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6.12 Classification of orthogonal Linear Operators

Classification of orthogonal Linear Operators

Let \(V\) be a real inner product space.

Lemma

\(f\) is an orthogonal map and let \(\lambda\) be an eigenvalue in \(\mathbb{R}\) of \(f\), then \(\lambda = \pm 1\).

\(f\) is an orthogonal map and let \(S \subseteq V\) be an \(f\)-invariant subspace. Then \(S^\perp \subseteq V\) is \(f\)-invariant.

Definition

If \(n=2\):

\(V = \mathbb{R}^2\) and \(f: \mathbb{R}^2 \to \mathbb{R}^2\) is an orthogonal operator.

  • \(f\) is a rotation if \(\det f = 1\)​
  • Let \(H \subseteq \mathbb{R}^2\) be a subspace of \(\dim 1\).

\(f\) is reflection (symmetry) with respect to \(H\) if \(f|_H = id_H\) and \(f|_{H^\perp} = -id_{H^\perp}\)

  1. \([f]_B = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\) is a rotation. (\(B\) is orthonormal).

  2. \([f]_B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\) is a symmetry respect to \(\langle v_1 \rangle\) where \(B = \{v_1, v_2\}\)

Definition

Let \(V\) be a real inner product space. Two bases \(B\) and \(B'\) are said to have the same orientation if \(\det(C(B, B')) > 0\).

Exercise

  1. Consider \(\mathbb{R}^2\) with inner product \(\langle -, - \rangle\), and let \(B = \{v_1, v_2\}\) be an oriented basis for \(\mathbb{R}^2\) s.t. \(||v_1|| = 1\) and \(||v_2|| = \sqrt{2}\).
    Find all orthogonal operators \(f\) on \(\mathbb{R}^2\) s.t. \(f(v_1) = \frac{7}{5}v_1 + \frac{4}{5}v_2\).

    Answer
    Since \(f\) is orthogonal, then \(\lang u_1,u_2\rang =\lang f(u_1),f(u_2)\rang\)
    Then \([\lang-,-\rang]_\mathcal{B}=[f^t]_\mathcal{B}[\lang -,-\rang]_\mathcal{B}[f]_\mathcal{B}\) where \(([\lang-,-\rang ]_\mathcal{B})_{ij}=\lang v_j,v_i\rang\)
    ​\([\lang-,-\rang]_\mathcal{B}=\begin{pmatrix} 1&b\\b&2 \end{pmatrix}\) and \([f]_{\mathcal{B}}= \begin{pmatrix} \frac{7}{5} & \alpha \\ \frac{4}{5} & \beta \end{pmatrix}\)
    Also, \(1=||v_{1}||^{2}=||f(v_{1})||^{2}=||\frac{7}{5}v_{1}+\frac{4}{5}v_{2}||^{2}\)
    Then \(\left(\frac{7}{5},\frac{4}{5}\right)\begin{pmatrix} 1&b\\b&2 \end{pmatrix}\begin{pmatrix} \frac{7}{5}\\ \frac{4}{5} \end{pmatrix}=\left(\frac{7}{5}+\frac{4}{5}b,\frac{7}{5}b+\frac{8}{5}\right)\begin{pmatrix} \frac{7}{5}\\ \frac{4}{5} \end{pmatrix}=\frac{81+56b}{25}=1\) with \(b=-1\)​

    \(\begin{pmatrix} 1&-1\\ -1&2 \end{pmatrix}=\begin{pmatrix} \frac{7}{5}&\frac{4}{5}\\ \alpha&\beta \end{pmatrix}\begin{pmatrix} 1&-1\\ -1&2 \end{pmatrix}\begin{pmatrix} \frac{7}{5}&\alpha\\ \frac{4}{5}&\beta \end{pmatrix}=\begin{pmatrix} 1&\frac{3\alpha+\beta}{5}\\ \frac{3\alpha+\beta}{5}&\alpha^2+2\beta^2-2\alpha\beta \end{pmatrix}\)
    Then \(\begin{cases} 3\alpha+\beta=-5 \\ \alpha^{2}+2\beta^{2}-2\alpha\beta=2 \end{cases}\Rightarrow \begin{cases} \alpha=-\frac{8}{5}\\ \beta=-\frac{1}{5} \end{cases}\text{ or }\begin{cases} \alpha=-\frac{6}{5}\\\beta=-\frac{7}{5} \end{cases}\)

