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Math429
ๆจกๅ—
Math429 L30

Lecture 30

Chapter VII Operators on Inner Product Spaces

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

Self adjoint and Normal Operators 7A

Definition 7.1

Suppose TโˆˆL(V,W)T\in \mathscr{L}(V,W). The adjoint is the function Tโˆ—:Wโ†’VT^*:W\to V such that

โŸจTv,wโŸฉ=โŸจv,Tโˆ—wโŸฉ,โˆ€vโˆˆV,wโˆˆW\langle Tv,w \rangle=\langle v,T^*w \rangle, \forall v\in V, w\in W

Theorem 7.4

Suppose TโˆˆL(V,W)T\in \mathscr{L}(V,W) then Tโˆ—โˆˆL(W,V)T^*\in \mathscr{L}(W,V)

Proof:

Additivity, let w1,w2โˆˆWw_1,w_2\in W. We want to show Tโˆ—(w1+w2)Tโˆ—w1+Tโˆ—w2T^*(w_1+w_2)T^*w_1+T^*w_2

Let vโˆˆVv\in V, then

โŸจv,Tโˆ—(w1+w2)โŸฉ=โŸจv,Tโˆ—(w1+w2)โŸฉ=โŸจTv,w1+w2โŸฉ=โŸจTv,w1โŸฉ+โŸจTv,w2โŸฉ=โŸจv,Tโˆ—w1โŸฉ+โŸจv,Tโˆ—w2โŸฉ=โŸจv,Tโˆ—w1+Tโˆ—w2โŸฉ\begin{aligned} \langle v,T^*(w_1+w_2) \rangle &=\langle v,T^*(w_1+w_2) \rangle\\ &=\langle Tv,w_1+w_2 \rangle\\ &=\langle Tv,w_1 \rangle+\langle Tv,w_2 \rangle\\ &=\langle v,T^*w_1 \rangle+\langle v,T^* w_2 \rangle\\ &=\langle v,T^*w_1 +T^* w_2 \rangle\\ \end{aligned}

Note: If โŸจv,uโŸฉ=โŸจv,uโ€ฒโŸฉ\langle v,u \rangle=\langle v,u'\rangle, forall vโˆˆVv\in V then u=uโ€ฒu=u'

Homogeneity: same as idea above.

Theorem 7.5

Suppose S,TโˆˆL(V,W)S,T\in \mathscr{L}(V,W), and ฮปโˆˆF\lambda\in \mathbb{F}, then

(a) (S+T)โˆ—=Sโˆ—+Tโˆ—(S+T)^*=S^*+T^*
(b) (ST)โˆ—=Tโˆ—Sโˆ—(ST)^*=T^* S^*
(c) (ฮปT)โˆ—=ฮปห‰Sโˆ—(\lambda T)^*=\bar{\lambda}S^*
(d) Iโˆ—=II^*=I
(e) (Tโˆ—)โˆ—=T(T^*)^*=T
(f) If TT is invertible, then (Tโˆ—)โˆ’1=(Tโˆ’1)โˆ—(T^*)^{-1}=(T^{-1})^*

Proof:

(d) โŸจ(ST)v,uโŸฉ=โŸจS(Tv),uโŸฉ=โŸจTv,Sโˆ—uโŸฉ=โŸจv,Tโˆ—Sโˆ—uโŸฉ\langle (ST)v,u \rangle=\langle S(Tv),u \rangle=\langle Tv,S^*u \rangle=\langle v,T^*S^*u \rangle

Theorem 7.6

Suppose TโˆˆL(V,W)T\in\mathscr{L}(V,W), then

(a) nullย Tโˆ—=(rangeย T)โŠฅnull\ T^*=(range\ T)^\perp
(b) nullย T=(rangeย Tโˆ—)โŠฅnull\ T=(range\ T^*)^\perp
(c) rangeย Tโˆ—=(nullย T)โŠฅrange\ T^*=(null\ T)^\perp
(d) rangeย T=(nullย Tโˆ—)โŠฅrange\ T=(null\ T^*)^\perp

