Non-degenerate form induces adjoint linear operators
Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.9 Solution: We can always define an inner product on a finite-dimensional vector space $V$. Hence we shall assume that…
Relation between non-degenerate forms and linear functionals
Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.8 Solution: We can always define an inner product on a finite-dimensional vector space $V$. Hence we shall assume that…
Form is left non-degenerate if and only if it is right non-degenerate
Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 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…
Form is non-degenerate if and only if the associated linear operator is non-singular
Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 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…
Diagonalize a symmetric matrix associated to a form
Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 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…
Symmetric sesqui-linear form over $\mathbb C$ is zero
Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 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 $f$…
Check if a form is an inner product
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…
Find the matrix of a form with respect to a basis
Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.2 Solution: It is not a form! I think it should be $$f((x_1,y_1),(x_2,y_2))=x_1x_2+y_1y_2.$$Then the matrices are\[\begin{pmatrix}1 & 0\\ 0 &…
Verify if they are sesqui-linear forms
Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 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…