base so that the matrix of endomorphism $\varphi$ is equal to the matrix of $\varphi^*$

How to show that there exists a base of $V$ so that the matrix $A$ of an endomorphism $\varphi: V \to V$ is equal to the matrix $A^*$ of the dual endomorphism $\varphi^*: V \to K$? ($V$ is a $K:=\mathbb{C}$ vector space)

Since $A = {}^t(A^*)$ I need to show that there exists a base so that $A$ is symmetric, right?

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Try to do this for $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. –  Mariano Suárez-Alvarez May 21 '12 at 16:44
In this case someone could just exchange the two base vectors to obtain $\left(\begin{smallmatrix}1&0\\0&0\\\end{smallmatrix}\right)$? –  steltjen May 21 '12 at 18:05
Maybe someone should actually try do that to see if it works :) –  Mariano Suárez-Alvarez May 21 '12 at 18:20
I think you need to make precise exactly what you want to prove. It is true that every matrix is similar to its transpose (and this has been asked in this site before: search for that) –  Mariano Suárez-Alvarez May 22 '12 at 3:31
Well, can you do that for my example? –  Mariano Suárez-Alvarez May 22 '12 at 4:09