A = P^{-1}DP$, must $P$ consists of eigenvectors of $A$?

Answer 1

Note that $PA=DP$, or if $P_{i}$ is the $i^{th}$ column of $P$ and $d_i$, the $i^{th}$ diagonal element of $D$, then $AP_i=d_iP_i$

Answer 2

Firstly, $P^{-1}$ must consist of eigenvectors of A, not $P$. Each column of $P^{-1}$ is an eigenvector of $A$: $$ P^{-1}= \begin{bmatrix} e_1& e_2& \ldots& e_n\end{bmatrix} $$ To prove this, we can multiply the equality $A=P^{-1}DP$ by $P^{-1}$: $$ AP^{-1}=P^{-1}D $$ or $$ A\begin{bmatrix} e_1& e_2& \ldots& e_n\end{bmatrix}= \begin{bmatrix} e_1& e_2& \ldots& e_n\end{bmatrix} \begin{bmatrix} \lambda_1&0&\ldots&0\\ 0&\lambda_2&\ldots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\ldots&\lambda_n \end{bmatrix} $$ which expands to $$ Ae_1=\lambda_1 e_1,\; Ae_2=\lambda_2 e_2,\;\ldots\;,\; Ae_n=\lambda_n e_n. $$