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How to prove that "If $A \in\mathcal M_{n,n}(\mathbb C)$ is Hermitian and $A^2= 0_{n,n}$, then $A$ must be a null matrix"?

This is probably easy but I am not able to get this, I know that for any Hermitian matrix the diagonal element must be purely real, does this help here?

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Hermitian (not Hermition) matrices are diagonalizable. – Chris Eagle Jan 16 '12 at 19:44
up vote 4 down vote accepted

Do you know that every Hermitian matrix is diagonalizable (for example by the spectral theorem)? If yes, write $A=S^{-1}DS$ with $D$ diagonal and see what you can conclude from $A^2=0$.

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Well you could write: $0=A^2=S^{-1}D^2S$ Then $D^2$ must be $0$, and since $D$ is diagonal, $D$ must also be $0$. Therefore $A=0$. But if you learned about diagonalizability today, you probably have not yet proved that Hermitian matrices can be diagonalized and Davides solution might be more suitable. – user20388 Jan 16 '12 at 20:06
I did it :-) $\space$ – Quixotic Jan 17 '12 at 14:28

Let $\langle \cdot,\cdot\rangle$ the canonical inner product in $\mathbb C^n$. Then for $x\in\mathbb C^n$: $$\langle Ax,Ax\rangle =\langle A^*Ax,x\rangle=\langle A^2x,x\rangle =0$$ so $Ax=0$. In particular for $x=e_j$, $j$-th vector of the canonical basis of $\mathbb C^n$, we get $A=0$.

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+1: It is much simpler than applying diagonalizability. – Jonas Meyer Jan 16 '12 at 19:55

Assuming you know about minimal polynomials or when you first learn of them...$A$ is Hermitian $\Rightarrow$ $A$ is diagonalizable (by the complex spectral theorem) and so its minimal polynomial, $m_t(A)$, splits with distinct linear factors (indeed $A$ diagonalizable $\Leftrightarrow$ $m_t(A)$ splits with distinct linear terms). Now, let $f(t) = t^2$ then $f(A) = 0$ $\Rightarrow$ $f$ is an annihilating polynomial of $A$. Hence, $m_t(A) | f(t)$; i.e., we have that either $m_t(A) = t^2$ (in which case $A$ is not diagonalizable) or $m_t(A) = t$ (in which case $A$ is the null matrix). Therefore, since $A$ is diagonalizable $m_t(A) = t$ and we see that $A$ is the null matrix.

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