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I know that a non-zero symmetric 2×2 matrix can't have only zero eigenvalues ( a zero eigenvalue with algebraic multiplicity 2), since such a matrix should have complex off diagonal entries to satisfy both trace and determinant being zero. I'm guessing if this is the case for the general case of any non-zero n×n symmetric matrix.

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    $\begingroup$ It's also true for any symmetric matrix. It's similar to a diagonal matrix with it's eigenvalues put at the diagonal. $\endgroup$ – Sergei Golovan May 27 '17 at 10:14
  • $\begingroup$ Thanks. Helping me to look at the problem from the perspective of spectral decomposition made it completely clear for me. $\endgroup$ – sanaz mat May 27 '17 at 10:25
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No, because then it is a nilpotent symmetric matrix, and since symmetric matrics are diagonalizable and the only diagonalizable nilpotent matrix is the zero matrix, we'd get a contradicition to "non-zero matrix"

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A big classic is to show that if $A$ nilpotent then $\operatorname{tr}(A^k)=0$.

Another classic is to show that $\mathcal N(A)=\operatorname{tr}(^tAA)$ is a norm on square matrices.

If $A$ is nilpotent and symmetric, these two combine and $\mathcal{N}(A)=\operatorname{tr}(^tAA)=\operatorname{tr}(A^2)=0$ so $A=0$.

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    $\begingroup$ The first one is an iff argument if we add " for all $\;k\in\Bbb N\;$" $\endgroup$ – DonAntonio May 27 '17 at 12:02

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