Exercise 6.7.3

Do the same to $L_A$ as that in Exercise 6.7.2. But the $\alpha$ here must be the standard basis. And the matrix consisting of column vectors $\beta$ and $\gamma$ is $V$ and $U$ respectly.

1.

We have $A^{\text{∗}}A=\left(\begin{array}{cc}\hfill 3\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 3\hfill \end{array}\right)$. So its eigenvalue is $6,0$ with eigenvectors

$$\beta =\left\{\frac{1}{\sqrt{2}}(1,1),\frac{1}{\sqrt{2}}(1,-1)\right\}.$$

Extend $L_A(\beta )$ to be an orthonormal basis

$$\gamma =\left\{\frac{1}{\sqrt{3}}(1,1,-1),\frac{1}{\sqrt{2}}(1,-1,0),\frac{1}{\sqrt{6}}(1,1,2)\right\}.$$

So we know that

$$U=\left(\begin{array}{ccc}\hfill \frac{1}{\sqrt{3}}\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill \frac{1}{\sqrt{6}}\hfill \\ \hfill \frac{1}{\sqrt{3}}\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill & \hfill \frac{1}{\sqrt{6}}\hfill \\ \hfill -\frac{1}{\sqrt{3}}\hfill & \hfill 0\hfill & \hfill \frac{2}{\sqrt{6}}\hfill \end{array}\right),\mathrm{\Sigma }=\left(\begin{array}{cc}\hfill \sqrt{6}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),V=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right).$$

2.

We have

$$U=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right),\mathrm{\Sigma }=\left(\begin{array}{ccc}\hfill \sqrt{2}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \sqrt{2}\hfill & \hfill 0\hfill \end{array}\right),V=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right).$$

3.

We have

$$U=\frac{1}{\sqrt{2}}\left(\begin{array}{cccc}\hfill \frac{2}{\sqrt{10}}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -\frac{3}{\sqrt{15}}\hfill \\ \hfill \frac{1}{\sqrt{10}}\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill & \hfill \frac{1}{\sqrt{3}}\hfill & \hfill \frac{1}{\sqrt{15}}\hfill \\ \hfill \frac{1}{\sqrt{10}}\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill \frac{1}{\sqrt{3}}\hfill & \hfill \frac{1}{\sqrt{15}}\hfill \\ \hfill \frac{2}{\sqrt{10}}\hfill & \hfill 0\hfill & \hfill -\frac{1}{\sqrt{3}}\hfill & \hfill \frac{2}{\sqrt{15}}\hfill \end{array}\right),\mathrm{\Sigma }=\left(\begin{array}{cc}\hfill \sqrt{5}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),V=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right).$$

4.

We have

$$U=\left(\begin{array}{ccc}\hfill \frac{1}{\sqrt{3}}\hfill & \hfill \frac{\sqrt{2}}{\sqrt{3}}\hfill & \hfill 0\hfill \\ \hfill \frac{1}{\sqrt{3}}\hfill & \hfill -\frac{1}{\sqrt{6}}\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill \\ \hfill \frac{1}{\sqrt{3}}\hfill & \hfill -\frac{1}{\sqrt{6}}\hfill & \hfill \frac{1}{\sqrt{2}}\hfill \end{array}\right),\mathrm{\Sigma }=\left(\begin{array}{ccc}\hfill \sqrt{3}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \sqrt{3}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right),V=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill \frac{1}{\sqrt{2}}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill \end{array}\right).$$

5.

We have

$$U=\frac{1}{2}\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill -1+i\hfill \\ \hfill 1-i\hfill & \hfill 1+i\hfill \end{array}\right),\mathrm{\Sigma }=\left(\begin{array}{cc}\hfill \sqrt{6}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),V=\left(\begin{array}{cc}\hfill \frac{\sqrt{2}}{\sqrt{3}}\hfill & \hfill \frac{1}{\sqrt{3}}\hfill \\ \hfill \frac{i+1}{\sqrt{6}}\hfill & \hfill \frac{-i-1}{\sqrt{3}}\hfill \end{array}\right).$$

6.

We have

$$U=\left(\begin{array}{ccc}\hfill \frac{1}{\sqrt{3}}\hfill & \hfill \frac{1}{\sqrt{6}}\hfill & \hfill \frac{1}{\sqrt{2}}\hfill \\ \hfill \frac{1}{\sqrt{3}}\hfill & \hfill -\frac{2}{\sqrt{6}}\hfill & \hfill 0\hfill \\ \hfill \frac{1}{\sqrt{3}}\hfill & \hfill \frac{1}{\sqrt{6}}\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill \end{array}\right),\mathrm{\Sigma }=\left(\begin{array}{cccc}\hfill \sqrt{6}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \sqrt{6}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \sqrt{2}\hfill & \hfill 0\hfill \end{array}\right),V=\left(\begin{array}{cccc}\hfill \frac{1}{\sqrt{2}}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{1}{\sqrt{2}}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill \end{array}\right).$$