NoteNextra-origin/pages/Math429/Math429_L33.md

Lecture 33

Chapter VII Operators on Inner Product Spaces

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

Positive Operators 7C

Definition 7.34

An operator T\in \mathscr{L}(V) is positive if T is self adjoint and \langle Tv, v\rangle\geq 0

Examples:

• I is positive.
• O\in \mathscr{L}(V) is positive if T\in\mathscr{L}(V) is self adjoint and b<4c then T^2+bT+cI is positive.

Definition 7.36

Let TR\in \mathscr{L}(V) then R is a square root of T if R^2=T.

Example:

Let T(x,y,z)=(z,0,0), R(x,y,z)=(y,z,0) R(R(x,y,z))=R(y,z,0)=(z,0,0), then R is a square root of T.

Theorem 7.38

Let T\in \mathscr{L}(V), then the following statements are equal:

(a) T is a positive operator
(b) T is self adjoint with all eigenvalues non-negative
(c) With respect to some orthonormal basis, T has a diagonal matrix.
(d) T has a positive square root. (stronger condition) (e) T has a self adjoint square root.
(f) T=R^*R for some R\in \mathscr{L}(V)

Proof:

d\implies e,e\implies f,b\implies c are all clear.

a\implies b: Let \lambda be an eigenvalue. Let v\in V be an eigenvector with eigenvalue \lambda, then 0\leq \langle Tv,v\rangle =\langle \lambda v, v\rangle =\lambda||v||^2\implies \lambda \geq 0

c\implies d Let M(T)=\begin{pmatrix}\lambda_1 &\dots & 0 \\&\ddots& \\0& \dots & \lambda_n\end{pmatrix}

with respect to some orthonormal basis and \lambda_1,...,\lambda_n\geq 0. Let R be the operator with M(R)=\begin{pmatrix}\sqrt{\lambda_1 }&\dots & 0\\&\ddots& \\0& \dots & \sqrt{\lambda_n}\end{pmatrix}

and \sqrt{\lambda_1},...,\sqrt{\lambda_n}\geq 0.

f\implies a: \langle R^*Rv,v\rangle=\langle Rv,Rv\rangle =||Rv||^2\geq 0

Theorem 7.39

Every positive operator on V has a unique positive square root

Proof:

Let e_1,...,e_n be an orthonormal basis, such that M(T,(e_1,...,e_n))=\begin{pmatrix}\sqrt{\lambda_1 }&\dots & 0\\&\ddots& \\0& \dots & \sqrt{\lambda_n} \end{pmatrix} with \lambda_1,...,\lambda_n\geq 0. Let R be a positive square root of T then R^2e_k=\lambda e_k. Then M(R^2)=\begin{pmatrix}\lambda_1 &\dots & 0 \\&\ddots& \\0& \dots & \lambda_n\end{pmatrix} so \lambda_1,...,\lambda_n are the eigenvalues with eigenvectors e_1,...,e_n

So R is unique because positive square root s are unique.

for better proof, you shall set up two square root of T and shows that they are the same.

Theorem 7.43

Suppose T is a positive operator and \langle Tv,v\rangle=0 then Tv=0

Proof:

\langle Tv,v\rangle=\langle \sqrt{T}\sqrt{T}v,v\rangle=\langle \sqrt{T}v,\sqrt{T}v\rangle=||\sqrt{T}v||^2. So \sqrt{T}v=0. So Tv=\sqrt{T}\sqrt{T}v=0

Isometries, Unitary Operators, and Matrix Factorization 7D

Definition 7.44

A linear map T\in\mathscr{L}(V,W) is an isometry if ||Tv||=||v||

Definition 7.51

A linear operator T\in\mathscr{L}(V) is unitary if it is an invertible isometry.

Note: n dimensional unitary matrices U(n)\subseteq n dimensional invertible matrices GL(n)\subseteq group of n\times n matrices \mathbb{F}^{n,n} (This is a starting point for abstract algebra XD)