Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.3
Solution: Because $A=A^*$, we have\[\overline{g(X,Y)}=(Y^*AX)^*=X^*AY=g(Y,X).\]It is also clear that $g$ defines a form. Hence we only need to check that $g(X,X)>0$ if $X\ne 0$.
Let $X=(\alpha,\beta)$, then\begin{align*}g(X,X)&=(\bar\alpha,\bar \beta)\begin{pmatrix}1 & i\\ -i & 2\end{pmatrix}\begin{pmatrix}\alpha\\ \beta\end{pmatrix}\\&=\alpha\bar{\alpha}+2\beta\bar\beta-i(\alpha\bar\beta-\bar\alpha\beta)\\&=(\alpha-i\beta)\overline{(\alpha-i\beta)}+\beta\bar\beta\\&=|\alpha-i\beta|^2+|\beta|^2 > 0.\end{align*}Suppose $g(X,X)=0$, then $\alpha-i\beta=0$ and $\beta=0$. Hence $X=0$. Therefore, we are done.
