5.19 Linear Functionals and Adjoints

Exercise: Let \(V\) be the \(\R\)-vs and \(V=\{(x_{1},x_{2},...):x_i\in \R,\forall i\in \N\text{ and }\{i:x_i\neq 0\}\text{ is finite}\}\)

Inner product: \((l^{2}\text{ inner product})\)

\(\lang (x_i)_{i\in \N},(y_i)_{i\in \N}\rang=\sum^\infty_{i=1}x_iy_i\) this is finite sum since \(x_i\neq 0\) is finite

Let \(W\) be the subspace: \(W=\{(x_i)_{i\in \N}\in V:\sum^\infty_{i=1}x_i=0\}\). \(W\) is not finite dimensional

Let \(v=(1,0,0,...)\notin W\), does the best approximation of \(v\) to \(W\) exist? That is, does it exists a vector \(w\in W:||v-w||\leq ||v-w'||,\forall w'\in W\)

Solution

\(||v-w'||=||(\frac{1}{m},...,\frac{1}{m},0,...)||=\sqrt{\frac{1}{m}}\), when \(m\to \infty,||v-w'||\to 0\), then \(v=w\) but \(v\notin W\)

Thus there doesn't exist best approximation

Hint: Consider the vertors \(w_m\in W\) where \(w_{m}=(1-\frac{1}{m},-\frac{1}{m},...,-\frac{1}{m},0,...),\forall m\in \N\)

Exercise2: Bessel's inequality

Let \((V,\lang,\rang)\) be a k-IPVS, orthogonal set \(\{v_1,...,v_m\}\subseteq V\), \(v_i\neq 0_v,\forall i\)

Prove that \(\forall v\in V,\sum^{m}_{i=1}\frac{|\lang v,v_{i}\rang|^{2}}{||v_i||^2}\leq ||v||^2\)

Extend that set to the basis of \(V\): \(B_V=\{v_1,...,v_m,v_{m+1},...,v_n\}\)

We know \(v=\sum_{i=1}^{n}\frac{\langle v,v_{i}\rangle}{||v_{i}||^{2}}v_{i}\), then \(\Vert v\Vert=\sum_{i=1}^{n}\frac{\langle v,v_{i}\rangle}{||v_{i}||^{2}}\left|\right |v_{i}\Vert=\sum_{i=1}^{n}\frac{\langle v,v_{i}\rangle}{||v_{i}||}\)

Thus \(||v||^{2}=\sum_{i=1}^{n}\frac{\langle v,v_{i}\rangle^{2}}{||v_{i}||^{2}}\geq\sum _{i=1}^{m}\frac{\langle v,v_{i}\rangle^{2}}{||v_{i}||^{2}}\)

Linear Functionals and Adjoints

Let \((V,\lang,\rang)\) be a k-IPVS, \(w\in V\) be a fixed vector. Consider the \(k\)-linear map \(\lang,w\rang =f_{w}:V\overset{\lang\cdot ,w\rang}{\longrightarrow}k\) where \(f_{w}(v)=\langle_{\cdot}v,w\rangle\)

Question: Is it true that given \(\phi:V\to k\) be k-linear, then \(\exists w\in V:\phi=f_w\)?

Recall \(\{g:V\to k,k\text{-linear}\}\) is a k-vector space called \(V^*\) the dual space

We will prove that if \(\dim V<\infty\), then the answer is YES

Theorem

Let \((V,\lang,\rang)\) be s k-IPVS, \(\dim V<\infty\). Given \(\phi:V\to k\) linear, \(\exists !w\in V:\phi=\lang ,w\rang\), that is \(\forall v\in V,\phi(v)=\lang v,w\rang\)

Proof

Existence:

Let \(\{v_1,...,v_m\}\) be an Orthonormal basis of \(V\), let \(w=\sum^{m}_{j=1}\overline{\phi(v_j)}v_j\) and consider \(f_w:V\to k\) where \(f_{w}(v)=\lang v,w\rang,\forall v\in V\)

Given \(i,1\leq i\leq m\), \(f_{w}(v_{i})=\langle v_{i},w\rangle=\langle v_{i},\sum_{j=1}^{m}\overline{\phi(v_{j})} v_{j}\rangle=\sum_{j=1}^{m}\overline{\overline{\phi(v_{j})}}\langle v_{i},v_{j}\rangle =\phi(v_{i})||v_{i}||^{2}=\phi(v_{i})\)

So \(f_w\) coincide with \(\phi\) on a basis of \(V\), so \(f_w=\phi\)

Uniqueness:

Suppose that \(w'\in W:f_{w'}=\phi\), then given \(v\in V\), \(f_w(v)=\phi(v)=\lang v,w\rang =f_{w'}(v)=\lang v,w'\rang\)

Thus \(\lang v,w\rang =\lang v,w'\rang ,\forall v\in V\), then \(\lang v,w-w'\rang =0,\forall v\in V\), taking \(v=w-w'\), we get \(w-w'=0_V\), then \(w=w'\)

Example

Show that the theorem is not true when vector space is not finite dimensional

Let \(V=\mathbb{R}[x]\), consider inner product \(\displaystyle \lang p(x),q(x)\rang =\int^{1}_{0}p(t)q(t)dt\). Fix \(z_0\in \R\), let \(f:V\to \R\) be defined by \(f(p(x))=p(z_0)\)

Suppose \(\exists q(x)\in \R[x]:f=\lang ,q(x)\rang\), that is \(\forall p(x)\in \R[x]\), \(\lang p(x),q(x)\rang =f(p(x))=p(z_0)\)

