Introduction E9F5FC

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Orthogonal matrices are defined by {$QQ^T=I$}

What form do they take?

{$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$}

{$a_{11}^2+a_{12}^2=1$}

{$a_{11}a_{21} + a_{12}a_{22}=0$}

{$a_{21}^2+a_{22}^2=1$}

Solve

{$a_{21}=\cos\theta,a_{22}=\sin\theta$}

{$a_{11}=-\frac{a_{12}a_{22}}{a_{21}}$}

{$\frac{a_{12}^2a_{22}^2}{a_{21}^2}+a_{12}^2=1$}

{$a_{12}^2[\frac{a_{22}^2}{a_{11}^2}+1]=1$}

{$a_{22}^2[\frac{\sin^2\theta}{\cos^2\theta}+1]=a_{22}^2[\frac{\sin^2\theta+cos^2\theta}{\cos^2\theta}]=a_{22}^2\frac{1}{\cos^2\theta}=1$}

{$a_{22}^2=\cos^2\theta, a_{22}=\pm\cos\theta$}

{$\cos\theta=-\frac{\sin\theta \pm\cos\theta}{a_{21}}$}

{$a_{21}=\mp\sin\theta$}

{$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$} and {$\begin{pmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$}

Thus we have two kinds of solutions:

{$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} u \\ v \end{pmatrix}=\begin{pmatrix} u\cos\theta - v\sin\theta \\ u\sin\theta + v\cos\theta \end{pmatrix}$} rotation

{$\begin{pmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} u \\ v \end{pmatrix}=\begin{pmatrix} -u\cos\theta + v\sin\theta \\ u\sin\theta + v\cos\theta \end{pmatrix}$} rotation and reflection of x-coordinate across y-axis

We can analyze this by noting that these maps are norm-preserving. For example:

{$(-u\cos\theta + v\sin\theta)^2+(u\sin\theta + v\cos\theta)^2 = u^2(\cos^2\theta + \sin^2\theta) + 2uv(-\cos\theta\sin\theta + \cos\theta\sin\theta) + u^2(\cos^2\theta + \sin^2\theta) = u^2+v^2$}.

Note that they differ in their determinants

{$\cos^2\theta -(-\sin^2\theta)=1$} and {$-\cos^2\theta-\sin^2\theta=-1$}

And then consider what happens to the unit vectors

{$\begin{pmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} -\cos\theta \\ \sin\theta \end{pmatrix}$} This is a rotation with a reflection of the x-coordinate across the y-axis

{$\begin{pmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} \sin\theta \\ \cos\theta \end{pmatrix}$} This is a rotation clockwise, not counterclockwise. Thus it is a rotation counterclockwise that has then been reflected across the y-axis.

Eigenvectors and their geometric interpretation