Show that $f$ and $f^*$ have the same eigenvalues. 
Show that the linear transformations $f$ and $f^*$ have the same eigenvalues.

Let $\vec a$ be an eigenvector of $f$ with eigenvalue $\lambda$.  Then $\lambda \vec a \cdot \vec b= f(\vec a) \cdot \vec b = \vec a \cdot f^*(\vec b) \iff \vec a \cdot (\lambda I - f^*)(\vec b)=0$, where $I$ is the identity function.  Thus $\vec a \bot (\lambda I - f^*)(\vec b)$ and $\vec a \neq 0$ because $\vec a$ is an eigenvector.  How do I know that this implies that $(\lambda I - f^*)(\vec b)=0$, as would be required to finish this proof?  Couldn't it just be that $(\lambda I - f^*)(\vec b)$ is orthogonal to $\vec a$ without being $0$?

As $b$ is arbitrary, the condition
$$a\bot (\lambda I-f^*(b))$$ 
implies that $a\not \in (\lambda I-f^*)(\mathbb R^n)$ thus $\lambda I-f^*$ isnt surjective and hence,due to Rank–nullity theorem,it isnt inyective, so
$$\exists b_0\not =0 \mbox{  s.t.  } (\lambda I-f^*)(b_0) = 0$$
