Diagonalization

Prerequisites

• matrix

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 $A=PD{P}^{-1}$.

$P=(v_1,v_2,\cdots ,v_n)\in F^{n\times n}$

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

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

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 $\lambda$.

if $v\ne 0$, 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

$\exists v\ne 0,v\in \text{Ker}(A-\lambda I)⟺\exists v\ne 0,(A-\lambda I)v=0⟺\det (A-\lambda I)=0$

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

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

If $F$ is algebraically closed, then $A$ will have $n$ eigenvalues in $F$. (Maybe with multiplicity)

Let $d_i=dim(Ker(A-{\lambda }_{i}I))$. If ${\sum }_{i=1}^n{d}_i=n$, then $A$ is diagonalizable.

3. Real symmetric matrix

If $A\in {\mathbb{R}}^{n\times n}$ is a real symmetric matrix, then $A$ is diagonalizable. Moreover, $A=PD{P}^T$ where $D$ is a diagonal matrix and $P$ is an orthogonal matrix.