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6.4 Theorem of normal and Classification

Definition

\(A\in M_n(k)\), \(k=\R\) or \(\mathbb{C}\). \(A\) is normal\(\iff\)​\(A\cdot A^{*}=A^{*}\cdot A\)

Example

Let \(A\in M_{3}(\mathbb{C})\), \(A= \begin{pmatrix} a_{1} & a_{4} & a_{5} \\ 0 & a_{2} & a_{6} \\ 0 & 0 & a_{3} \end{pmatrix}\). When is \(A\) normal?

\(A^{*}= \begin{pmatrix} \bar{a}_{1} & 0 & 0 \\ \bar{a}_{4} & \bar{a}_{2} & 0 \\ \bar{a}_{5} & \bar{a}_{6} & \bar{a}_{3} \end{pmatrix}\), then \(A\cdot A^{*}= \begin{pmatrix} a_{1}\bar{a}_{1}+a_{4}\bar{a}_{4}+a_{5}\bar{a}_{5} & a_{4}\bar{a}_{2}+a_{5}\bar{a}_{6} & a_{5}\bar{a}_{3} \\ a_{2}\bar{a}_{4}+a_{6}\bar{a}_{5} & a_{2}\bar{a}_{2}+a_{6}\bar{a}_{6} & a_{6}\bar{a}_{3} \\ a_{3}\bar{a}_{5} & a_{3}\bar{a}_{6} & a_{3}\bar{a}_{3} \end{pmatrix}\)

\(A^* \cdot A = \begin{pmatrix} \bar{a}_{1}a_{1} & \bar{a}_{1}a_{4} & \bar{a}_{1}a_{5} \\ \bar{a}_{4}a_{1} & \bar{a}_{4}a_{4}+\bar{a}_{2}a_{2} & \bar{a}_{4}a_{5}+\bar{a}_{2}a_{6} \\ \bar{a}_{5}a_{1} & \bar{a}_{5}a_{4}+\bar{a}_{6}a_{2} & \bar{a}_{5}a_{5}+\bar{a}_{6}a_{6}+\bar{a}_{3}a_{3} \end{pmatrix}\)

Then in \(A\cdot A^*\), \(a_{1}\bar a_{1}+a_{4}\bar a_{4}+a_{5}\bar a_{5}=|a_{1}|^{2}+|a_{4}|^{2}+|a_{5}|^{2}\)

In \(A^*A\), \(a_1\bar a_1=|a_1|^2\), then if \(A\cdot A^{*}=A^{*}\cdot A\), \(a_4=a_5=0\)

Then \(A\cdot A^{*}= \begin{pmatrix} a_{1}\bar{a}_{1} & 0 & 0 \\ 0 & a_{2}\bar{a}_{2}+a_{6}\bar{a}_{6} & a_{6}\bar{a}_{3} \\ 0 & a_{3}\bar{a}_{6} & a_{3}\bar{a}_{3} \end{pmatrix}\), \(A^{*}\cdot A= \begin{pmatrix} \bar{a}_{1}a_{1} & 0 & 0 \\ 0 & a_{2}\bar{a}_{2} & \bar{a}_{2}a_{6} \\ 0 & \bar{a}_{6}a_{2} & \bar{a}_{3}a_{3} \end{pmatrix}\)

Also we get \(a_6=0\), then \(A\cdot A^{*}= \begin{pmatrix} a_{1}\bar{a}_{1} & 0 & 0 \\ 0 & a_{2}\bar{a}_{2} & 0 \\ 0 & 0 & a_{3}\bar{a}_{3} \end{pmatrix}\), \(A^{*}\cdot A= \begin{pmatrix} \bar{a}_{1}a_{1} & 0 & 0 \\ 0 & a_{2}\bar{a}_{2} & 0 \\ 0 & 0 & \bar{a}_{3}a_{3} \end{pmatrix}\)

Thus \(A\) is normal\(\iff\)​\(A\) diagonal

Theorem 6

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

Let \(\mathcal{B}=\{v_1,...,v_n\}\) be an orthonormal basis of \(V\). If \([f]_\mathcal{B}\) is upper triangular, then \(f\) is normal\(\iff\)​\([f]_\mathcal{B}\) is diagonal

