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7.1 Orthogonal Matrices: Class Notes

Definition

A square matrix $A$ is said to be orthogonal if its transpose is the same as its inverse, that is, if $A^{-1}=A^T$ or, equivalently, if $A{A}^T=A^{T}A=I$. A matrix transformation $T_A:R^n\to R^n$ is said to be an orthogonal transformation or an orthogonal operator if $A$ is an orthogonal matrix.

Consider the matrix in Exercise 6.

$$A=\left[\begin{array}{ccc}1/3& 2/3& 2/3\\ 2/3& -2/3& 1/3\\ -2/3& -1/3& 2/3\end{array}\right]$$

Multiply $A$ by its transpose. Each row of $A$ becomes the matching column of $A^T$, shown in the same color:

$$A{A}^T=\left[\begin{array}{ccc}1/3& 2/3& 2/3\\ 2/3& -2/3& 1/3\\ -2/3& -1/3& 2/3\end{array}\right]\left[\begin{array}{ccc}1/3& 2/3& -2/3\\ 2/3& -2/3& -1/3\\ 2/3& 1/3& 2/3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$

A row of $A$ times the matching column of $A^T$ (the same row again) gives each 1 on the diagonal, and every other product is 0. So $A{A}^T=I$, and $A$ is orthogonal.

Theorem 7.1.1

Tip: The rows (or columns) must be orthonormal, not just orthogonal.

Theorem 7.1.2

Proof of (d)

$A$ orthogonal $\Rightarrow A{A}^T=I$, so

$$\begin{array}{cc}\hfill {\displaystyle det(A{A}^T)=det(I)}& {\displaystyle \Rightarrow det(A)det(A^T)=1}\hfill \end{array}$$

Since $det(A)=det(A^T)$,

$$\begin{array}{cc}\hfill {\displaystyle {[det(A)]}^2=1}& {\displaystyle \Rightarrow det(A)=\pm 1}\hfill \end{array}$$

#8

Let $T_A:R^3\to R^3$ be multiplication by the orthogonal matrix in Exercise 6. Find $T_A(𝐱)$ for the vector $𝐱=(0,1,4)$, and confirm $\Vert T_A(𝐱)\Vert =\Vert 𝐱\Vert$ relative to the Euclidean inner product on $R^3$.

$$A=\left[\begin{array}{ccc}1/3& 2/3& 2/3\\ 2/3& -2/3& 1/3\\ -2/3& -1/3& 2/3\end{array}\right]$$

Solution

$$\Vert 𝐱\Vert =\sqrt{0^2+1^2+4^2}=\sqrt{17}$$

$$T_A(𝐱)=\left[\begin{array}{ccc}1/3& 2/3& 2/3\\ 2/3& -2/3& 1/3\\ -2/3& -1/3& 2/3\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 4\end{array}\right]=\left[\begin{array}{c}10/3\\ 2/3\\ 7/3\end{array}\right]$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \Vert T_A(𝐱)\Vert }& {\displaystyle =\frac{1}{3}\sqrt{100+4+49}=\frac{1}{3}\sqrt{153}=\frac{1}{3}\sqrt{9·17}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\sqrt{17}}\hfill \end{array}}$$

So $\Vert T_A(𝐱)\Vert$ matches $\Vert 𝐱\Vert =\sqrt{17}$. This is always the case if $A$ is orthogonal.

Theorem 7.1.3

Proof of (a) ⇒ (b)

$A$ orthogonal $\Rightarrow A{A}^T=A^{T}A=I$, so

$$\Vert A𝐱\Vert =(A𝐱·A𝐱{)}^{1/2}=({𝐱}^T\underset{I}{\underbrace{A^{T}A}}𝐱{)}^{1/2}=(𝐱·𝐱{)}^{1/2}=\Vert 𝐱\Vert$$

Example 4: Rotations in the xy-plane

Sometimes a problem is more easily approached in $R^2$ by considering a set of rotated axes, $x\prime y\prime$. Then a fixed point $(x,y)$ has coordinates $(x\prime ,y\prime )$ relative to the rotated axes.

