105 lines
3.0 KiB
Markdown
105 lines
3.0 KiB
Markdown
# Lecture 30
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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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### Self adjoint and Normal Operators 7A
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#### Definition 7.1
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Suppose $T\in \mathscr{L}(V,W)$. The adjoint is the function $T^*:W\to V$ such that
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$$
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\langle Tv,w \rangle=\langle v,T^*w \rangle, \forall v\in V, w\in W
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$$
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#### Theorem 7.4
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Suppose $T\in \mathscr{L}(V,W)$ then $T^*\in \mathscr{L}(W,V)$
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Proof:
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Additivity, let $w_1,w_2\in W$. We want to show $T^*(w_1+w_2)T^*w_1+T^*w_2$
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Let $v\in V$, then
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$$
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\begin{aligned}
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\langle v,T^*(w_1+w_2) \rangle &=\langle v,T^*(w_1+w_2) \rangle\\
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&=\langle Tv,w_1+w_2 \rangle\\
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&=\langle Tv,w_1 \rangle+\langle Tv,w_2 \rangle\\
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&=\langle v,T^*w_1 \rangle+\langle v,T^* w_2 \rangle\\
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&=\langle v,T^*w_1 +T^* w_2 \rangle\\
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\end{aligned}
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$$
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Note: If $\langle v,u \rangle=\langle v,u'\rangle$, forall $v\in V$ then $u=u'$
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Homogeneity: same as idea above.
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#### Theorem 7.5
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Suppose $S,T\in \mathscr{L}(V,W)$, and $\lambda\in \mathbb{F}$, then
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(a) $(S+T)^*=S^*+T^*$
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(b) $(ST)^*=T^* S^*$
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(c) $(\lambda T)^*=\bar{\lambda}S^*$
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(d) $I^*=I$
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(e) $(T^*)^*=T$
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(f) If $T$ is invertible, then $(T^*)^{-1}=(T^{-1})^*$
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Proof:
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(d) $\langle (ST)v,u \rangle=\langle S(Tv),u \rangle=\langle Tv,S^*u \rangle=\langle v,T^*S^*u \rangle$
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#### Theorem 7.6
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Suppose $T\in\mathscr{L}(V,W)$, then
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(a) $null\ T^*=(range\ T)^\perp$
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(b) $null\ T=(range\ T^*)^\perp$
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(c) $range\ T^*=(null\ T)^\perp$
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(d) $range\ T=(null\ T^*)^\perp$
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Proof:
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$(a)\iff (c)$ since we can use **Theorem 7.5** (c) while replacing $T$ with $T^*$. Same idea give $(b)\iff (d)$. Also $(a)\iff (d)$ Since $(V^\perp)^\perp=V$
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Now we prove (a). Suppose $w\in null\ T^*\iff T^*w=0$
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$T^*w=0\iff \langle v,T^* w\rangle=0\forall v\in V\iff \langle Tv,w\rangle =0\forall v\in V\iff w\in (range\ T)^\perp$
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#### Definition 7.7
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The **conjugate transpose** of a $m\times n$ matrix $A$ is the $n\times m$ matrix denoted $A^*$ given the conjugate of the transpose.
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ie. $(A^*)_{j,k}=A_{j,k}$
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#### Theorem 7.9
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Let $T\in \mathscr{L}(V,W)$ and $e_1,..,e_n$ an orthonormal basis of $V$, $f_1,...,f_m$ be an orthonormal basis of $W$. Then $M(T^*,(f_1,...,f_m),(e_1,..,e_n))=M(T,(f_1,...,f_m),(e_1,..,e_n))^*$
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Proof:
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The k-th column of $T$ is given by writing $Te_k$. in the basis $f_1,...,f_m$. ie. the $k,j$ entry of $M(T)$ is $\langle Te_k,f_j \rangle$, but then the $j,k$ entry of $M(T^*)$ is $\langle T^*f,e_k \rangle$. But $\langle Te_k,f_j\rangle=\langle e_k,T^*f_j\rangle=\overline{\langle T^*f_j,e_k\rangle}$
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Example:
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Suppose $T(x_1,x_2,x_3)=(x_2+3x_3,2x_1)$
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$$
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\begin{aligned}
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\langle T(x_1,x_2,x_3),(y_1,y_2)\rangle&=\langle (x_2+3x_3,2x_1),(y_1,y_2)\rangle\\
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&=(x_2+3x_3,2x_1)\bar{y_1},(x_2+3x_3,2x_1)\bar{y_2}\\
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&=\bar{y_1}x_2+3\bar{y_1}x_3+2\bar{y_2}x_1\\
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&=\langle (x_1,x_2,x_3),(2y_2,y_1,3y_1)\rangle
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\end{aligned}
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$$
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So $T^*(y_1,y_2)=(2y_2,y_1,3y_1)$
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Idea: Reisz Representation gives a function from $V$ to $V'$ (3.118) tells us given $T\in \mathscr{L}(V,W)$, we have
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