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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?

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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.

So the Gram Schmidt process actually will produce orthogonal vectors, but they will not be able to diagonalize the matrix.

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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.

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  • $\begingroup$ Thank you for the counterexample but I still don't totaly get what wrong in the process. Indeed the process is fine since the two vectors are orthogonal and correctly gives the same two vectors $\endgroup$ – Dac0 Feb 16 '16 at 16:23
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    $\begingroup$ If $x$ is a complex vector, then $x^Tx=0$ does not imply that $x=0$. $\endgroup$ – Chris Godsil Feb 16 '16 at 16:27
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    $\begingroup$ With $x=(1,i)^T$ we have $x^Tx=0$, but not $x=0$. $\endgroup$ – Dietrich Burde Feb 16 '16 at 16:28
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    $\begingroup$ So the Gram Schmidt process itself would work, but the matrix would not be diagonal anyway? $\endgroup$ – Dac0 Feb 16 '16 at 16:29

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