NoteNextra-origin/content/Math429/Math429_L29.md

Lecture 29

Chapter VI Inner Product Spaces

Orthogonal Complements and Minimization Problems 6C

Minimization Problems

Theorem 6.61

Suppose U is a finite dimensional subspace of V. Let v\in V, u\in U. Then ||v-P_u v||\leq|| v-u||. with equality if and only if u=P_u v

Proof:

Using triangle inequality

\begin{aligned}
||v-P_u v||^2 &\leq ||v-P_u v||^2+||P_u v-u||^2\\
&=||(v-P_u v)+(P_u v-u)||^2\\
&=||v-u||^2
\end{aligned}

Example:

Find $u(x)\in \mathscr{P}_5(\mathbb{R}) minimizing

\int^{\pi}_{-\pi}|sin(x)-u(x)|^2 dx

V=C([-\pi,\pi])= continuous (real valued) function on [-\pi,\pi]

u=\mathscr{P}_5(\mathbb{R}). Note U\subseteq V and u is finite dimensional.

\langle f,g \rangle=\int^{\pi}_{-\pi}fg gives an inner product on V.

Minimize ||sin-u||^2, choose an orthonormal basis e_0,...,e_5 of \mathscr{P}_5(\mathbb{R}), so u=P_u(sin)=\langle e_0,sin\rangle e_0+...+\langle e_5,sin \rangle e_5

Pseudo inverses

Idea: Want to (approximately) solve Tx=b.

• If T is invertible x=T^{-1}b
• If T is not invertible, want T^{T} such that y=T^{T}b is the "best solution"

Lemma 6.67

If V is a finite dimensional vector space, T\in \mathscr{L}(V,W) then T\vert_{{null\ T}^\perp} is one to one onto range\ T.

Proof:

Note (null\ T)^\perp \simeq V/(null\ T)

Exercise, prove this...

If v\in null(T\vert_{{null}^\perp})\implies v\in null\ T, and v\in (null\ T)^\perp\implies v=0

If w\in range\ T so \exists v\in V such that Tv=w write v as v=u+x sor u\in null\ T,x\in (null\ T)^\perp.

Definition 6.68

V is a finite dimensional space T\in \mathscr{L}(V,W). The pseudo-inverse denoted T^\dag\in \mathscr{L}(W,V) is given by

T^\dag w=(T\vert_{{null\ T}^\perp})^{-1}P_{range\ T}w

Some explanation:

Let T\in \mathscr{L}(V,W).Since there exists isomorphism between (null\ T)^\perp\subseteq V and range\ T\subseteq W.We can always map W to V using T^\dag\in \mathscr{L}(W,V). P_{range\ T} is the map that W\mapsto range\ T and (T\vert_{{null\ T}^\perp})^{-1} is a linear map that map $w\in W$

Proposition 6.69

V is a finite dimensional vector space. T\in\mathscr{L}(V,W), then

(a) If T is invertible, then T^\dag=T^{-1}.
(b) TT^\dag=P_{range\ T}.
(c) T^\dag T=P_{(null\ T)^\perp}.

Theorem 6.70

V is a finite dimensional vector space. T\in\mathscr{L}(V,W), for b\in W, then

(a) If x\in V, then ||T(T^* b)-b||\leq ||Tx-b|| with equality if and only if x\in T^\dag b+null\ T (T^\dag is the best solution we can have as "inverse" for non-invertible linear map)

(b) If x\in T^\dag b+null\ T then

||T^\dag b ||\leq ||x||

Proof:

(a) Tx-b=(Tx-TT^\dag b)+(TT^\dag b-b)

Using pythagorean theorem, we have

||Tx-b||\geq ||TT^\dag b-b||

Chapter VII Operators on Inner Product Spaces

Self adjoint and Normal Operators 7A

Definition 7.1

Let T\in \mathscr{L}(V,W), then the adjoint of T denoted T^* is the function T^*:W\to V such that \langle Tv,w \rangle =\langle v,T^* w \rangle

For euclidean inner product T^* is given by the conjugate transpose.