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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: $U^*A^3U=(U^*AU)^3$ – Brett Frankel Apr 28 '12 at 21:01
@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: $A^{3}=B^{3}\Rightarrow A=B$? – M.Krov Apr 28 '12 at 21:22
That won't be true in general. – Brett Frankel Apr 28 '12 at 21:24
But using Jordan form, you can show that if $A^3$ is diagonal, $A$ is diagonalizable over $\mathbb{C}$. – Brett Frankel Apr 28 '12 at 21:28
@ Brett Frankel: interesting! – M.Krov Apr 28 '12 at 21:30

This is becoming a bit long for a discussion in comments, so here's the whole thing put together:

As you pointed out, $(U^*AU)^3=U^*A^3U$ is diagonal for some unitary $U$. Thus $(U^*AU)^3$ is diagonal. Now, think about the Jordan blocks of $A$. If $A$ has a nontrivial Jordan block (ie. a block with a 1 above the diagonal), then that Jordan block's cube will not be diagonal. So we conclude that $A$ has no nontrivial Jordan blocks, ie. $A$ is diagonal.

Note that just because $(U^*AU)^3$ is diagonal, we cannot conclude that $U^*AU$ is diagonal. For instance, rotation by $2\pi/3$ is not diagonal, but its cube is.

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Another approach is the following:

Lemma. Matrix $A$ is diagonalizable if and only if the minimal polynomial of $A$ can be factorized into different linear factors, i.e. $m_A(x) = \Pi (x - \lambda_i)$ with distinct $\lambda_i$ 's.

Now from the assumptions we know that $A^3$ possesses the property above, therefore $m_{A^3}(x) = \Pi (x - \lambda_i)$. Note that $\lambda_i$ are distinct and nonzero (since A is invertible, from the fact that $A^3$ is unitary and invertible.)

Also $\Pi (A^3 - \lambda_i I) = 0$. Thus the polynomial $p(x) = \Pi(x^3-\lambda_i)$ is an annihilator of the matrix $A$.

Now factor the polynomial $p(x)$ into linear factors over $\mathbb{C}[x]$ using 3rd roots of unity. Note that the resulting polynomial is a product of distinct linear factors $x-z_i$ (Again using the properties: $\lambda_i$ are nonzero and distinct.) Thus using the lemma at the top of the article, one can conclude that the minimal polynomial of $A$ which divides $p(x)$, satisfies the desiring property and we are done.

strong text

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