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Problem:

I want to prove that if $A^3$ is unitary, then $A$ is diagonalizable. Definitely, since $A^3$ is unitary, then it is normal, and we know that every normal matrix is diagonalizable. So, there exists $U$ unitary such that: $U^{*}A^{3}U=D=\begin{bmatrix} \lambda_{1}^{3}&0 &0 &0 \\ 0&\lambda_{2}^{3} &0 &0 \\ 0 & 0 & \lambda_{3}^{3}&0 \\ 0& 0 &0 & \lambda_{n}^{3} \end{bmatrix}$

Where $\lambda_{i}$ are eigenvalues of $A$. I am thinking to prove that $A$ is also normal, which implies that $A$ is diagonalizable, but I have no idea how to approach it.

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    Hint: $U^*A^3U=(U^*AU)^3$2012-04-28
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    @Brett Frankel: Do you mean that $U^{*}AU=D^{'}=\begin{bmatrix} \lambda_{1}&0 &0 &0 \\ 0&\lambda_{2} &0 &0 \\ 0 & 0 & \lambda_{3}&0 \\ 0& 0 &0 & \lambda_{n} \end{bmatrix}$? Is the following assertion true in general: $A^{3}=B^{3}\Rightarrow A=B$?2012-04-28
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    That won't be true in general.2012-04-28
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    But using Jordan form, you can show that if $A^3$ is diagonal, $A$ is diagonalizable over $\mathbb{C}$.2012-04-28
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    @ Brett Frankel: interesting!2012-04-28

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