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6.9 Classification for n

Classification for \(n=3\)

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

Definition

Let \(f:\R^3\to \R^3\) be an orthogonal map and \(f\) is rotation if \(\det f=1\)

Let \(H\subseteq \R^3\) a subspace of dimension \(2\), \(f\) is symmetry with respect to \(H\) if \(f|_H=Id_H\) and \(f|_{H^{\perp}}=-Id_{H^\perp}\)

\(\chi_f(x)\in \R[x]\), \(\deg \chi_f(x)=3\), so \(\chi_f(x)\) has at least one root.

By previous lemma, we have that \(\lambda =\pm 1\)

  • Case 1: \(\lambda =1\) is an eigenvalue of \(f\)

Let \(v_1\) be an eigenvector with eigenvalue \(1\) and \(||v_1||=1\).
Let \(S=\lang v_1\rang\), consider \(S^{\perp}\), recall since \(S\) is \(f\)-invariant, then \(S^\perp\) is also \(f^*\)-invariant, then \(f\)-invariant since
We know \(\dim S^\perp = 2\), consider \(f|_{{S}^{\perp}}:S^\perp \to S^\perp\) is orthogonal, and \((S^\perp,\lang,\rang)\) is \(\R\)-IPVS
There are two possibilities \(\tilde B=\{v_{2},v_{3}\}\) O.N. basis of \(S^\perp\)

  • \([f]_{\tilde{\mathcal{B}}}=\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}\), then \([f]_{\mathcal{B}}=\)image and \(\det f =1\), then \(f\) is a rotation
  • \([f]_{\tilde{\mathcal{B}}}=\begin{pmatrix} 1&0\\0&1 \end{pmatrix}\), then \([f]_{\mathcal B}=\)image and \(\det f=-1\), then \(f\) is a symmetry with respect to \(H=\lang v_1,v_2\rang\)
  • Case 2: \(\lambda =1\) is not eigenvalue of \(f\) but \(\lambda =-1\) is an eigenvalue of \(f\)

Let \(v_1\) be an eigenvector of eigenvalue \(-1\) s.t. \(||v_1||=1\) and \(S=\lang v_1\rang\) and \(\dim S^{\perp} =2\)​

And \(S^\perp\) is \(f\)-invariant, \(f|_{S^\perp}:S^\perp \to S^\perp\) orthogonal map again, there are two cases for \(\tilde{B}=\{v_2,v_3\}\) be an O.N. basis of \(S^\perp\)

  • \([f_{S^\perp}]_{\tilde{\mathcal{B}}}=\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}\), then \([f]_{\mathcal{B}}=\)image\(=\)image​
  • \([f|_{S^\perp}]_{\tilde{\mathcal{B}}}=\begin{pmatrix} 1&\\&-1 \end{pmatrix}\), then \([f]_{\mathcal{B}}=\)image, this is impossible since we have eigenvalue \(1\)
Example

Define a rotation \(f:\R^3\to \R^3\) s.t. \(f(1,1,0)=(0,1,1)\) and the axis of the rotation is orthogonal to \((1,1,0)\) and \((0,1,1)\)

Let \(H=\lang (1,1,0),(0,1,1)\rang\) be subspace of \(\dim =2\), then \(H^{\perp}=\{(x_{1},x_{2},x_{3}):x_{1}+x_{2}=0=x_{2}+x_{3}\}=\lang \underbrace{(1,-1,1)}_{w_1}\rang\)

Then let \(v_{1}=\frac{w_{1}}{||w_{1}||}=(\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3})\) and construct O.N. basis of \(H\): \(v_{2}=\frac{(1,1,0)}{||(1,1,0)||}=(\frac{1}{\sqrt2},\frac{1}{\sqrt2},0)\) and \(v_{3}=\frac{(1,1,2)}{\sqrt{1+1+4}}=(-\frac{1}{\sqrt6},\frac{1}{\sqrt6},\frac{2}{\sqrt6})\)

\(f(v_1)=v_1\) since the axis of rotation is fixed since it is orthogonal to two vectors in \(\R^3\)