Proof

Let \(A=[f]_\mathcal{B}\) (\(A\) is upper triangular). Since \(\mathcal{B}\) is orthonormal, we have \([f^{*}]_{\mathcal{B}}=\overline{[f]_{\mathcal{B}}}^t=\bar A^t=A^*\)

\(\Leftarrow\)) Since \(A\) is diagonal, then \(A^{*}=\bar A^t=\bar A\) , then \(A\cdot A^{*}=A\cdot \bar A=\bar A\cdot A=A^*\cdot A\)

Thus \([f\circ f^{*}]_{\mathcal{B}}=[f^*\circ f]_{\mathcal{B}}\), then \(f\circ f^{*}=f^{*}\circ f\), then \(f\) is normal

\(\Rightarrow\)) \(f\) is normal and \(A=[f]_\mathcal{B}\) is upper triangular, then \(A= \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \placeholder{} & a_{22} & \cdots & a_{2n} \\ \placeholder{} & \placeholder{} & \ddots & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & a_{nn} \end{pmatrix}\), \(A^{*}= \begin{pmatrix} \overline{a_{11}} & 0 & \cdots & 0 \\ \overline{a_{12}} & \overline{a_{22}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ \overline{a_{1n}} & \overline{a_{2n}} & \cdots & \overline{a_{nn}} \end{pmatrix}\)

Then \(f(v_1)=a_{11}v_1\), then \(v_1\) is an eigenvector of \(f\) of eigenvalue \(a_{11}\)

By Theorem 5, \(v_1\) is an eigenvector of \(f^*\) of eigenvalue \(\overline{a_{11}}\), on the other hand, \(\overline{a_{11}}v_1=f^{*}(v_{1})=\sum_{j=1}^{n}\overline{a_{1j}}v_{j}\), then \(a_{1j}=0,\forall 2\leq j\leq n\)

Now \(f(v_{2})=a_{12}v_{1}+a_{22}v_{2}=a_{22}v_{2}\), then \(v_2\) is an eigenvector of \(f\) of eigenvalue \(a_{22}\)

Then \(v_2\) is an eigenvector of \(f^*\) with eigenvalue \(\overline{a_{22}}\)

We can conclude as before, that \(a_{2j}=0\), \(\forall j,3\leq j\leq n\) and \(a_{21}=0\)

We can finfish the proof by induction on \(n\)

Theorem 7

Let \((V,\lang,\rang)\) be \(\mathbb{C}\)-IPVS, finite dimension. \(f:V\to V\) be \(\mathbb{C}\)-linear

There exists an orthonormal basis \(\mathcal{B}=\{v_{1},...,v_{n}\}\) of \(V\) s.t. \([f]_\mathcal{B}\) is upper triangular

Remark

From theorem \(6\) and \(7\), we will conclude that if \(f\) is normal and \(k=\mathbb{C}\), then \(\exists B\) orthonormal basis of \(V\) s.t. \([f]_\mathcal{B}\) is diagonal

Proof

\(\dim V=n\). Induction on \(n\)

The case \(n=1\), the result is trivially true

Assume that the result is true for \(\mathbb{C}\)-IPVS of dimension \(n-1\). Since \(V\) is a finite-dimensional vector space

Then \(\exists v\in V,\exists \lambda \in \mathbb{C}\) s.t. \(f^*(v)=\lambda v\), let \(W=\lang v\rang^\perp\). Consider \(\tilde{f}:W\to W\), by Theorem 3 we know that \(W\) is \(f\)-invariant, then let \(\tilde{f}=f|_W\), moreover \(\dim W=n-1\)

By induction hypothesis, we know that \(\exists\) an orthonormal basis \(\tilde{B}=\{v_2,...,v_n\}\) s.t. \([\tilde{f}]_{\tilde{\mathcal{B}}}\) is upper triangular

Take \(v_1=\frac{v}{||v||}\), let \(\mathcal{B}=\tilde{B}\cup\{v_{1}\}\), then \([f]_{\mathcal{B}}= \begin{pmatrix} & & & a_{1n} \\ & [\tilde{f}]_{\tilde{B}} & \placeholder{} & a_{2n} \\ & \placeholder{} & & \vdots \\ 0 & \cdots & 0 & a_{nn} \end{pmatrix}\) is also upper triangular since \(f(v_{1})\) is a linear combination of previous vectors

