What's the role of the inner product here Consider the following statement:
If $V$ is a finite dimensional vector space over $\mathbb R$ and $T:V \to V$ is linear then there is a basis for $V$ of eigen vectors of $T$ if and only if there is an inner product on $V$ with respect to which $T$ is self adjoint.
My thoughts on this are: If $T=T^\ast$ for some inner product on $V$ then by the spectral theorem it follows that there is an eigen basis for $V$. In the other direction: let there be an eigen basis of $V$. Then with respect to this basis $T$ has a diagonal matrix $M(T)$. Then $T=M(T)=(M(T))^\ast=T^\ast$. But where is the inner product  used here? 
 A: The first thing here was explained in the comments: $T^*$ is defined via inner product as a matrix such that $\langle Tx, y \rangle = \langle x, T^*y \rangle$ for all $x,y \in V$.
However, I wanted to add an answer to comment on this:

If $T = T^*$ for some inner product on $V$ then by the spectral theorem it follows that there is an eigen basis for $V$.

You're missing the key part here: an orthogonal basis of the eigenvectors of $T$. How can we define orthogonality, if we cannot say $\langle v_i, v_j \rangle = 0$ for $i \ne j$?
A: One direction, as you already know, is just part of the spectral theorem. So let us turn to the other direction, that is, let us take a diagonalizable linear operator $T$ and let us find an inner product $\langle, \rangle$ that makes $T$ self-adjoint. 

To do this let us recall that to assign an inner product on $V$ it suffices to take a basis $\{e_1\ldots e_n\}$ and declare it to be orthonormal: 
  $$\tag{1} \langle e_i, e_j\rangle=\begin{cases} 1 & i=j \\ 0 & i\ne j\end{cases}.$$
  The relation (1) completely determines the scalar product $\langle, \rangle$ because if $v,\, w $ are vectors then they are uniquely written as
  \begin{align}
v&= v_1e_1+\cdots +v_n e_n \\
w&= w_1 e_1 +\cdots + w_n e_n,
\end{align}
  from which it follows that 
  $$ 
\langle v, w \rangle= \sum_{i,j=1}^n v_i w_j \langle e_i, e_j\rangle =\sum_{i=1}^n v_i w_i.$$

This said, let $\{e_1\ldots e_n\}$ be a basis of $V$ which diagonalizes $T$. We need to construct an inner product on $V$. What is the first thing that comes into mind?
