5.26 Unitary and Normal

Example

1. Given a finite dimensional k-IPVS, \((V,\lang,\rang)\). Let \(n=\dim V\), then \(V\cong k^n\) with the standard inner product where is isomorphism of k-IPVS

   Let \(B=\{v_1,...,v_n\}\) orthonormal basis of \(V\), define the map \(f:V\to k^{n}\) where \(v_i\mapsto e_i\) where \(f\) send an orthonormal basis to an orthonormal basis, so \(f\) preserves inner product, so it is an isomorphism of k-IPVS

   Given \(v=\sum^n_{i=1}\lambda_iv_i\), \(f(v)=\sum^n_{i=1}\lambda_ie_i\), \(f(v)=(\lambda_1,...,\lambda_n)=[v]_B\)

2. \(M_{3}(\R)\supseteq W=\{A\in M_3(\R):A^t=-A\}\) with inner product \(\lang A,B\rang =\frac{1}{2}\text{tr}(AB^*)\)​

   \(\dim W=3\) since the diagonal is zero and symmetry, then we have 3 free variables

   Let \(V=\R^3\) with the standard inner product. Let \(f:V\to W\) be the map defined by \(f(x_{1},x_{2},x_{3})=\begin{pmatrix} 0&-x_3&x_2\\ x_3&0&-x_1\\ -x_2&x_1&0 \end{pmatrix}\)​

   Prove that \(f\) preserves inner product

3. Let \(P\in Gl_{n}(\mathbb{C})\) and \(G=P^*P\), \(G\) is invertible sincce \(\det P^{*}=\det(\bar P^{t})=\det \bar P=\overline{\det P}\), then \(\det P=0\iff\det P^*=0\) with inner product \(\lang X,Y\rang_1 =Y^{*}X\)​

   Let \(W\) be the \(\mathbb{C}\)-IPVS such that its vector are elements of \(\mathbb{C}^{n\times 1}\) with inner product \(\lang X,Y\rang_2=Y^*GX\)​

   Are \((V,\lang,\rang_1)\) and \((W,\lang,\rang_2)\) isomorphic of inner product vector space? Yes, they have the same dimension

   Let \(f:W\to V\) be defined by \(f(X)=PX\) where \(\lang X,Y\rang_{2}=Y^{*}GX=Y^{*}P^{*}PX=(PY)^{*}PX=\lang PX,PY\rang_{1}=\lang f(X),f(Y)\rang_1\)​

4. \(V=\{f:[0,1]\to \R:f\text{ is continuous }\}=W\)​

   \(\displaystyle\lang f,g\rang_{V}=\int^{1}_{0}f(t)g(t)t^{2}dt,\quad\lang f,g\rang_{W}=\int^{1}_{0}f(t)g(t)dt\)​

   Define \(T:V\to W\) such that \(T(f(t))=t(f(t))\) where \(f:[0,1]\to \R\) is continuous.

   \(T\) preserves inner products, \(\displaystyle\lang T(f(t)),T(g(t))\rang=\lang tf(t),tg(t)\rang_W=\int^1_0t\cdot f(t)\cdot t\cdot g(t)dt=\int^1_0f(t)\cdot g(t)\cdot t^2dt=\lang f(t),g(t)\rang_V\)​

   But \(T\) is not an isomorphism because it is not surjective

   Take the constant function \(1\in W\), then \(1\notin \text{Im}T\)​

   This is because preserves inner product \(\Rightarrow\) injective. But we usually use dimension theorem to conclude it is surjective.