    • If \([f]_{\mathcal{B}}= \begin{pmatrix} \frac{7}{5} & -\frac{8}{5} \\ \frac{4}{5} & -\frac{1}{5} \end{pmatrix}\), then \(\det f=1\). Thus \(f\) is a rotation
      Also we know if \([f]_\mathcal{B}\sim [f]_\mathcal{B'}\), then \(\det([f]_\mathcal{B})=\det([f]_\mathcal{B'}),\text{tr}([f]_\mathcal{B})=\text{tr}([f]_\mathcal{B'})\)
      Then \(\text{tr}([f]_{\mathcal{B}})=\frac{6}{5}=2\cos\theta\Rightarrow\cos\theta =\frac{3}{5}\).
      Consider \(\mathcal{B}'=\{v_1,f(v_1)\}\) and \(\mathcal{B}=\{v_1,v_2\}\), then \(\det(C[\mathcal{B}',\mathcal{B}])=\det\begin{pmatrix} 1&\frac{7}{5}\\ 0&\frac{4}{5} \end{pmatrix}=\frac{4}{5}>0\)
      Then \(\theta=\arccos\frac{3}{5}\), then \(\exists O.N.\mathcal{B}''\) s.t. \([f]_\mathcal{B''}=\begin{pmatrix} \frac{3}{5}&-\frac{4}{5}\\ \frac{4}{5}&\frac{3}{5} \end{pmatrix}\)
    • If \([f]_{\mathcal{B}}= \begin{pmatrix} \frac{7}{5} & -\frac{6}{5} \\ \frac{4}{5} & -\frac{7}{5} \end{pmatrix}\), then \(\det f=-1\). Then \(f\) is symmetry respect to \(E_1=\ker (f-Id)\)
      After calculating, \(E_1=\lang(3,1)_\mathcal{B}\rang=\lang 3v_1+v_2\rang\)

Case \(n=3\):
Def: If \(V = \mathbb{R}^3\) with \(\left\langle\cdot, \cdot\right\rangle\), and \(f\) is an orthogonal op. on \(V\). We say

  • \(f\) is a rotation if \(\det f = 1\)​
  • If \(H \subseteq V\) is a subspace of \(\dim 2\), \(f\) is a symmetry respect to \(H\) if \(f|_H = \mathrm{id}_H\) and \(f|_{H^\perp} = -\mathrm{id}_{H^\perp}\).

If \(\lambda = 1\): \(B = \{v_1, v_2, v_3\}\) an O.N. basis for \(V\).

  • \([f]_{B} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{bmatrix}\)​\(\implies \det(f)_{B} = 1\), then \(\operatorname{tr}(f)_{B} = 1 + 2\cos\theta\).

And \(\theta \in [0, \pi]\), \(V = \mathbb{R}^3\), \(f\) is a rotation, \(E_1\) is the rotation axis. * \([f]_{B} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}\), \(\begin{cases} \det f = -1 \\ \operatorname{tr} f = 1 \end{cases}\). WE DON'T NEED O.N. BASIS
\(V = \mathbb{R}^3\), \(f\) is a symmetry with respect to \(H =\text{ span}\{v_{1}, v_{2}\}\).

\(\lambda = -1: 1 \text{ is not a root, } \exists B \text{ an O.N basis s.t. } f(v_1) = -v_1\)
\([f]_B = \begin{pmatrix} -1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix} \quad \theta \in (0, \pi]\) \(= \underbrace{\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}}_{\text{symmetry}} \underbrace{\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix}}_{\text{rotation}}\)

\(f\) is a comp. of a rotation with a symmetry.

Ex 2: Let \(V\) be a RIPS. Let \(B=\{v_1, v_2, v_3\}\) be an ordered basis that defines the positive orientation.
If \([\langle -,-\rangle]_B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix}\), find \([f]_B\) s.t. \(f\) is a rotation with \(\alpha = \pi/3\) respect to \((1,0,0)_B\).

Answer

Let \(B'=\{w_1, w_2, w_3\}\) O.N. s.t. \(w_1=(1,0,0)_B\).
Consider \(\widetilde{w}_2=(x,y,z)_B\). And \(0 = \langle w_{1}, \widetilde{w}_{2} \rangle = (x,y,z)_{B} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{pmatrix}_{B} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}_{B} = x\).
\(\implies x=0\)

If \(\widetilde{w}_2=(0,1,0)_B\), let \(\widetilde{w}_3=(x,y,z)_B\) s.t. \(x=0\). \(\widetilde{w}_3=(0,y,z)\).

