# Do complex symmetric matrices with zero row sums have linearly independent eigenvectors?

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?

• See my edited post Aug 10 '15 at 12:52
• Thank you, Michael. That settles it. Aug 10 '15 at 13:30

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 $$\begin{pmatrix} 0 & 0 & 0\\ 0 & i\sqrt3 & 1\\ 0 & 0 & i\sqrt3 \end{pmatrix}$$
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 $$\begin{pmatrix} 0 & 0 & 0\\ 0 & 3 + i\sqrt3 & 1\\ 0 & 0 & 3 + i\sqrt3 \end{pmatrix}.$$