   But this is finite dimension, then we cannot conclude that it is surjective

Theorem

Let \(\lang V,\lang,\rang_V)\) and \((W,\lang,\rang_W)\) be k-IPVS. Let \(f:V\to W\) be a k-linear map

\(f\) preserves inner product\(\iff\)​\(||f(v)||_W=||v||_V\), \(\forall v\in V\)

Proof

\(\Rightarrow\)) We already know from this

\(\Leftarrow\)) Polarization Identities

If \(k=\R\), then \(\forall x,y\in V,\lang x,y\rang =\frac{1}{4}(||x+y||^2-||x-y||^2)\)

If \(k=\mathbb{C}\), then \(\lang x,y\rang=\frac{1}{4}(||x+y||^{2}-||x-y||^{2}-i||x+iy||^{2}+i||x-iy||^{2})\)

Definition

Let \(f:V\to W\) be an isomorphism of k-IPVS, we say that \(f\) is a unitary operator

Remark

\(\{f:V\to V:f\text{ is unitary}\}\) is closed under composition and inverse. Why sum not? Since \(f(x)+(-f(x))=0\) is not isomorphism

Proposition

Let \(U:V\to V\) be k-linear. \(U\) is unitary\(\iff\)​\(\exists U^*\) and \(U^*\circ U=Id_V=U\circ U^*\)

Proof

\(\Rightarrow\)) \(\exists U^{-1}\) and \(\forall v,w\in V\) s.t. \(\lang U(v),w\rang =\lang U(v),U(U^{-1}(w))\rang\). Since \(U\) preserves inner product, then \(=\lang v,U^{-1}(w)\rang\), then \(U^*\) exists and \(U^*=U^{-1}\)

\(\Leftarrow\)) Suppose \(U^*\) exists and \(U^*=U^{-1}\). Given \(v,w\in V\), \(\lang U(v),U(w)\rang=\lang v,U^*(U(w))\rang=\lang v,U^{-1}(U(w))\rang=\lang v,w\rang\)

Example

\(V=\mathbb{C}^{n\times 1}\), \(\lang X,Y\rang =Y^*X\). Let \(A\in M_n(\mathbb{C})\), \(U:V\to V\) be defined as \(U(X)=AX\)

\(\lang U(X),U(Y)\rang=\lang AX,AY\rang=(AY)^*AX=Y^*A^*AX\), so \(U\) is unitary\(\iff\)​\(Y^*A^*AX=Y^*X,\forall X,Y\in \mathbb{C}^{n\times 1}\)

\(\iff A^*A=Id\). For example \(Y=e_i,X=e_j\), then \(e_{i}^{*}A^*Ae_j=(A^*A)_{ij}\) and \(e^*_ie_j=\delta_{ij}\)

Definition

Let \(A\in M_n(\mathbb{C})\) is called unitary\(\iff A^{*}A=Id_{n}\). In particular \(A\) is unitary, then \(A\) is invertible

Proposition

Let \((V,\lang,\rang)\) be a finite dimension k-IPVS. Let \(f:V\to V\) be a k-linear map.

\(f\) is unitary\(\iff\)​\(\exists B\) orthonormal basis of \(V:[f]_B\) is unitary

Proof

Let \(B\) be an orthonormal basis of \(V\), let \(A=[f]_B\), then \(A^*A=Id\)(that is \(A\) is unitary)

\(A^{*}=[f^*]_{B}\) since \(B\) is orthonormal basis. \(\iff\)​\(f^{*}\circ f=Id\iff f\) is unitary

---

Let's look more carefully at the condition \(A\in M_n(\mathbb{C})\) is unitary

\(\delta_{ij}=(A^{*}\cdot A)_{ij}=\sum^{n}_{r=1}(A^{*})_{ir}\cdot (A)_{rj}=\sum^{n}_{r=1}\overline {(A)_{ri}}\cdot A_{rj}=\sum^{n}_{i=1}\overline{a_{ri}}\cdot a_{rj}\)

Thus \(\delta_{ij}=\lang c_{j},c_{i}\rang=c_i^*\cdot c_j\) standard inner product

​

Thus \(A\in M_n(\mathbb{C})\) is unitary\(\iff\) The set of columns of \(A\) is an orthonormal basis of \((C^n,\lang,\rang)\)

Remark

\(A\) is unitary, \(\det A\) ?

