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)