\(0 = \langle \widetilde{w}_2, \widetilde{w}_3 \rangle = (0,1,0)_B \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{pmatrix}_B \begin{pmatrix} 0 \\ y \\ z \end{pmatrix}_B = 2y+z\)​\(\implies z=-2y\).
Let \(\widetilde{w}_3=(0,1,-2)_B\).

Let \(\tilde{B}' = \{(1,0,0)_B, (0,1,0)_B, (0,-1,2)_B\}\), then \(\tilde{B}'\) is an Orthogonal basis for V.
\(\det(C(B',B)) = \det \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{bmatrix} = 2 > 0\). \(\checkmark\)

\(||(0,1,0)||^{2} = \langle (0,1,0), (0,1,0) \rangle = (0,1,0) \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = (0,2,1) \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = 2\).

Define \(B^{\prime}=\left\{\frac{(1,0,0)_{B}}{||(1,0,0)_{B}||},\frac{(0,1,0)_{B}}{||(0,1,0)_{B}||} ,\frac{(0,-1,2)_{B}}{||(0,-1,2)_{B}||}\right\}=\left\{(1,0,0)_{B},(0,1/\sqrt{2},0 )_{B},(0,-1/\sqrt{2},2/\sqrt{2})_{B}\right\}\).

Therefore, \([f]_{B^{\prime}}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\pi/3 & -\sin\pi/3 \\ 0 & \sin\pi/3 & \cos\pi/3 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/2 & -\sqrt{3}/2 \\ 0 & \sqrt{3}/2 & 1/2 \end{bmatrix}\).

\([f]_{B} = \left( C(B', B) [f]_{B'}C(B, B') \right) \\= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} \\ 0 & 0 & 2/\sqrt{2} \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/2 & -\sqrt{3}/2 \\ 0 & \sqrt{3}/2 & 1/2 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} \\ 0 & 0 & 2/\sqrt{2} \end{bmatrix}^{-1}\\= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1-\sqrt{3}}{2} & -\sqrt{3}/2 \\ 0 & \sqrt{3} & \frac{1+\sqrt{3}}{2} \end{bmatrix}\)

Consider \(\mathbb{R}^3\) with the standard inner product.
Find \([f]_E\) if \(f\) is a symmetry respect \(H=\text{span}\{(1,1,1), (2,0,1)\}\).

\(f: \mathbb{R}^3 \to \mathbb{R}^3\) s.t. \(f|_H = id_H\) and \(f|_{H^\perp} = -id_{H^\perp}\)

\(H^\perp = \{(x,y,z) \in \mathbb{R}^3 | (x,y,z) \cdot h = 0 \forall h \in H\}\)
Thus \(\begin{cases} x+y+z=0 \\ 2x+z=0 \end{cases} \implies \begin{cases} y=x \\ z=-2x \end{cases}\). Then \(H^\perp = \text{span}\{(1,1,-2)\}\).

\(\text{Let }B = \{(1,1,1), (2,0,1), (1,1,-2)\}\) and \([f]_{B} = \begin{bmatrix} 1&& \\ &1 &\\ &&-1 \end{bmatrix}\)
\([f]_E = C(B,E) [f]_B C(E,B)\)
\(= \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & -2 \end{bmatrix}^{-1} = \begin{bmatrix} 2/3 & -1/3 & 2/3 \\ -1/3 & 2/3 & 2/3 \\ 2/3 & 2/3 & -1/3 \end{bmatrix}\)

Ex b) Find an orthogonal op. s.t. \([f]_B = \begin{bmatrix} 1 \\ & -1 \\ & & -1 \end{bmatrix}\) for some basis B containing \((1,1,1)\) s.t. \(f(1,1,1) = (1,1,1)\).

Ex 4) Consider \([f]_E = \begin{bmatrix} 1/\sqrt{2} & -\sqrt{2}/2 & -1/2 \\ -\sqrt{2}/2 & 0 & -\sqrt{2}/2 \\ -1/2 & -\sqrt{2}/2 & 1/2 \end{bmatrix}\). Is f a rotation, a symmetry or a comp. of both?