Since \(A^*A=Id\), then \(\det(A^{*}A)=\det(Id)=1\), and \(\det(A^*A)=\overline{\det(A)}\det(A)\), then \(|\det A|^2=1\)

Suppose \(A\in M_n(\R)\), then \(\det A=\pm 1\). If \(A\in M_n(\mathbb{C})\), then \(|\det A|=1\)

Also, if \(k=\R\), \(A^*A=Id\iff A^tA=Id\). If \(k=\mathbb{C}\), \(A^{*}A=Id\iff\bar A^{t}\cdot A=Id\)

We say that \(A\) is orthogonal

Example

\(A= \begin{pmatrix} \cos a & \sin a \\ -\sin a & \cos a \end{pmatrix}\) where \(a\in [0,2\pi)\). \(A\) is orthogonal \(\forall a\in [0,2\pi)\)

\(A^{t}A= \begin{pmatrix} \cos a & -\sin a \\ \sin a & \cos a \end{pmatrix}\cdot \begin{pmatrix} \cos a & \sin a \\ -\sin a & \cos a \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

Let \(f:\R^2\to \R^2\), \(f(X)=AX\), we will say that \(f\) is a rotation

Remark

Let \((V,\lang,\rang)\) k-IPVS, \(\dim V=n\). Let \(B=\{v_1,...,v_n\}\) and \(B'=\{v_1',...,v_n'\}\) be orthonormal basis of \(V\)

Given \(v\in V\) we have \([v]_B\) and \([v]_{B'}\), let \([v]_{B'}=C_{BB'}[v]_{B}\)

Exercise: Prove that \(C_{BB'}=(P_{B}^{B'})^{-1}\) is unitary, write it and column orthonormal

---

\(A\) is unitary\(\iff\)​\(A\cdot A^{*}=Id=A^*\cdot A\). In particular, \(AA^{*}=A^{*}A\)

\(f\) is unitary\(\iff\)​\(f\circ f^*=Id=f^*\circ f\). ...

Problem: \(f:V\to V\) and \(V\) finite dimension k-IPVS, when is \(f\) is diagonalize?

Suppose \(f\) is diagonalizable and \(f^{*}\circ f=f\circ f^*\), then \(\exists\) a orthonormal basis \(B:[f]_{B}= \begin{pmatrix} \lambda_{1} & & 0 \\ & \ddots & \\ 0 & & \lambda_{n} \end{pmatrix}\) and \([f^{*}]_{B}=\overline{[f]}_{B}^{t}= \begin{pmatrix} \bar\lambda_{1} & & 0 \\ & \ddots & \\ 0 & & \bar\lambda_{n} \end{pmatrix}\)

We will prove that \(f^{*}\circ f=f\circ f^*\), then \(f\) and \(f^*\) will be diagonalizable in an orthonormal basis for \(k=\mathbb{C}\)

Definition

\((V,\lang,\rang)\) k-IPVS, \(f:V\to V\) is normal\(\iff f^*\circ f=f\circ f^*\) be k-linear

Examples: \(Id_V,0:V\to V,\lambda\cdot Id_V,\forall \lambda\in k\)

Let \(f:V\to V\), given \(B\) orthonormal basis \(\{v_1,...,v_n\}\) and \(f(v_{i})= \lambda_{i}v_i\), then \([f]_{B}= \begin{pmatrix} \lambda_{1} & & 0 \\ & \ddots & \\ 0 & & \lambda_{n} \end{pmatrix}\) and \([f^{*}]_{B}=\overline{[f]}_{B}^{t}= \begin{pmatrix} \bar\lambda_{1} & & 0 \\ & \ddots & \\ 0 & & \bar\lambda_{n} \end{pmatrix}\)

Self-adjoint maps (\(f^{*} =f\))

Exercise
Find \(g, f: \mathbb{C}^2 \longrightarrow \mathbb{C}^2\) such that \(f\) and \(g\) are normal but \(f+g\) is not normal.

Remark

Unitary is says \(f\) commutes with \(f^*\) and equals to \(Id\)

But normal is just that \(f\) commutes with \(f^*\)