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 $0$?
 A: 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
$$\exists b_0\not =0 \mbox{  s.t.  } (\lambda I-f^*)(b_0) = 0$$
A: Suppose $f:\mathbb{R}^n \to \mathbb{R}^n$, you can get equivalence between eigenvalues and eigenvectors using the Rayleigh quotient.
Let 
$$R_f:\mathbb{R}^n \to \mathbb{R}:x \mapsto \frac{x\cdot f(x)}{x\cdot x},$$
let us denote by $\nabla g(x)$ the gradient of the function $g$, then
$$\nabla R_f(x) = 0 \iff (x\cdot x)\nabla(x\cdot f(x)) =(x\cdot f(x))\nabla (x\cdot x) \iff \nabla (x\cdot f(x))=R(x)x$$
Since $f$ is linear we assumed that the dimension is finite, there is some matrix $A$ such that $f(x)=Ax$, in particular this imply that $\nabla (f(x)\cdot x)= A^T$. It follows that $$\nabla R_f(x) = 0 \iff A^Tx =R(x)x,$$
this shows that $x$ is a eigenvector of $A$ associated to the eigenvalue $\lambda=R_f(x)$ if and only $\nabla R_f(x)=0$. Finally note that $\nabla (x\cdot f^*(x))= A$ and for every $x$,
$$R_f(x)=\frac{f(x)\cdot x}{x\cdot x}=\frac{x\cdot f^*(x)}{x\cdot x}=R_{f^*}(x),$$
i.e. $\nabla R_f(x)=0 \iff \nabla R_{f^*}(x)=0 $.
A: A different way to prove the claim is to prove the stronger statement 

$f$ and $f^*$ have the same characteristic polynomial

If $M$ is the matrix of $f$ is a given basis, then $M^T$ is the matrix of $f^*$.
Hence $\chi_f=\chi_M=\det(M-xI)=\det((M-xI)^T)=\det(M^T-xI)=\chi_{M^T}=\chi_{f^*}$
