show a matrix is diagonalizable Lets Say I have a $A_{n\times n}$ . I also know that every eigenvector of $A$ is also eigenvector of $A^*$ the Conjugate transpose matrix
According to my understating, it seems that $A$ should be diagonalizable. but I have troubles to show it. 
 A: Let $v$ be an eigenvector of $A$ corresponding to the eigenvalue $\lambda \in \mathbb{C}$. Then $Av = \lambda v$ and, by applying conjugate tranpose, $v^{*}A^{*} = \overline{\lambda} v^{*}$. Since $v$ is an eigenvector of $A^{*}$, we have $A^{*}v = \mu v$ for some $\mu \in \mathbb{C}$ and
$$ \mu ||v||^2 = \mu (v^{*} \cdot v) = v^{*} \cdot (\mu v) = v^{*} \cdot (A^{*} v) = (v^{*} A^{*}) \cdot v = \overline{\lambda}(v^{*} \cdot v) = \overline{\lambda} ||v||^2 $$
so $\mu = \overline{\lambda}$. That is, $v$ is an eigenvector of $A^{*}$ corresponding to the eigenvalue $\overline{\lambda}$. 
To show that $A$ is diagonalizable, mimic the proof that a normal matrix is orthogonally diagonalizable. Namely, pick some eigenvector $v$ of $A$, let $U := \mathrm{span} \{ v \}$, show that $U^{\perp}$ is $A$ and $A^{*}$ invariant and argue inductively.
In fact, if $A$ is a normal matrix, a typical step of the proof shows that if $v$ is an eigenvector of $A$ associated to the eigenvalue $\lambda$ then $v$ is also an eigenvector of $A^{*}$ associated to the eigenvalue $\overline{\lambda}$ so your condition is equivalent to requiring that $A$ is normal, or, alternatively, orthogonally diagonalizable.