We start with ${𝐮}_1$, ${𝐮}_2$ (in particular, ${𝐞}_1$, ${𝐞}_2$), unit vectors along the $x$- and $y$-axes, and consider ${𝐮}_1\prime$, ${𝐮}_2\prime$ along the $x\prime$- and $y\prime$-axes. We'll treat the rotation as a change from an old basis $B=\{{𝐮}_1,{𝐮}_2\}$ to $B\prime =\{{𝐮}_1\prime ,{𝐮}_2\prime \}$.

Unit circle with the $x\prime$- and $y\prime$-axes drawn horizontal and vertical. The $x$- and $y$-axes sit at angle $\theta$ clockwise from them, since the $x\prime y\prime$-axes are the $xy$-axes rotated counterclockwise through $\theta$. Unit vectors ${𝐮}_1$ and ${𝐮}_2$ point along the $x$- and $y$-axes, and ${𝐮}_1\prime$ and ${𝐮}_2\prime$ point along the $x\prime$- and $y\prime$-axes.

From 4.7,

$$\left[\begin{array}{c}x\prime \\ y\prime \end{array}\right]=P_{B\to B\prime }\left[\begin{array}{c}x\\ y\end{array}\right],\text{where}{P}_{B\to B\prime }=\left[\begin{array}{cc}\hfill {\displaystyle [{𝐮}_1{]}_{B\prime }}\hfill & \hfill {\displaystyle [{𝐮}_2{]}_{B\prime }}\hfill \end{array}\right]$$

Relative to $x\prime$ and $y\prime$,

$${𝐮}_1=(\cos \theta ,-\sin \theta ),{𝐮}_2=(\sin \theta ,\cos \theta )$$

so

$$[{𝐮}_1{]}_{B\prime }=\left[\begin{array}{c}\cos \theta \\ -\sin \theta \end{array}\right],[{𝐮}_2{]}_{B\prime }=\left[\begin{array}{c}\sin \theta \\ \cos \theta \end{array}\right]$$

So

$$\left[\begin{array}{c}x\prime \\ y\prime \end{array}\right]=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$$

$$\Rightarrow x\prime =x\cos \theta +y\sin \theta ,y\prime =-x\sin \theta +y\cos \theta$$

These are the rotation equations for $R^2$.

For $R^3$ (rotating about the $z$-axis):

$$P=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$$

This differs from the $R_{\theta }$ operator in 1.8 because $R_{\theta }$ rotates a vector to a new location. Here, the point is fixed; the axes are rotated.

Note that the rotation matrices are orthogonal matrices:

$$A^T=A^{-1}\Rightarrow {𝐜}_i·{𝐜}_j=\{\begin{array}{cc}{\displaystyle 1}\hfill & {\displaystyle \text{if }i=j}\hfill \\ {\displaystyle 0}\hfill & {\displaystyle \text{if }i\ne j}\hfill \end{array}$$

$$A{A}^T=I\Rightarrow {𝐫}_i·{𝐫}_j=\{\begin{array}{cc}{\displaystyle 1}\hfill & {\displaystyle \text{if }i=j}\hfill \\ {\displaystyle 0}\hfill & {\displaystyle \text{if }i\ne j}\hfill \end{array}$$

#14

Consider the rectangular $x\prime y\prime$-coordinate system obtained by rotating a rectangular $xy$-coordinate system counterclockwise through the angle $\theta =3\pi /4$.

a. Find the $x\prime y\prime$-coordinates of the point whose $xy$-coordinates are $(-2,6)$.

b. Find the $xy$-coordinates of the point whose $x\prime y\prime$-coordinates are $(5,2)$.

Solution

The $xy$-axes with the $x\prime$-axis (dashed) at angle $\theta =3\pi /4$ counterclockwise from the positive $x$-axis, pointing up and to the left, and the $y\prime$-axis (dashed) pointing down and to the left. The point $(-2,6)$ is in quadrant II of the $xy$-system.

