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Is rank of symmetric or skew-symmetric matrix equal to number of non zero eigenvalues of the matrix?

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Notice that over $\mathbb{Z}_2$ the answer is no. Notice that the matrix $\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}$ is symmetric and skew-symmetric, if we consider the coefficients in $\mathbb{Z}_2($since $1=-1$ in $\mathbb{Z}_2)$. Now the eingevalues of this matrix in $\mathbb{Z}_2$ are both equal to $0$, but this matrix has rank 1.

For symmetric matrices over the complex field the answer is also No. Consider $\begin{pmatrix}i & 1\\ 1 & -i\end{pmatrix}$. This matrix has both eigenvalues equal to $0$, but its rank is $1$.

Over the reals the answer is yes, because symmetric matrices with real coefficients and skew-symmetric matrices with real coefficients multiplied by $i$ are Hermitian matrices. Therefore symmetric matrices and skew-symmetric matrices with real coefficients are diagonalizable. For diagonalizable matrices the rank coincides with the number of non zero eigenvalues.

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