# Why a complex symmetric matrix is not diagonalizible?

I know an Hermitian matrix is diagonalizable, and similarly a real symmetric matrix is diagonalizable, but what's wrong in a complex symmetric matrix.

Why does the Gram-Schmidt process fail?

As Chris Godsil and Dietrich Burde pointed out it's because $\langle x,y \rangle =x^*y=0$ which is the orthogonality condition on complex vectors does not imply that $x^Ty=0$ which is the complex symmetry condition.
First of all, there is an easy counterexample. The complex symmetric matrix $$\begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix}$$ is not diagonalizable, because trace and determinant are zero, but the matrix is not zero. Now try the Gram-Schmidt process in this example.
• If $x$ is a complex vector, then $x^Tx=0$ does not imply that $x=0$. – Chris Godsil Feb 16 '16 at 16:27
• With $x=(1,i)^T$ we have $x^Tx=0$, but not $x=0$. – Dietrich Burde Feb 16 '16 at 16:28