Lecture 32

Chapter VII Operators on Inner Product Spaces

Assumption: $V,W$ are finite dimensional inner product spaces.

Spectral Theorem 7B

Recall

Definition 7.10

An operator $T\in \mathscr{L}(V)$ is self adjoint if $T=T^*$

Definition 7.18

AN operator $T\in\mathscr{L}(V)$ is normal if $TT^*=T^*T$

Theorem 7.20

Suppose $T\in \mathscr{L}(V)$, $T$ is normal $\iff ||Tv||=||T^*v||$

Lemma 7,26

Suppose $T\in\mathscr{L}(V)$ is self adjoint operator and $b,c\in \mathbb{R}$ such that $b^2<4c$, then

is invertible.

Proof:

Prove $T^2+bT+cI$ is injective by showing $\langle(T^2+bT+cI),v\rangle\neq 0$ (for $v\neq 0$)

Theorem 7.27

Suppose $T\in \mathscr{L}(V)$ is self adjoint. Then the minimal polynomial is of the form $(z-\lambda_1)...(z-\lambda_m)$ for some $\lambda_1,...,\lambda_m\in\mathbb{R}$

Proof:

$\mathbb{F}=\mathbb{C}$ clear from previous results

$\mathbb{F}=\mathbb{R}$ assume for contradiction $q(z)$, where $b^2\leq 4c$. Then $P(T)=0$ but $q(T)\neq 0$. So let $v\in V$ such that $q(T)v\neq 0$.

then $(T^2+bT+cI)(q(T)v)=0$ but $T^2+bT+cI$ is invertible so $q(T)v=0$ this is a contradiction so $p(z)=(z-\lambda_1)...(z-\lambda_m)$

Theorem 7.29 Real Spectral theorem

Suppose $V$ is a finite dimensional real inner product space and $T\in \mathscr{L}(V)$ then the following are equivalent.

(a) $T$ is self adjoint.
(b) $T$ has a diagonal matrix with respect to same orthonormal basis.
(c) $V$ has an orthonormal basis of eigenvectors of $T$

Proof:

$b\iff c$ clear by definition

$b\implies a$ because the transpose of a diagonal matrix is itself.

$a\implies b$ by (Theorem 7.27) there exists an orthonormal basis such that $M(T)$ is upper triangular. But $M(T^*)=M(T)$ and $M(T^*)=(M(T))^*$

but this $M(T)$ is both upper and lower triangular, so $M(T)$ is diagonal.

Theorem 7.31 Complete Spectral Theorem

Suppose $V$ is a complex finite dimensional inner product space. $T\in \mathscr{L}(V)$, then the following are equivalent.

(a) $T$ is normal
(b) $T$ has a diagonal matrix with respect to an orthonormal basis
(c) $V$ has an orthonormal basis of eigenvectors of $T$.

$a\implies b$

with respect to an appropriate basis $e_1,...,e_n$

Then $||Te_1||^2=|a_{1,1}|^2$, $||Te_1||^2=||T^*e_1||^2=|a_{1,1}|^2+|a_{1,2}|^2+...+|a_{1,n}|^2$. So $a_{1,2}=...=a_{1,n}=0$, without loss of generality, $||Te_2||^2=0$. Repeating this procedure we have $M(T)$ is diagonal.

Example:

$T\in \mathscr{L}(\mathbb{C}^2)$ $M(T)=\begin{pmatrix} 2&-3\\ 3&2 \end{pmatrix}$

$M(T,(f_1,f_2))=\begin{pmatrix} 2+3c&0\\ 0&2-3c \end{pmatrix}$