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Math429Linear Algebra (Lecture 30)

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 T∈L(V,W)T\in \mathscr{L}(V,W). The adjoint is the function T∗:W→VT^*:W\to V such that

⟨Tv,w⟩=⟨v,T∗w⟩,∀v∈V,w∈W\langle Tv,w \rangle=\langle v,T^*w \rangle, \forall v\in V, w\in W

Theorem 7.4

Suppose T∈L(V,W)T\in \mathscr{L}(V,W) then T∗∈L(W,V)T^*\in \mathscr{L}(W,V)

Proof:

Additivity, let w1,w2∈Ww_1,w_2\in W. We want to show T∗(w1+w2)T∗w1+T∗w2T^*(w_1+w_2)T^*w_1+T^*w_2

Let v∈Vv\in V, then

⟨v,T∗(w1+w2)⟩=⟨v,T∗(w1+w2)⟩=⟨Tv,w1+w2⟩=⟨Tv,w1⟩+⟨Tv,w2⟩=⟨v,T∗w1⟩+⟨v,T∗w2⟩=⟨v,T∗w1+T∗w2⟩\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 ⟨v,u⟩=⟨v,u′⟩\langle v,u \rangle=\langle v,u'\rangle, forall v∈Vv\in V then u=u′u=u'

Homogeneity: same as idea above.

Theorem 7.5

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

(a) (S+T)∗=S∗+T∗(S+T)^*=S^*+T^*
(b) (ST)∗=T∗S∗(ST)^*=T^* S^*
(c) (λT)∗=λˉS∗(\lambda T)^*=\bar{\lambda}S^*
(d) I∗=II^*=I
(e) (T∗)∗=T(T^*)^*=T
(f) If TT is invertible, then (T∗)−1=(T−1)∗(T^*)^{-1}=(T^{-1})^*

Proof:

(d) ⟨(ST)v,u⟩=⟨S(Tv),u⟩=⟨Tv,S∗u⟩=⟨v,T∗S∗u⟩\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 T∈L(V,W)T\in\mathscr{L}(V,W), then

(a) null T∗=(range T)⊥null\ T^*=(range\ T)^\perp
(b) null T=(range T∗)⊥null\ T=(range\ T^*)^\perp
(c) range T∗=(null T)⊥range\ T^*=(null\ T)^\perp
(d) range T=(null T∗)⊥range\ T=(null\ T^*)^\perp

Proof:

(a)  ⟺  (c)(a)\iff (c) since we can use Theorem 7.5 (c) while replacing TT with T∗T^*. Same idea give (b)  ⟺  (d)(b)\iff (d). Also (a)  ⟺  (d)(a)\iff (d) Since (V⊥)⊥=V(V^\perp)^\perp=V

Now we prove (a). Suppose w∈null T∗  ⟺  T∗w=0w\in null\ T^*\iff T^*w=0

T∗w=0  ⟺  ⟨v,T∗w⟩=0∀v∈V  ⟺  ⟨Tv,w⟩=0∀v∈V  ⟺  w∈(range T)⊥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

Definition 7.7

The conjugate transpose of a m×nm\times n matrix AA is the n×mn\times m matrix denoted A∗A^* given the conjugate of the transpose.

ie. (A∗)j,k=Aj,k(A^*)_{j,k}=A_{j,k}

Theorem 7.9

Let T∈L(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(T∗,(f1,...,fm),(e1,..,en))=M(T,(f1,...,fm),(e1,..,en))∗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 ⟨Tek,fj⟩\langle Te_k,f_j \rangle, but then the j,kj,k entry of M(T∗)M(T^*) is ⟨T∗f,ek⟩\langle T^*f,e_k \rangle. But ⟨Tek,fj⟩=⟨ek,T∗fj⟩=⟨T∗fj,ek⟩‾\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)

⟨T(x1,x2,x3),(y1,y2)⟩=⟨(x2+3x3,2x1),(y1,y2)⟩=(x2+3x3,2x1)y1ˉ,(x2+3x3,2x1)y2ˉ=y1ˉx2+3y1ˉx3+2y2ˉx1=⟨(x1,x2,x3),(2y2,y1,3y1)⟩\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 T∗(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 V′V' (3.118) tells us given T∈L(V,W)T\in \mathscr{L}(V,W), we have

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