82 lines
3.1 KiB
Markdown
82 lines
3.1 KiB
Markdown
# Lecture 33
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## Chapter VII Operators on Inner Product Spaces
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**Assumption: $V,W$ are finite dimensional inner product spaces.**
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### Positive Operators 7C
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#### Definition 7.34
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An operator $T\in \mathscr{L}(V)$ is **positive** if $T$ is self adjoint and $\langle Tv, v\rangle\geq 0$
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Examples:
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* $I$ is positive.
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* $O\in \mathscr{L}(V)$ is positive if $T\in\mathscr{L}(V)$ is self adjoint and $b<4c$ then $T^2+bT+cI$ is positive.
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#### Definition 7.36
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Let $TR\in \mathscr{L}(V)$ then $R$ is a square root of $T$ if $R^2=T$.
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Example:
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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 $T$.
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#### Theorem 7.38
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Let $T\in \mathscr{L}(V)$, then the following statements are equal:
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(a) $T$ is a positive operator
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(b) $T$ is self adjoint with all eigenvalues non-negative
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(c) With respect to some orthonormal basis, $T$ has a diagonal matrix.
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(d) $T$ has a positive square root. (stronger condition)
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(e) $T$ has a self adjoint square root.
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(f) $T=R^*R$ for some $R\in \mathscr{L}(V)$
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Proof:
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$d\implies e,e\implies f,b\implies c$ are all clear.
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$a\implies b$: Let $\lambda$ be an eigenvalue. Let $v\in V$ be an eigenvector with eigenvalue $\lambda$, then $0\leq \langle Tv,v\rangle =\langle \lambda v, v\rangle =\lambda||v||^2\implies \lambda \geq 0$
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$c\implies d$ Let $M(T)=\begin{pmatrix}\lambda_1 &\dots & 0 \\&\ddots& \\0& \dots & \lambda_n\end{pmatrix}$
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with respect to some orthonormal basis and $\lambda_1,...,\lambda_n\geq 0$. Let $R$ be the operator with $M(R)=\begin{pmatrix}\sqrt{\lambda_1 }&\dots & 0\\&\ddots& \\0& \dots & \sqrt{\lambda_n}\end{pmatrix}$
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and $\sqrt{\lambda_1},...,\sqrt{\lambda_n}\geq 0$.
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$f\implies a$: $\langle R^*Rv,v\rangle=\langle Rv,Rv\rangle =||Rv||^2\geq 0$
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#### Theorem 7.39
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Every positive operator on $V$ has a unique positive square root
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Proof:
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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 $e_1,...,e_n$
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So $R$ is unique because positive square root s are unique.
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_for better proof, you shall set up two square root of $T$ and shows that they are the same._
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#### Theorem 7.43
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Suppose $T$ is a positive operator and $\langle Tv,v\rangle=0$ then $Tv=0$
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Proof:
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$\langle Tv,v\rangle=\langle \sqrt{T}\sqrt{T}v,v\rangle=\langle \sqrt{T}v,\sqrt{T}v\rangle=||\sqrt{T}v||^2$. So $\sqrt{T}v=0$. So $Tv=\sqrt{T}\sqrt{T}v=0$
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### Isometries, Unitary Operators, and Matrix Factorization 7D
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#### Definition 7.44
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A linear map $T\in\mathscr{L}(V,W)$ is an **isometry** if $||Tv||=||v||$
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#### Definition 7.51
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A linear operator $T\in\mathscr{L}(V)$ is **unitary** if it is an invertible isometry.
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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)
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