\(f(v_{2})=\frac{1}{2}v_{2}+\frac{\sqrt{3}}{2}v_{3}\), then \(\mathcal{B}=\{v_{1},v_{2},v_{3}\}\) is O.N. basis of \(\R^3\) and it contains an O.N. basis of \(H^\perp\) and an O.N. basis of \(H\)

Then \([f]_{\mathcal{B}}=\begin{pmatrix} 1&0&0\\ 0&\frac{1}{2}&-\frac{\sqrt3}{2}\\ 0&\frac{\sqrt3}{2}&\frac{1}{2} \end{pmatrix}\) and \(\cos\theta =\frac{1}{2}\), \(\sin\theta=\frac{\sqrt3}{2}\), then \(\theta =\frac{\pi}{3}\)

General Case

Let \((V,\lang,\rang)\) be \(\R\)-IPVS, finite dimension. \(f:V\to V\) be orthogonal map and \(\dim V=n\)

Theorem

In this situation, \(\exists\) an O.N. basis \(\mathcal{B}\) of \(V\) s.t.

\[ [f]_{\mathcal{B}}= \begin{pmatrix} (Id_{n_1\times n_1}) & & & & \\ & (-Id_{n_2\times n_2}) & & & \\ & & \begin{pmatrix}\cos\theta_{1}&-\sin\theta_{1}\\ \sin\theta_{1}&\cos\theta_{1}\end{pmatrix} & & \\ & & & \ddots & \\ & & & & \begin{pmatrix}\cos\theta_{r}&-\sin\theta_{r}\\ \sin\theta_{r}&\cos\theta_{r}\end{pmatrix} \end{pmatrix} \]

where \(n=n_1+n_2+2r\)

Proof

By induction on \(n\). We've already proved the case \(n=2\).

Let \(n>2\). Suppose that the result is true for all orthogonal linear maps in dimension that are strictly smaller than \(n\)

Let \(f:V\to V\) orthogonal, \(\dim V=n\). If \(\exists\) an eigenvalue \(\lambda\) of \(f\), then \(\lambda =1\) or \(\lambda = -1\)

  • Case 1: \(\lambda =1\) is an eigenvalue of \(f\)​

Let \(v_1\) be an eigenvector of eigenvalue \(1\) and \(||v_1||=1\). Let \(S=\lang v_1\rang\), \(S\) is \(f\)-invariant.

\(\dim S^{\perp}=n-1\geq 2\) and \(S^\perp\) is also \(f\)-invariant, let \(\tilde{f}=f|_{S^\perp}:S^\perp\to S^\perp\) is orthogonal and \(\dim S^\perp<n%\)​

By inductive hypothesis we get a basis \(\tilde{B}=\{v_2,...,v_n\}\) of \(S^\perp\)s.t. \([\tilde{f}]_{\tilde{\mathcal{B}}}= \begin{pmatrix} (Id_{n_1}) & & & & \\ & (-Id_{n_2}) & & & \\ & & \begin{pmatrix}\cos\theta_{1}&-\sin\theta_{1}\\ \sin\theta_{1}&\cos\theta_{1}\end{pmatrix} & & \\ & & & \ddots & \\ & & & & \begin{pmatrix}\cos\theta_{r}&-\sin\theta_{r}\\ \sin\theta_{r}&\cos\theta_{r}\end{pmatrix} \end{pmatrix}\)

Then take \(B=\{v_1,v_2,...,v_n\}\), we get \(\begin{pmatrix} \begin{array}{c|ccc}(1) & & & \\\hline & & & \\ & & [\tilde{f}]_{\tilde{\mathcal{B}}} & \\ & & & &\end{array} \end{pmatrix}\) * Case 2: \(\lambda =1\) is not eigenvalue of \(f\) but \(\lambda =-1\) is an eigenvalue of \(f\)​

Let \(v_1\) be an eigenvector of eigenvalue \(-1\) and \(||v_1||=1\). Let \(S=\lang v_1\rang\), \(S\) is \(f\)-invariant.

