Diagonalization

Prerequisites

• rank

1. Diagonalization

$A\in F^{n\times n}$ is a matrix.

$A$ is diagonalizable if there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PD{P}^{-1}$.

$P=[v_1\cdots v_n]\in F^{n\times n}$, where $v_i\in F^n$ for every $i\in [n]$.

$D=\mathrm{diag}({\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n)\in F^{n\times n}$

$$A=PD{P}^{-1}⟺AP=PD⟺\forall i\in [n],A{v}_i={\lambda }_i{v}_i$$

If $\exists v\ne 0$ s.t. $Av=\lambda v$, then $\lambda$ is an eigenvalue of $A$ and $v$ is an eigenvector of $A$ corresponding to $\lambda$.

For every $v\in F^n$, we have $Av=\lambda v⟺(A-\lambda I)v=0⟺v\in \ker (A-\lambda I)$.

$\ker (A-\lambda I)$ is the eigenspace of $A$ corresponding to $\lambda$.

2. Characteristic polynomial

$\exists v\ne 0,v\in \ker (A-\lambda I)⟺\exists v\ne 0,(A-\lambda I)v=0⟺\det (\lambda I-A)=0$

$\det (\lambda I-A)$ is a polynomial of $\lambda$ with degree $n$, called the characteristic polynomial of $A$.

The roots of the characteristic polynomial are the eigenvalues of $A$.

Let ${\lambda }_1,\dots ,{\lambda }_k$ be the distinct eigenvalues of $A$, and let $d_i=\dim \ker (A-{\lambda }_{i}I)$ be the geometric multiplicity of ${\lambda }_i$. Then $A$ is diagonalizable if and only if ${\sum }_{i=1}^k{d}_i=n$.

3. Real symmetric matrix

By the spectral theorem, every real symmetric matrix $A\in {\mathbb{R}}^{n\times n}$ is orthogonally diagonalizable: there exist an orthogonal matrix $P$ and a real diagonal matrix $D$ such that $A=PD{P}^T$.