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Let $Y \in \mathbb{C}^{N \times N}$ be a complex symmetric (but not Hermitian) matrix such that $Y = Y^\mathrm{T}$ , and it is known that $Y\mathbb{1}_{N} = 0$, where $\mathbb{1}_{N}$ is the vector of all ones.

Is this matrix $Y$ always diagonalizable? I know in general complex symmetric matrices are not.

EDIT: Michael provided a counterexample to this. However, if it is further known that all off-diagonal elements have non-positive real-parts and the zero eigenvalue corresponding to the eigenvector $\mathbb{1}_{N}$ is simple (i.e. algebraic and geometric multiplicity 1), is Y diagonalizable in that case?

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  • $\begingroup$ See my edited post $\endgroup$ – Michael Galuza Aug 10 '15 at 12:52
  • $\begingroup$ Thank you, Michael. That settles it. $\endgroup$ – Mohit Sinha Aug 10 '15 at 13:30
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No. Here is counterexample: $$ Y = \begin{pmatrix} 0 & 1 & -1\\ 1 & -1 + i\sqrt3 & -i\sqrt3\\ -1 & -i\sqrt3 & 1 + i\sqrt3 \end{pmatrix} $$ Jordan normal form of $Y$ is $$ \begin{pmatrix} 0 & 0 & 0\\ 0 & i\sqrt3 & 1\\ 0 & 0 & i\sqrt3 \end{pmatrix} $$

EDIT

Unfortunately, no. Jordan normal form of $$ \begin{pmatrix} 2 & 0 & -2\\ 0 & 1 + i\sqrt3 & -1 - i\sqrt3\\ -2 & -1 - i\sqrt3 & 3 + i\sqrt3 \end{pmatrix} $$ is $$ \begin{pmatrix} 0 & 0 & 0\\ 0 & 3 + i\sqrt3 & 1\\ 0 & 0 & 3 + i\sqrt3 \end{pmatrix}. $$

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