Exercise 6.8.6

Answers

We have

H ((\begin{matrix} a_{1} \\ a_{2} \end{matrix}), (\begin{matrix} b_{1} \\ b_{2} \end{matrix})) = {(\begin{matrix} a_{1} \\ a_{2} \end{matrix})}^{t} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} b_{1} \\ b_{2} \end{matrix}) .

Hence we find the matrix $A$ . And the form $x^{t} Ay$ is a bilinear form.

jephianlin

2011-06-27 00:00