NoteNextra-origin/content/Math429/Math429_L30.md

Lecture 30

Chapter VII Operators on Inner Product Spaces

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

Self adjoint and Normal Operators 7A

Definition 7.1

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

\langle Tv,w \rangle=\langle v,T^*w \rangle, \forall v\in V, w\in W

Theorem 7.4

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

Proof:

Additivity, let w_1,w_2\in W. We want to show T^*(w_1+w_2)T^*w_1+T^*w_2

Let v\in V, then

\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 \langle v,u \rangle=\langle v,u'\rangle, forall v\in V then u=u'

Homogeneity: same as idea above.

Theorem 7.5

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

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

Proof:

(d) \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\in\mathscr{L}(V,W), then

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

Proof:

(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

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

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\times n matrix A is the n\times m matrix denoted A^* given the conjugate of the transpose.

ie. (A^*)_{j,k}=A_{j,k}

Theorem 7.9

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))^*

Proof:

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}

Example:

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

\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^*(y_1,y_2)=(2y_2,y_1,3y_1)

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