# Complex Spectral Theorem

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• dim null $T+$ dim range $T=$ dim $V$
• Solution to $Ax=b$ (if it exists): $u+$ null$(A)$
• Four fundamental subspaces of a matrix
• Column space, row space, null space, left null space
• Eigenvalue $\lambda$ and Eigenvector $v$: $Tv=\lambda v$
• There is a basis w.r.t. which a linear map is upper-triangular
• If there is a basis of eigenvectors, linear map is diagonal w.r.t. it
• Inner products, norms, orthogonality and orthonormal basis
• There is an orthonormal basis w.r.t. which a linear map is upper-triangular
• Orthogonal projection: distance from a subspace
• Adjoint of a linear map: $\langle Tv,w\rangle=\langle v,T^*w\rangle$
• null $T=$ $($range $T^*)^{\perp}$
• Self-adjoint: $T=T^*$, Normal: $TT^*=T^*T$
• Eigenvectors corresponding to different eigenvalues are orthogonal

## Diagonalizable operators

$T:V\to V$, diagonalizable

Eigenvector basis of $V$: $\{v_1,\ldots,v_n\}$

$Tv_i=\lambda_iv_i$

Matrix of $T$ w.r.t. eigenvector basis

$D=\begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}$ (diagonal)

## Matrices of diagonalizable operators

$A$: $n\times n$ matrix, represents $T$ in standard basis

Matrix with $v_i$ (coordinates in standard basis) as $i$-th column
$S=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ v_1&\cdots&v_i&\cdots&v_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}$

Let $S^{-1}=\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&u^T_i&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}$, matrix with $u^T_i$ as $i$-th row

$A=SDS^{-1}=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ v_1&\cdots&v_i&\cdots&v_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}\begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&u^T_i&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}$

$A=\lambda_1v_1u^T_1+\cdots+\lambda_nv_nu^T_n$

## General form of matrices of diagonalizable operators

$T:V\to V$: diagonalizable

There exists a basis s.t. matrix of $T$ has the following form:
$A=\lambda_1v_1u^T_1+\cdots+\lambda_nv_nu^T_n$

$v_i$: eigenvectors in the basis, $u_i$: rows of inverse matrix

Suppose rank $T=r$.

There exists a basis s.t. matrix of $T$ has the following form:
$A=\lambda_1v_1u^T_1+\cdots+\lambda_rv_ru^T_r$

$\lambda_i$: nonzero eigenvalues with corresponding eigenvectors $v_i$, $1\le i\le r$

## Examples

1. $A=\begin{bmatrix} 1&0\\ 0&2 \end{bmatrix}$ (self-adjoint, diagonal)

$A=\begin{bmatrix} 1\\ 0 \end{bmatrix}\begin{bmatrix} 1&0 \end{bmatrix}+2\begin{bmatrix} 0\\ 1 \end{bmatrix}\begin{bmatrix} 0&1 \end{bmatrix}$

2. $A=\begin{bmatrix} 1&2\\ 3&2 \end{bmatrix}$ (not self-adjoint)

$\lambda=-1,4$

$S=\begin{bmatrix} 1&2\\ -1&3 \end{bmatrix}$, $S^{-1}=\begin{bmatrix} 3/5&-2/5\\ 1/5&1/5 \end{bmatrix}$

$A=-\begin{bmatrix} 1\\ -1 \end{bmatrix}\begin{bmatrix} 3/5&-2/5 \end{bmatrix}+4\begin{bmatrix} 2\\ 3 \end{bmatrix}\begin{bmatrix} 1/5&1/5 \end{bmatrix}$

## Example: Normal operator

$A=\begin{bmatrix} 2&-3\\ 3&2 \end{bmatrix}$ (normal)

$\lambda=2+3i,2-3i$

$S=\frac{1}{\sqrt{2}}\begin{bmatrix} i&-i\\ 1&1 \end{bmatrix}$, $S^{-1}=\frac{1}{\sqrt{2}}\begin{bmatrix} -i&1\\ i&1 \end{bmatrix}$

$A=\frac{2+3i}{2}\begin{bmatrix} i\\ 1 \end{bmatrix}\begin{bmatrix} -i&1 \end{bmatrix}+\frac{2-3i}{2}\begin{bmatrix} -i\\ 1 \end{bmatrix}\begin{bmatrix} i&1 \end{bmatrix}$

Observations

Eigenvector basis: Orthonormal

## Complex spectral theorem

$V$: inner product space over $C$, $T:V\to V$

The following are equivalent.

1. $T$ is normal.
2. $V$ has an orthonormal basis of eigenvectors of $T$.
3. $T$ is diagonal w.r.t. an orthonormal basis.

Proof

Clearly, (2) implies (3), and (3) implies (2)

Proof of (3) implies (1)

$B$: orthonormal basis s.t. $M(T,B)$ is diagonal

Then, $M(T^*,B)$: conjugate-transpose of $M(T,B)$, which is diagonal

So, $M(T,B)M(T^*,B)=M(T^*,B)M(T,B)$, or $T$ commutes with $T^*$

## Proof (continued)

Proof of (1) implies (3)

Schur’s theorem: $T$ is upper-triangular w.r.t. an orthonormal basis

Orthonormal basis: $\{e_1,\cdots,e_n\}$

$M(T,B)=\begin{bmatrix} a_{11}&\cdots&a_{1n}\\ &\ddots&\vdots\\ 0 & &a_{nn} \end{bmatrix}$, $M(T^*,B)=\begin{bmatrix} \overline{a_{11}}& &0\\ \vdots &\ddots& \\ \overline{a_{1n}}&\cdots&\overline{a_{nn}} \end{bmatrix}$

We will show that $M(T,B)$ is, in fact, diagonal

$\lVert Te_1\rVert^2=\lvert a_{11}\rvert^2=\lVert T^*e_1\rVert^2=\lvert a_{11}\rvert^2+\cdots+\lvert a_{1n}\rvert^2$

So, $a_{12}=\cdots=a_{1n}=0$

$\lVert Te_2\rVert^2=\lvert a_{22}\rvert^2=\lVert T^*e_2\rVert^2=\lvert a_{22}\rvert^2+\cdots+\lvert a_{2n}\rvert^2$

So, $a_{23}=\cdots=a_{2n}=0$

and so on…

## Matrices of normal operators

$A$: $n\times n$ matrix, represents a normal operator $T$ in standard basis

$T$ is diagonal w.r.t. orthonormal eigenvector basis $\{e_1,\cdots,e_n\}$

Corresponding eigenvalues: $\{\lambda_1,\ldots,\lambda_n\}$

Matrix with $e_i$ (coordinates in standard basis) as $i$-th column
$S=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ e_1&\cdots&e_i&\cdots&e_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}$

$S^{-1}=\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&\overline{e^T_i}&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}$, matrix with $\overline{e^T_i}$ as $i$-th row

$A=SDS^{-1}=\lambda_1e_1\overline{e^T_1}+\cdots+\lambda_n e_n \overline{e^T_n}$