Proof:

(a)โ€…โ€ŠโŸบโ€…โ€Š(c)(a)\iff (c) since we can use Theorem 7.5 (c) while replacing TT with Tโˆ—T^*. Same idea give (b)โ€…โ€ŠโŸบโ€…โ€Š(d)(b)\iff (d). Also (a)โ€…โ€ŠโŸบโ€…โ€Š(d)(a)\iff (d) Since (VโŠฅ)โŠฅ=V(V^\perp)^\perp=V

Now we prove (a). Suppose wโˆˆnullย Tโˆ—โ€…โ€ŠโŸบโ€…โ€ŠTโˆ—w=0w\in null\ T^*\iff T^*w=0

Tโˆ—w=0โ€…โ€ŠโŸบโ€…โ€ŠโŸจv,Tโˆ—wโŸฉ=0โˆ€vโˆˆVโ€…โ€ŠโŸบโ€…โ€ŠโŸจTv,wโŸฉ=0โˆ€vโˆˆVโ€…โ€ŠโŸบโ€…โ€Šwโˆˆ(rangeย T)โŠฅT^*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

Definition 7.7

The conjugate transpose of a mร—nm\times n matrix AA is the nร—mn\times m matrix denoted Aโˆ—A^* given the conjugate of the transpose.

ie. (Aโˆ—)j,k=Aj,k(A^*)_{j,k}=A_{j,k}

Theorem 7.9

Let TโˆˆL(V,W)T\in \mathscr{L}(V,W) and e1,..,ene_1,..,e_n an orthonormal basis of VV, f1,...,fmf_1,...,f_m be an orthonormal basis of WW. Then M(Tโˆ—,(f1,...,fm),(e1,..,en))=M(T,(f1,...,fm),(e1,..,en))โˆ—M(T^*,(f_1,...,f_m),(e_1,..,e_n))=M(T,(f_1,...,f_m),(e_1,..,e_n))^*

Proof:

The k-th column of TT is given by writing TekTe_k. in the basis f1,...,fmf_1,...,f_m. ie. the k,jk,j entry of M(T)M(T) is โŸจTek,fjโŸฉ\langle Te_k,f_j \rangle, but then the j,kj,k entry of M(Tโˆ—)M(T^*) is โŸจTโˆ—f,ekโŸฉ\langle T^*f,e_k \rangle. But โŸจTek,fjโŸฉ=โŸจek,Tโˆ—fjโŸฉ=โŸจTโˆ—fj,ekโŸฉโ€พ\langle Te_k,f_j\rangle=\langle e_k,T^*f_j\rangle=\overline{\langle T^*f_j,e_k\rangle}

Example:

Suppose T(x1,x2,x3)=(x2+3x3,2x1)T(x_1,x_2,x_3)=(x_2+3x_3,2x_1)

โŸจT(x1,x2,x3),(y1,y2)โŸฉ=โŸจ(x2+3x3,2x1),(y1,y2)โŸฉ=(x2+3x3,2x1)y1ห‰,(x2+3x3,2x1)y2ห‰=y1ห‰x2+3y1ห‰x3+2y2ห‰x1=โŸจ(x1,x2,x3),(2y2,y1,3y1)โŸฉ\begin{aligned} \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\\ &=(x_2+3x_3,2x_1)\bar{y_1},(x_2+3x_3,2x_1)\bar{y_2}\\ &=\bar{y_1}x_2+3\bar{y_1}x_3+2\bar{y_2}x_1\\ &=\langle (x_1,x_2,x_3),(2y_2,y_1,3y_1)\rangle \end{aligned}

So Tโˆ—(y1,y2)=(2y2,y1,3y1)T^*(y_1,y_2)=(2y_2,y_1,3y_1)

Idea: Reisz Representation gives a function from VV to Vโ€ฒV' (3.118) tells us given TโˆˆL(V,W)T\in \mathscr{L}(V,W), we have