## 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…

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…

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…

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…

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…