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If $A^3$ is an Hermitian matrix, and $A$ is a normal matrix ($ A^{*}A = AA^*$), is $ A=A^* $?

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Hint: Since $A$ is normal, it is diagonalizable by unitary transformation. The cubes of its eigenvalues are real. But real numbers can have non-real cube roots...

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In particular, try to look at $1 \times 1$ matrices. – Phira Nov 9 '11 at 18:08
Okay, and what if we also add $ A = 2A^3 - A^* $? – dan Nov 9 '11 at 18:36
I'll try to answer myself - let $Q$ the matrix that diagonalizes $A$ into $D$, then from $ A + A^* = 2A^3 $ we get $ QDQ^{-1} + (QDQ^{-1})^* = 2Q(D^3)Q^{-1} $ and then $ QDQ^{-1} + QDQ^{-1} = 2Q(D^3)Q^{-1} $ so that $ D=D^3 $. This forces $ A=A^3 $ so that $ A $ gets to be Hermitian. – dan Nov 9 '11 at 19:18
@dan, your argument is circular. $(QDQ^{-1})^{} = QDQ^{-1}$ relies on $A$ being Hermitian. If $A$ is normal, then it may have complex eigenvalues, so $D^{} \neq D$. – rcollyer Nov 9 '11 at 22:04
If $A = 2 A^3 - A^*$, any eigenvalue $\lambda$ of $A$ satisfies $\lambda = 2 \lambda^3 - \overline{\lambda}$. Now if $\lambda^3$ is real but $\lambda$ is not real, $\lambda = r \omega$ where $r > 0$ and $\omega$ is a primitive cube root of either $1$ or $-1$. But in either case, $\lambda + \overline{\lambda}$ has the opposite sign to $\lambda^3$, so this is impossible. – Robert Israel Nov 9 '11 at 22:48

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