\(\dim S^{\perp}=n-1\geq 2\) and \(S^\perp\) is also \(f\)-invariant, let \(\tilde{f}=f|_{S^\perp}:S^\perp\to S^\perp\) is orthogonal and \(\dim S^\perp<n%\)

By inductive hypothesis we get a basis \(\tilde{B}=\{v_2,...,v_n\}\) of \(S^\perp\)s.t. \([\tilde{f}]_{\tilde{\mathcal{B}}}= \begin{pmatrix} (-Id_{n_2}) & & & \\ & \begin{pmatrix}\cos\theta_{1}&-\sin\theta_{1}\\ \sin\theta_{1}&\cos\theta_{1}\end{pmatrix} & & \\ & & \ddots & \\ & & & \begin{pmatrix}\cos\theta_{r}&-\sin\theta_{r}\\ \sin\theta_{r}&\cos\theta_{r}\end{pmatrix} \end{pmatrix}\) here \(n_1=0\) since \(1\) is not eigenvalue of \(f\)​

Then take \(B=\{v_1,v_2,...,v_n\}\), we get \(\begin{pmatrix} \begin{array}{c|ccc}(-1) & & & \\\hline & & & \\ & & [\tilde{f}]_{\tilde{\mathcal{B}}} & \\ & & & &\end{array} \end{pmatrix}\) * Case 3: Netiher \(1\) nor \(-1\) is an eigenvalue

In this case \(\chi_f(x)=P_1(x)\cdot P_2(x)\cdot ...\cdot P_r(x):P_i(x)\in\R[x]\) irreducible \(\deg P_i(x)=2,\forall i\)​

Then \(m_{f}(x)=Q_{1}(x)\cdot ..\cdot Q_{r}(x)\) where \(Q_i(x)\) is one of the \(P_i(x)\)'s and \(m_f(x)\mid \chi_f(x)\)​

Let's define \(t(x)=Q_2(x)\cdot Q_3(x)\cdot ...\cdot Q_r(x)=\frac{m_f(x)}{Q_1(x)}\), then \(m_f(x)\nmid t(x)\), then \(t(f)\neq 0\)​

Then \(\exists w\in V:t(f)(w)=v\neq 0_{V}\), let \(S=\lang v,f(v)\rang\) is subspace spanned by \(v\) and \(f(v)\)​

We know there is no eigenvalues, then \(\dim S=2\)​

\(v=t(f)(w)=Q_2(f)\circ ...\circ Q_r(f)(w)\) and \(\dim Q_{1}(x)=2\), \(Q_{1}(f)(v)=Q_1(f)\circ Q_2(f)\circ ...\circ Q_r(f)(w)=m_f(f)(w)=0\)
Then \(f^{2}(v)+af(v)+b=0\), then \(f^2(v)=-af(v)-b\)

Then \(f^2(v)\in S\), then \(S\) is \(f\)-invariant, then \(\dim S^\perp = n-2\), then \(S^\perp\) is \(f\)-invariant

Consider \(\tilde{f}=f|_{S^\perp}:S^\perp\to S^\perp\), by inductive hypothesis \(\exists\) a basis \(\tilde{B}\) of \(S^{\perp}:[\tilde{f}]_{\tilde{\mathcal{B}}}= \begin{pmatrix} \begin{pmatrix}\cos\theta_{2}&-\sin\theta_{2}\\ \sin\theta_{2}&\cos\theta_{2}\end{pmatrix} & & \\ & \ddots & \\ & & \begin{pmatrix}\cos\theta_{r}&-\sin\theta_{r}\\ \sin\theta_{r}&\cos\theta_{r}\end{pmatrix} \end{pmatrix}\)​

Let \(\{v_{1},v_{2}\}\) be an O.N. basis of \(S\) s.t. \([f|_{S}]_{\{v_1,v_2\}}= \begin{pmatrix} \cos\theta_{2} & -\sin\theta_{2} \\ \sin\theta_{2} & \cos\theta_{2} \end{pmatrix}\) take \(B=\{v_{1},v_{2}\}\cup\tilde{B}\)​