Consider \(p(x)\in \R[x]:p(z_{0})=0\), then \(\displaystyle 0=p(z_{0})=\int^{1}_{0}p(t)q(t)dt\). Given \(s(x)\in \R[x]\), cosider \(\widetilde{p(x)}=p(x)s(x)\), then \(\displaystyle\widetilde{p(z_0)}=p(z_{0})s(z_{0})=0=\int^{1}_{0}p(t)s(t)q(t)dt\)

Fix now \(s(x)=(x-z_0)q(x)\), take \(p(x)=x-z_0\), so \(\displaystyle 0=\int^{1}_{0}p(t)(t-z_{0})q(t)q(t)dt=\int^1_0(t-z_0)^2(q(t))^2dt\) here \((t-z_0)^2\) is not zero function

Thus \(q(x)=0\), then \(f=0\) which is a contradiction

Theorem

Let \((V,\lang,\rang)\) be a finite k-IPVS, and \(f:V\to V\) be a \(k\)-linear map, \(\exists!\underbrace{f^*}_{\text{the adjoint of f}}:V\to V:\forall v,w\in V,\lang f(v),w\rang=\lang v,f^*(w)\rang\)

Example

Suppose \(f:V\to V\) be defined as \(f(v)=\alpha\cdot v\) where \(\alpha\in k\)

Given \(v,w\in V\), \(\lang f(v),w\rang=\lang \alpha v,w\rang=\alpha\lang v,w\rang=\lang v,\bar\alpha w\rang\)

Consider \(f^*:V\to V\) where \(f^*(v)=\bar\alpha\cdot v,\forall v\in V\)

Proof

Let \(w\in V\), consider the map from \(V\to k\) where \(v\mapsto \lang f(v),w\rang\) it is \(k\)-linear because this is composition of two linear map: \(f\) and \(\lang,\rang\)

By previous theorem, \(\exists!w'\in V:\lang f(v),w\rang = \lang v,w'\rang\), \(\forall v\in V\), that is \(\lang f(),w\rang=\lang,w'\rang\)

Let \(f^*:V\to V\) be defined by \(f^*(w)=w'\)

Note that \(\forall v,w\in V,\lang f(v),w\rang =\lang v,f^*(w)\rang\). NTP: \(f^*\) is linear

Let \(w_1,w_2\in V,\alpha\in k\). Is \(f^{*}(w_{1}+\alpha w_{2})=f^*(w_1)+\alpha f^*(w_2)\)

NTP: the inner product in all vector are equal

For this, given \(v\in V\), \(\lang v,f^{*}(v_{1}+v_{2})\rang=\lang f(v),w_{1}+\alpha w_{2}\rang=\lang f(v),w_1\rang+\bar\alpha\lang f(v),w_2\rang=\lang v,f^*(w_1)\rang +\bar \alpha\lang v,f^*(w_2)\rang=\lang v,f^*(w_1)\rang +\lang v,\alpha f^*(w_2)\rang=\lang v,f^*(w_1)+\alpha f^*(w_2)\rang\)

Then \(f^{*}(w_{1}+\alpha w_{2})=f^{*}(w_{1})+\alpha f^{*}(w_{2})\), here we use the fact that if \(z,z'\in V:\lang v,z\rang =\lang v,z'\rang,\forall v\in V\), then \(z=z'\)

Uniqueness: Clearly, by previous theorem, we have unique \(w'\), then \(f^*\) is unique

Theorem

Let \((V,\lang,\rang)\) be a finite dimensional k-IPVS. Let \(B=\{v_1,...,v_m\}\) be an ordered orthonormal basis of \(V\)

Let \(f:V\to V\) be a k-linear map, call \(A=[f]_B\) and the coefficient is \(a_{ij}\)

Then \(a_{ij}=\lang f(v_j),v_i\rang\)

Proof

\(\forall v\in V\), \(v=\sum^{m}_{l=1}\lang v,v_{l}\rang\cdot v_l\)

So \(f(v_{j})=\sum^{m}_{j=1}\lang f(v_{j}),v_{l}\rang\cdot v_l\), then \(a_{lj}=\lang f(v_{j}),v_{l}\rang\) here we let \(f(v_j)=v\)

Corollary

\(f=f^{**}\)

Proof

\(f,f^{**}:V\to V\) are k-linear. \(\forall v,w\in V,\lang f(v),w\rang=\lang v,f^{*}(w)\rang=\overline{\lang f^*(w),v\rang }=\overline{\lang w,f^{**}(v)\rang}=\lang f^{**}(v),w\rang\)

So \(\forall v\in V,f(v)=f^{**}(v)\Rightarrow f=f^{**}\)
\([f^{*}]_B=\overline{[f]}_B^t\), \(B\) must be orthonormal basis!

Proof

Let \(A=[f]_B\), \(A'=[f^*]_B\), \(a_{lj}=\lang f(v_j),v_l\rang\) and \(a'_{lj}=\lang f^*(v_j),v_l\rang =\lang v_j,f^{**}(v_l)\rang=\lang v_j,f(v_l)\rang =\overline{\lang f(v_l),v_j\rang}=\overline {a_{jl}}\)

Example

Let \((V,\lang,\rang)\) be k-IPVS, finite dimension. Let \(W\subseteq V\) be a subspace. Given \(v,v'\in V\), \(\lang \underbrace{P_{W}(v)}_{\text{orthogonal projection of v to W}},v'\rang=\lang P_{W}(v),\underbrace{v'-P_W(v')}_{\in W^\perp}+\underbrace{P_W(v')}_{\in W}\rang=\lang P_W(v),P_W(v')\rang\)

Now \(\lang v,P_{W}(v')\rang=\text{same argument on the 1st variable}=\lang P_W(v),P_W(v')\rang\)

Then \(\forall v,v'\in V,\lang v,P_W(v')\rang =\lang P_W(v),v'\rang\), thus \((P_W)^*=P_W\)