3.0 KiB
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
T^2+bT+cI
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)
\begin{aligned}
\langle(T^2+bT+cI),v\rangle&=\langle T^2v,v\rangle+\langle bTv,v\rangle+c\langle v,v\rangle\\
&=\langle Tv,Tv\rangle+b\langle Tv,v\rangle +c||v||^2\\
&\geq ||Tv||^2-|b|\ ||Tv||\ ||v||+c||v||^2 \textup{ by cauchy schuarz}\\
&=\left(||Tv||-\frac{b||v||}{2}\right)^2+\left(c-\frac{b^2}{4}\right)||v||^2>0
\end{aligned}
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
M(T)=\begin{pmatrix}
a_{1,1}&\dots&a_{1,n}\\
&\ddots &\vdots\\
0& & a_{n,n}
\end{pmatrix}
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) $M(T)=\begin{pmatrix}
2&-3\
3&2
\end{pmatrix}$
$M(T,(f_1,f_2))=\begin{pmatrix} 2+3c&0\ 0&2-3c \end{pmatrix}$