Matrix diagonalization - eigenvalues on diagonal

Question

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.

Answer 1

you don't need a theorem for this. $$A = PDP^{-1} \implies AP = PD = P \, diag(d_1, d_2, \cdots, d_n)$$ let the columns of $P$ are $u_1, u_2, \cdots, u_n.$ then the matrix multiplication gives you $$Au_1 = d_1 u_1, Au_2 = du_2, \cdots, Au_n = d_nu_n $$

now argue that nonsingular matrix have nonzero columns, therefore $u_j$ is an eigenvector of $A$ corresponding to the eigenvalue $d_j$ for $j = 1, 2, \cdots, n.$

Answer 2

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 $v_i$.

Answer 3

Change basis, taking as new basis the eigenvectors $(v_1, \dots, v_n)$. In this basis, the matrix $A$ becomes the diagonal matrix $D(\lambda_1,\dots,\lambda_n)$ since, by definition: $Av_1=\lambda_1v_1,\dots,Av_n=\lambda_nv_n$.

Now the change of basis matrix (from the canonical basis to the basis of eigenvectors) $P$, by definition, is made up of the coordinates of the eigenvectors in the canonical basis. This matrix is invertible, since we can go back from the basis of eigenvectors to the canonical basis.

It can be shown that, if an endomorphism has matrix $M$ in the canonical basis, its matrix $M'$ in the basis of eigenvectors of $A$ is given by the formula: $$ M'=PMP^{-1}.$$

This applies in particular to $A$ itself; but we already know its matrix $D$ in the basis of eigenvectors. Hence: $$D=PAP^{-1}.$$