Matrix diagonalization - eigenvalues on diagonal Diagonalization of a square matrix $A$ consists in finding matrices $P$ and $\Delta$ such that $A=PD P^{-1}$ where $D$ is a diagonal matrix.
What theorem tells us that $P$ is a matrix composed of the eigenvectors of $A$, $D$ is the diagonal matrix constructed from the corresponding eigenvalues, and $P^{-1}$ is the matrix inverse of $P$? I'm also interested in the proof.

Let $f$ the endomorphism of $\Bbb R^n$ which $A$ is its representation in the standard basis $B$ and let $B'=(v_1,\ldots,v_n)$  the basis wherein  $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ is the matrix of $f$ then we have $f(v_i)=\lambda_i v_i$ so $v_i$ is the eigenvector of $f$ associated to $\lambda_i$ and the change matrix from the standard basis to the basis of the eigenvectors is $P=(v_1\cdots v_n)$ i.e. the $i^{th}$ column of $P$ is $v_i$.
