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 $V$ is a finite-dimensional inner product space.
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, see Exercise 9.2.6 and Exercise 9.2.7. 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.