#### Exercise 9.2.1

Solution:

(a) No. Since $f(0,\beta)\ne 0$.

(b) No. Since $f((0,0),(1,0))\ne 0$.

(c) Yes. Since $f(\alpha,\beta)=4x_1\bar y_1$.

(d) No. Because of $\bar x_2$ there, it is not linear on $\alpha$.

#### Exercise 9.2.2

Solution: It is not a form! I think it should be $$f((x_1,y_1),(x_2,y_2))=x_1x_2+y_1y_2.$$The matrices are\[\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix},\begin{pmatrix}2 & 0\\ 0& 2\end{pmatrix},\begin{pmatrix}5 & 11\\ 11 & 25\end{pmatrix},\]respectively.

#### 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.

#### Exercise 9.2.4

Solution: Since $f$ is a form, we have $f$ is linear on $\alpha$. Since $f(\alpha,\beta)=f(\beta,\alpha)$, we also have $f$ is linear on $\beta$. Therefore, $f$ is a form which is also a bilinear form. Then we have\[-if(\alpha,\beta)=f(\alpha,i\beta)=f(i\beta,\alpha)=if(\beta,\alpha)=if(\alpha,\beta).\]Thus $f(\alpha,\beta)=0$. That is $f=0$.

#### Exercise 9.2.5

Solution: The matrix of $f$ in the standard basis is given by $\begin{pmatrix}1 & 2\\2& 4\end{pmatrix}$. Its eigenvalues are zero and 5. The corresponding eigenvectors are $(2,-1)$ and $(1,2)$. Now it is easy to check that the matrix of $f$ in the basis $(2,-1)$ and $(1,2)$ is diagonal $\mathrm{diag}(0,25)$.

#### Exercise 9.2.6

Solution: If $f$ is non-degenerate, we show that $T_f$ is non-singular. Suppose $T_f\alpha=0$ for some $\alpha\in V$, then we have $$f(\alpha,\beta)=(T\alpha|\beta)=0$$ for all $\beta$. Since $f$ is non-degenerate, we must have $\alpha=0$. Hence $T_f$ is non-singular.

Conversely, suppose $f(\alpha,\beta)=0$ for all $\beta$, then we have $$f(\alpha,\beta)=(T_f\alpha|\beta)=0,$$for all $\beta$. In particular, we have $0=(T_f\alpha|T_f\alpha)$. Hence $T_f\alpha=0$. But $T_f$ is non-singular, therefore $\alpha=0$. Thus $f$ is non-degenerate.

#### Exercise 9.2.7

Solution:

Call the form $f$ **right non-degenerate** if $0$ is the only vector $\alpha$ such that $f(\beta,\alpha)=0$ for all $\beta$.

Suppose $f$ is left non-degenerate, we show that $f$ is right non-degenerate. Suppose $f(\beta,\alpha)=0$ for all $\beta$. Hence $(T_f\beta|\alpha)=0$ for all $\beta$. Because $f$ is left non-degenerate, $T_f$ is non-singular by Exercise 9.2.6. Since $V$ is finite-dimensional, we have $T_f$ is invertible. Hence there exists $\beta$ such that $T_f\beta=\alpha$. Therefore, we conclude that $$0=(T_f\beta|\alpha)=(\alpha|\alpha).$$Thus $\alpha=0$, namely $f$ is right non-degenerate.

Similar to Exercise 9.2.6, one can show that $f$ is right non-generate if and only if $T_f^*$ is non-singular. Then repeat our argument, we can show the other direction.

#### Exercise 9.2.8

Solution: By Theorem 6 of page 291, there exists a vector $\beta_0$ such that $L(\alpha)=(\alpha|\beta_0)$. On the other hand, we also have $$f(\alpha,\beta)=(T_f\alpha|\beta)=(\alpha|T_f^*\beta).$$Since $f$ is non-degenerate, $T_f$ is non-singular. Because $V$ is finite-dimensional, we conclude that $T_f$ is invertible. So is $T_f^*$. Therefore, there exists a unique $\beta\in V$ such that $T_f^*\beta=\beta_0$. We have\[L(\alpha)=(\alpha|\beta_0)=(\alpha|T_f^*\beta)=f(\alpha,\beta).\]We showed the existence.

Suppose $\beta_1$ and $\beta_2$ are vectors such that $$L(\alpha)=f(\alpha,\beta_1)=f(\alpha,\beta_2).$$Then $f(\alpha,\beta_1-\beta_2)=0$ for all $\alpha$. Since $f$ is non-degenerate, we have $\beta_1-\beta_2=0$, which shows the uniqueness.

#### Exercise 9.2.9

Solution: As we have seen in previous exercises, $T_f^*$ is invertible. We have\[f(S\alpha,\beta)=(T_fS\alpha|\beta)=(\alpha|S^*T_f^*\beta),\]\[f(\alpha,S'\beta)=(T_f\alpha|S'\beta)=(\alpha|T_f^*S'\beta).\]Hence it suffices to choose some $S'$ such that $S^*T_f^*=T_f^*S'$, that is $S'=(T_f^*)^{-1}S^*T_f^*$.

Indeed, such $S'$ is unique! Try to prove it.