Corollary

Let \(A\in M_n(\mathbb{C})\). There exists a unitary matrix \(U\in M_n(\mathbb{C})\) s.t. \(U^{-1}AU=U^{*}AU\) is upper triangular

Proof

Let \(f:\mathbb{C}^{n\times1}\to\mathbb{C}^{n\times1}\) defined by \(f(X)=AX\)

By theorem 7, \(\exists\) an orthonormal basis \(B\) of \(\mathbb{C}^{n\times 1}\) s.t. \([f]_\mathcal{B}\) is upper triangular with standard inner product

Also, we know that \(A=[f]_E\) where \(E\) is orthonormal basis for the \(\mathbb{C}^{n\times 1}\), standard inner product

\(A=[f]_{E}=P^{-1}[f]_{\mathcal{B}}P\) where \(P=C_{E,B}\) since both \(E\) and \(B\) are orthonormal basis

We know that \(P\) is unitary, so \([f]_{\mathcal{B}}=P\cdot A\cdot P^{-1}\) where \(P=U^{-1}\)

Theorem 8

Let \((V,\lang,\rang)\) be \(\mathbb{C}\)-IPVS, finite dimension. \(f:V\to V\) be \(k\)-linear and normal

There exists an orthonormal basis \(\mathcal{B}=\{v_1,...,v_n\}\) of \(V\) s.t. \(\forall i,1\leq i\leq n,v_i\) is an eigenvector of \(f\)

Proof

It follows from Theorem 6 and Theorem 7

Let \((V,\lang,\rang)\) be \(\R\)-IPVS, finite dimension. \(f:V\to V\) be \(\R\)-linear and unitary

Orthogonal map = Unitary in \(k=\R\), aim: obtain a classification of orthogonal maps (finite dimensional case)

Recall: If \(f:V\to V\) orthogonal and \(\lambda\) is an eigenvalue of \(f\), then \(\lambda =\pm 1\) (preserves the norm???)

If \(f:V\to V\) orthogonal, let \(W\subseteq V\) a subspace s.t. \(W\) is \(f\)-invariant, then \(W^\perp\) is \(f^*\)-invariant

Question: Is \(W^\perp\) \(f\)-invariant? We know that \(f\) is an isomorphism.

Let \(w\in W\), so \(\exists ! w'\) s.t. \(w=f(w')\). Since \(W\) is \(f\)-invariant, we can consider \(\tilde{f}=f|_W:W\to W\)

We know that \(\tilde{f}\) is monomorphism since it's the restriction of a monomorphism

As a consequence \(\tilde{f}\) is an isomorphism, let \(w\in W\), \(\exists !w'\in W\) s.t. \(\tilde{f}(w')=w=f(w')\)

Let \(z\in W^\perp\), then \(\langle f(z),w\rangle=\langle f(z),f(w^{\prime})\rangle=\langle z,f^{*}(f(w^{\prime} ))\rangle=\langle z,w^{\prime}\rangle\) since \(f\) is orthogonal

Since \(w'\in W\) and \(z\in W^\perp\), then \(=0\), then \(f(z)\in W^\perp\). Thus \(W^\perp\) is \(f\)-invariant

Classification for \(n=2\)

Let \((V,\lang,\rang)\) be \(\R\)-IPVS, finite dimension. \(f:V\to V\) be orthogonal

Let \(B=\{v_1,v_2\}\) be an orthonormal basis of \(V\). Then \(\{f(v_1),f(v_2)\}\) is an orthonormal basis of \(V\) since \(\lang f(v_1),f(v_2)\rang =\lang v_1,v_2\rang =0\)

Also, \(f(v_1)=\alpha v_1+\beta v_2\) and \(f(v_2)=\alpha' v_2+\beta' v_2\). Consider \(\R^2\) standard inner product.

