Given $ A \in M_n(\Bbb C) $ and $ <x,y>_A = x^TA\overline y $
I need to proof that if A is non diagonalizable then $<.,.>_A$ is not an inner product.
I thought about: Let A be non diagonalizable, then A is similar to a Jordan matrix and therefore the inner product, will contain $ x_1 ... x_{n-1} $ * $ y_2 ... y_{n} $ which is not symmetric.
But I think I am wrong..
I'm not really sure what the argument about $x_1,\ldots,x_{n-1}$ and $y_2,\ldots,y_n$ means. In addition, using the Jordan transformation is not really a happy choice since the similarity transformation matrix cannot be absorbed in $x$ and $y$.
What you actually want to prove is that if $A$ induces an inner product then $A$ is diagonalisable (equivalently, $A$ is not diagonalisable implies that $\langle\cdot,\cdot\rangle_A$ is not an inner product). It is easy to prove a stronger statement that $A$ is Hermitian positive definite matrix just by using the properties of an inner product.
Let $\langle x, y\rangle_A=y^*Ax$. Note that the $i,j$ component of $A=(a_{ij})$ is given by $a_{ij}=e_i^*Ae_j=\langle e_j,e_i\rangle_A$. Assume that $\langle\cdot,\cdot\rangle_A$ is an inner product, that is, the following properties hold with some implications on the properties of $A$:
Conjugate symmetry $\langle x,y\rangle_A=\overline{\langle y,x\rangle}_A$ implies that $a_{ij}=\bar{a}_{ji}$ for all $i,j=1,\ldots,n$ (by using standard basis vectors $e_i$ and $e_j$ as "special" $x$s and $y$s), hence $A$ is Hermitian. Note that we could already stop here, because all Hermitian matrices are (orthogonally) diagonalisable.
Linearity does not imply anything here. The form $\langle\cdot,\cdot\rangle_A$ is linear for any $A$.
Positive definiteness, that is, $\langle x,x\rangle_A\geq 0$ for any $x$ and $\langle x,x\rangle_A=0$ iff $x=0$, implies that $x^*Ax\geq 0$ for any $x$ and $x^*Ax=0$ iff $x=0$, that is, $A$ is positive definite.
Hence
Inverting this gives