Andrew Thangaraj
Aug-Nov 2020
\(V\): finite-dimensional inner product space over \(F=\mathbb{R}\) or \(\mathbb{C}\)
\(\phi:V\to F\) is a linear functional
Riesz representation theorem: There exists unique \(u\in V\) such that \[\phi(v)=\langle v,u\rangle\]
How to find \(u\) for a given \(\phi\)?
Orthonormal basis for \(V\): \(e_1,\ldots,e_n\)
\(u=\overline{\phi(e_1)}e_1+\cdots+\overline{\phi(e_n)}e_n\)
\(V,W\): finite-dimensional inner product space over \(F=\mathbb{R}\) or \(\mathbb{C}\)
\(T:V\to W\) is a linear map
Fix some \(w\in W\)
Let \(\phi_{T,w}(v)=\langle Tv,w\rangle\)
\(\phi_w:V\to F\) is a linear functional
Riesz: There exists unique \(u_{T,w}\in V\) s.t. \[\phi_{T,w}(v)=\langle Tv,w\rangle=\langle v,u_{T,w}\rangle\]
\(x=(x_1,x_2,x_3,x_4)\in\mathbb{R}^4\)
\(Tx=(x_1+2x_2+3x_3+4x_4,3x_1+4x_2+5x_3+6x_4)\in\mathbb{R}^2\)
\(\langle Tx,w\rangle=7x_1+10x_2+13x_3+16x_4\)
\(u_{T,w}=(7,10,13,16)\)
\(\langle Tx,w\rangle=(w_1+3w_2)x_1+(2w_1+4w_2)x_2+(3w_1+5w_2)x_3+(4w_1+6w_2)x_4\)
\(u_{T,w}=(w_1+3w_2,2w_1+4w_2,3w_1+5w_2,4w_1+6w_2)\in\mathbb{R}^4\)
\(V,W\): finite-dimensional inner product space over \(F=\mathbb{R}\) or \(\mathbb{C}\)
\(T:V\to W\) is a linear map
For \(w\in W\), let \(u_{T,w}\in V\) be s.t.
\[\langle Tv,w\rangle=\langle v,u_{T,w}\rangle\]
The adjoint of \(T\), denoted \(T^*\), is the function from \(W\) to \(V\) mapping \(w\) to \(u_{T,w}\)
\(T^*\): adjoint of \(T\) satisfies
\[\langle Tv,w\rangle=\langle v,T^*w\rangle\]
\(x=(x_1,x_2,x_3,x_4)\in\mathbb{R}^4\)
\(Tx=(x_1+2x_2+3x_3+4x_4,3x_1+4x_2+5x_3+6x_4)\in\mathbb{R}^2\)
\(w=(w_1,w_2)\in\mathbb{R}^2\)
\(\langle Tx,w\rangle=(w_1+3w_2)x_1+(2w_1+4w_2)x_2+(3w_1+5w_2)x_3+(4w_1+6w_2)x_4\)
\(T^*w=(w_1+3w_2,2w_1+4w_2,3w_1+5w_2,4w_1+6w_2)\in\mathbb{R}^4\)
\(V,W\): finite-dimensional inner product space over \(F=\mathbb{R}\) or \(\mathbb{C}\)
\(T:V\to W\) is a linear map
\(T^*\): adjoint of \(T\) is a linear map from \(W\) to \(V\)
Proof
\(\begin{align} \langle v,T^*(w_1+w_2)\rangle &= \langle Tv,w_1+w_2\rangle\\ &=\langle Tv,w_1\rangle+\langle Tv,w_2\rangle\\ &=\langle v,T^*w_1\rangle+\langle v,T^*w_2\rangle\\ &=\langle v,T^*w_1+T^*w_2\rangle \end{align}\)
\(\begin{align} \langle v,T^*(aw)\rangle &= \langle Tv,aw\rangle\\ &=\bar{a}\langle Tv,w\rangle\\ &=\bar{a}\langle v,T^*w\rangle\\ &=\langle v,aT^*w\rangle \end{align}\)