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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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    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.2011-11-09