Consider \(\{(\alpha,\beta),(\alpha ',\beta ')\}=\Gamma\), exercise: Prove that \(\Gamma\) is an orthonormal basis of \(\R^2\) standard inner product

Consider \([f]_{\mathcal{B}}= \begin{pmatrix} \alpha & \alpha^{\prime} \\ \beta & \beta^{\prime} \end{pmatrix}\), \(\alpha^2+\beta^{2}=1\), \(\alpha'^2+\beta'^2=1\), \(\alpha \alpha'+\beta\beta'=0\)

Two possibilities. 1. \((\alpha',\beta')=(\alpha,\beta)\) 2. \((\alpha',\beta')=(\beta,-\alpha)\)

Case 1: \((\alpha',\beta')=(-\beta,\alpha)\), \([f]_{B}=\begin{pmatrix} \alpha &-\beta\\ \beta&\alpha \end{pmatrix}\), \(\chi_{f}(x)=x^{2}-2\alpha x+(\alpha^{2}+\beta^{2})=x^2-2\alpha x+1\)

When \(\alpha =\pm 1\), then \(\lambda =\pm 1\) with multiplicity respectively 2

If \(\alpha^2\neq 1\), there is no roots. Then \([f]_{\mathcal{B}}= \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\) \(f\) is a rotation of angle \(\theta\)

If \(\alpha = 1\Rightarrow \beta = 0\), \([f]_{\mathcal{B}}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\), then \(f=Id\) angle 0

If \(\alpha =-1,\beta = 0\), \([f]_{B}= \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\), then \(f=-Id\) angle \(\pi\)

Case 2: \((\alpha',\beta')=(\beta,-\alpha)\), \([f]_{B}= \begin{pmatrix} \alpha & \beta \\ \beta & -\alpha \end{pmatrix}\), \(\chi_{f}(x)=x^{2}-(\alpha^{2}+\beta^{2})=\left(x-1\right)\left(x+1\right)\)

Then \(\exists\) basis \(\mathcal{B}'\) s.t. \([f]_{\mathcal{B'}}= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\), then \(f\) is symmetric with respect to the eigenspace of eigenvalue \(1\)

Let \(\mathcal{B}=\{v_1,v_2\}\), let \(H=\lang v_1\rang\), \(f|_{H}=Id,f|_{H^\perp}=-Id\), then \(f\) is symmetry with respect to \(H\)

Example

  1. Find the symmetry \(f:\R^2\to \R^2\) with respect to the line \(L\) with equation \(x_{1}+x_{2}=0\)​

    \(L=\lang (1,-1)=w_1\rang,L^{\perp}=\lang(1,1)=w_2\rang\), let \(v_{1}=\frac{w_{1}}{||w_{1}||},v_2=\frac{w_2}{||w_2||}\), then \(\mathcal{B}=\{v_1,v_2\}\) O.N. basis

    Then \([f]_{\mathcal{B}}=\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\) since \(f(v_2)=-v_2\)​

    In general we get that \(f(x_{1},x_{2})=(-x_{2},-x_{1})\)​

  2. Find a rotation \(f:\R^2\to \R^2\) s.t. \(f(2,1)=(1,2)\)
    We verify that \(||(2,1)||=\sqrt 5=||(1,2)||\), then such \(f\) exists since if \(f\) is orthogonal, then \(||f(v)||=||v||\)
    We will consider an adequate O.N. basis \(B\) of \(\R^2\)
    For this, take \(v_{1}=\frac{(2,1)}{||(2,1)||}=(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})\), now we choose \(v_2:v_2\perp v_1\) and \(||v_2||=1\)
    ​\(v_{2}=(a_{1},a_{2}):0=\lang (a_{1},a_{2}),(2,1)\rang=2a_1+a_2\Rightarrow a_2=-2a_1\), then \(v_{2}=(\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}})\).
    Then \(\mathcal{B}=\{v_1,v_2\}\) is O.N. basis of \(\R^2\) and \(f(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})=(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}) =\frac{4}{5}v_1+\frac{3}{5}v_2\)
    ​\([f]_\mathcal{B}=\begin{pmatrix} \frac{4}{5} & -\frac{3}{5}\\ \frac{3}{5} & \frac{4}{5} \end{pmatrix}\) which is correct since this satisfy \([f]_{\mathcal{B}}= \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\) and \(f(v_{2})=-\frac{3}{5}v_1+\frac{4}{5}v_2\)​

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