Exercise 6.8.17

Use the formula given in the previous exercise to find $H$. To diagonalize it, we may use the method in the paragraph after Theorem 6.35. Here the notation $\alpha$ is the standard basis in the corresponding vector spaces.

1.

We have

$$H\left(\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \end{array}\right),\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \end{array}\right)\right)=\frac{1}{2}[K\left(\begin{array}{c}\hfill a_1+b_1\hfill \\ \hfill a_2+b_2\hfill \end{array}\right)-K\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \end{array}\right)-K\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \end{array}\right)]$$

$$=-2a_1{b}_1+2a_1{b}_2+2a_2{b}_1+a_2{b}_2.$$

Also, we have ${\varphi }_{\alpha }=\left(\begin{array}{cc}\hfill -2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right)$. Hence we know that

$$\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill -2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill -2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 3\hfill \end{array}\right).$$

So the basis $\beta$ could be

$$\left\{(1,0),(1,1)\right\}.$$

2.

We have

$$H\left(\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \end{array}\right),\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \end{array}\right)\right)=7a_1{b}_1-4a_1{b}_2-4a_2{b}_1+a_2{b}_2$$

and

$$\beta =\left\{(1,0),(\frac{4}{7},1)\right\}.$$

3.

We have

$$H\left(\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \\ \hfill a_3\hfill \end{array}\right),\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill b_3\hfill \end{array}\right)\right)=3a_1{b}_1+3a_2{b}_2+3a_3{b}_3-a_1{b}_3-a_3{b}_1.$$

and

$$\beta =\left\{(1,0,0),(0,1,0),(\frac{1}{3},0,1)\right\}.$$