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Prove that if a $2011\times 2011$ real symmetric matrix $A$ satisfies $A^{2011}=0$, then $A=0$

What I know is that a real symmetric matrix always has real eigenvalue.

Can you help me?

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Real symmetric matrices are diagonalizable, i.e., $A=U^TDU$, where $$ D=\mathrm{diag}(d_1,\ldots,d_n), $$ is diagonal and $U$ an orthogonal matrix.

Then $A^{2011}=U^TD^{2011}U$, which implies that $$ 0=D^{2011}=\mathrm{diag}\big(d_1^{2011},\ldots,d_n^{2011}\big). $$ Therefore $$ d_i=0,\quad\text{for all}\,\,\, i=1,\ldots,2011, $$ which implies that $D=0$, and hence $A=0$.

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Hint: Since $A^{2011}$, it follows that $0$ is $A$'s only eigenvalue. But another hypothesis implies that $A$ is diagonalizable. Conclude.

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