Exercise 6.8.5

See the definition of the matrix representation.

1.

It’s a bilinear form since

$$H\left(\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \\ \hfill a_3\hfill \end{array}\right),\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill b_3\hfill \end{array}\right)\right)={\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \\ \hfill a_3\hfill \end{array}\right)}^t\left(\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right)\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill b_3\hfill \end{array}\right).$$

The matrix representation is

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

2.

It’s a bilinear form since

$$H\left(\left(\begin{array}{cc}\hfill a_1\hfill & \hfill a_2\hfill \\ \hfill a_3\hfill & \hfill a_4\hfill \end{array}\right),\left(\begin{array}{cc}\hfill b_1\hfill & \hfill b_2\hfill \\ \hfill b_3\hfill & \hfill b_4\hfill \end{array}\right)\right)={\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \\ \hfill a_3\hfill \\ \hfill a_4\hfill \end{array}\right)}^t\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill b_3\hfill \\ \hfill b_4\hfill \end{array}\right).$$

The matrix above is the matrix representation with respect to the standard basis.

3.

Let

$$f=a_1\cos <!--\; FUNCTION\; APPLICATION\; -->t+a_2\sin <!--\; FUNCTION\; APPLICATION\; -->t+a_3\cos <!--\; FUNCTION\; APPLICATION\; -->2t+a_4\sin <!--\; FUNCTION\; APPLICATION\; -->2t$$

and

$$g=b_1\cos <!--\; FUNCTION\; APPLICATION\; -->t+b_2\sin <!--\; FUNCTION\; APPLICATION\; -->t+b_3\cos <!--\; FUNCTION\; APPLICATION\; -->2t+b_4\sin <!--\; FUNCTION\; APPLICATION\; -->2t.$$

We compute that

$$H(f,g)=(a_2+2a_4)(-b_1-4b_3)$$

$$={\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \\ \hfill a_3\hfill \\ \hfill a_4\hfill \end{array}\right)}^t\left(\begin{array}{cccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill -4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill -4\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill b_3\hfill \\ \hfill b_4\hfill \end{array}\right).$$

Hence it’s a bilinear form. And the matrix is the matrix representation.