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 $T^*)^{\perp}$

$V$: finite-dimensional inner product space over $F=\mathbb{R}$ or $\mathbb{C}$

An operator $T:V\to V$ is said to be self-adjoint if $T=T^*$.

In other words, $\langle Tv,w\rangle=\langle v,Tw\rangle$ for $v,w\in V$.

In terms of matrices

$M(T)$: matrix of $T$ w.r.t. an orthonormal basis

$T$ is self-adjoint if $M(T)$ is equal to its conjugate-transpose.

• also called Hermitian in complex spaces and symmetric in real spaces

## Two properties

If $T$ is self-adjoint, null $T=$ $($range $T)^{\perp}$

Proof

null $T=$ $($range $T^*)^{\perp}=$ $($range $T)^{\perp}$

If $T$ is self-adjoint, $\langle Tv,v\rangle\in\mathbb{R}$

Proof

$\langle Tv,v\rangle=\langle v,T^*v\rangle=\langle v,Tv\rangle$

and

$\langle Tv,v\rangle=\overline{\langle v,Tv\rangle}$

## When are eigenvalues real?

$F=\mathbb{R}$

$v$: eigenvector of $T$ with eigenvalue $\lambda$

Since $v$ is real and $Tv$ is real, $\lambda$ has to be real.

$F=\mathbb{C}$

Eigenvalues can be complex.

Example

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

Eigenvalues: $i$, $-i$

Eigenvectors: $(1,i)$ for $i$, and $(-1,i)$ for $-i$

Every eigenvalue of a self-adjoint operator is real.

Proof

$T=T^*$ and $\lambda$ is an eigenvalue with eigenvector $v$

$\lambda\lVert v\rVert^2=\langle \lambda v,v\rangle=\langle Tv,v\rangle=\langle v,Tv\rangle=\langle v,\lambda v\rangle=\bar{\lambda}\lVert v\rVert^2$

$\lambda=\bar{\lambda}$

## When is $Tv$ always orthogonal to $v$?

Example

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

$\langle Ax,x\rangle=0$ for all $x\in\mathbb{R}^2$

In real spaces, there are non-trivial $T$ s.t. $\langle Tv,v\rangle=0$ for all $v$.

What about complex inner product spaces?

Example: $\langle Ax,x\rangle=\langle (i,-1),(1,i)\rangle=2i$ for $x=(1,i)$

There exists $x\in\mathbb{C}^2$ for which $\langle Ax,x\rangle\ne0$.

## Complex inner product spaces and $\langle Tv,v\rangle$

$V$: inner product space over $\mathbb{C}$.

If $\langle Tv,v\rangle=0$ for all $v$, then $T=0$.

Proof

$\langle Tv,v\rangle=0$ for all $v$

\begin{align} \langle Tu,w\rangle &= \dfrac{1}{4}\langle T(u+w),u+w\rangle-\dfrac{1}{4}\langle T(u-w),u-w\rangle\\ &+\dfrac{i}{4}\langle T(u+iw),u+iw\rangle-\dfrac{i}{4}\langle T(u-iw),u-iw\rangle \end{align}

So, $\langle Tu,w\rangle=0$ for all $u,w$

Set $w=Tu$ to get $\lVert Tu\rVert=0$ for all $u$

So, $T=0$

## Self-adjoint operators and $\langle Tv,v\rangle$: Result I

$V$: inner product space over $\mathbb{C}$.

$T$ is self-adjoint iff $\langle Tv,v\rangle\in\mathbb{R}$ for all $v$.

Proof

$\langle Tv,v\rangle-\overline{\langle Tv,v\rangle}=\langle Tv,v\rangle-\langle v,Tv\rangle=\langle (T-T^*)v,v\rangle$

$\langle Tv,v\rangle\in\mathbb{R}$ implies LHS = 0, which implies $T=T^*$

$T=T^*$ implies RHS = 0, which implies $\langle Tv,v\rangle\in\mathbb{R}$

## Self-adjoint operators and $\langle Tv,v\rangle$: Result II

$V$: inner product space over $\mathbb{R}$ or $\mathbb{C}$.

If $T$ is self-adjoint and $\langle Tv,v\rangle=0$ for all $v$, then $T=0$.

Proof

Over $\mathbb{C}$, result is true even if $T$ is not self-adjoint.

Over $\mathbb{R}$, when $T$: self-adjoint, we have

$\langle Tu,w\rangle = \dfrac{1}{4}\langle T(u+w),u+w\rangle-\dfrac{1}{4}\langle T(u-w),u-w\rangle$

So, $\langle Tu,w\rangle=0$ for all $u,w$

Set $w=Tu$ to get $\lVert Tu\rVert=0$ for all $u$

So, $T=0$