With $\theta =3\pi /4$, $\cos \theta =-1/\sqrt{2}$ and $\sin \theta =1/\sqrt{2}$.

a.

$$\left[\begin{array}{c}x\prime \\ y\prime \end{array}\right]=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$$

$$\left[\begin{array}{c}x\prime \\ y\prime \end{array}\right]=\left[\begin{array}{cc}-1/\sqrt{2}& 1/\sqrt{2}\\ -1/\sqrt{2}& -1/\sqrt{2}\end{array}\right]\left[\begin{array}{c}-2\\ 6\end{array}\right]=\left[\begin{array}{c}4\sqrt{2}\\ -2\sqrt{2}\end{array}\right]$$

b. Solve for $x$ and $y$ with the inverse. The rotation matrix is orthogonal, so its inverse is its transpose: its first column becomes the first row and its second column becomes the second row.

$$\left[\begin{array}{c}x\\ y\end{array}\right]={\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]}^{-1}\left[\begin{array}{c}x\prime \\ y\prime \end{array}\right]=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}x\prime \\ y\prime \end{array}\right]$$

$$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}-1/\sqrt{2}& -1/\sqrt{2}\\ 1/\sqrt{2}& -1/\sqrt{2}\end{array}\right]\left[\begin{array}{c}5\\ 2\end{array}\right]=\left[\begin{array}{c}-7/\sqrt{2}\\ 3/\sqrt{2}\end{array}\right]$$

Theorem 7.1.5

Theorem 7.1.4

#23

The set ${\displaystyle S=\{\frac{1}{\sqrt{3}},\frac{1}{\sqrt{2}}x,\sqrt{\frac{3}{2}}{x}^2-\sqrt{\frac{2}{3}}\}}$ is an orthonormal basis for $P_2$ with respect to the evaluation inner product at the points $x_0=-1$, $x_1=0$, $x_2=1$. Let $𝐩=p(x)=1+x+x^2$ and $𝐪=q(x)=2x-x^2$.

a. Find $(𝐩{)}_S$ and $(𝐪{)}_S$.

b. Use Theorem 7.1.4 to compute $\Vert 𝐩\Vert$, $d(𝐩,𝐪)$, and $⟨𝐩,𝐪⟩$.

Solution

Call the basis vectors, in order, ${𝐩}_1$, ${𝐩}_2$, ${𝐩}_3$.

Part of (a) Since $S$ is orthonormal,

$$(𝐩{)}_S=(⟨𝐩,{𝐩}_1⟩,⟨𝐩,{𝐩}_2⟩,⟨𝐩,{𝐩}_3⟩)$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ⟨𝐩,{𝐩}_1⟩}& {\displaystyle =p(-1)p_1(-1)+p(0)p_1(0)+p(1)p_1(1)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(1)\left(\frac{1}{\sqrt{3}}\right)+(1)\left(\frac{1}{\sqrt{3}}\right)+(3)\left(\frac{1}{\sqrt{3}}\right)=\frac{5}{\sqrt{3}}}\hfill \end{array}}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ⟨𝐩,{𝐩}_2⟩}& {\displaystyle =p(-1)p_2(-1)+p(0)p_2(0)+p(1)p_2(1)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(1)(-\frac{1}{\sqrt{2}})+(1)(0)+(3)\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2}}\hfill \end{array}}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ⟨𝐩,{𝐩}_3⟩}& {\displaystyle =(1)(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}})+(1)(-\sqrt{\frac{2}{3}})+(3)(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}})}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{4\sqrt{3}}{\sqrt{2}}-\frac{5\sqrt{2}}{\sqrt{3}}=\frac{12-10}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\frac{\sqrt{2}}{\sqrt{3}}}\hfill \end{array}}$$

Part of (b)

$${\displaystyle \Vert 𝐩\Vert =\Vert (𝐩{)}_S\Vert =\sqrt{\frac{25}{3}+2+\frac{2}{3}}=\sqrt{11}}$$