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Solution to Linear Algebra Hoffman & Kunze Chapter 9.2

Exercise 9.2.1


(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


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.


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

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