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Let $A^3$ be a (complex) unitary matrix. Show that $A$ is diagonalizable.

My attempt: Since $A^3$ is unitary, it is diagonalizable (over $\mathbb{C}$), so its minimal polynomial $p_{A^3}$ is a product of distinct linear factors $p_{A^3}(x)=(x-\lambda_1)...(x-\lambda_k)$. Since $A^3$ is unitary, we know $|\lambda_k|=1$ for each $k$. Now define $g(x)=(x^3-\lambda_1)...(x^3-\lambda_k)$. Let $p_A$ be the minimal polynomial for $A$. Since $g(A)=0$, $p_a|g$, so if we can show $g$ splits as a product of distinct linear factors, then $p_a$ is a product of distinct linear factors $\implies A$ is diagonalizable. But how would we show $g$ factors this way?

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$x^3 - \lambda_i$ splits into linear factors corresponding to the three cube roots of $\lambda_i$, and of course these are different for different $i$ (i.e. a complex number has only one cube).

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