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In the context of linear algebra and diagonalisable matrices, I need help with the following problem:

Let $A \in M_{n \times n}(\mathbb{R})$, with $n \geq 2$. Assume that $A$ is diagonalisable. Show that there exists a matrix $\Gamma \in M_{n \times n}(\mathbb{R})$ such that $\Gamma^{5} = A$.

I'm sorry for the lack of effort on my part but I honestly don't have a clue what to do. I can state the obvious by saying that, since $A$ is diagonalisable, it is similar to a diagonal matrix $D = S^{-1}AS$ but I don't see how this is helpful here.

As a side note, I have found a similar problem (without solution) where we're asked to show there exists a matrix $\Gamma \in M_{n \times n}(\mathbb{R})$ such that $\Gamma^{\color{red}{7}} = A$. Does this statement generalises to any power of the matrix $\Gamma$?

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    what is the answer if $A$ is already diagonal? Example, $$ A = \left( \begin{array}{rr} 32 & 0 \\ 0 & -243\end{array}\right) $$ – 2017-02-07
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    I don't think this works for even exponent $r$. Hint : for all $r$ _odd_ and real number $\lambda$, there is a real $\mu$ with $\mu^r = \lambda$. – 2017-02-07
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    @N.H. It works for even $n$ if $A$ is positive semidefinite (or we're working over the complex numbers). – 2017-02-07
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    @N.H. you are confusing the exponent with the dimension of the (square) matrix – 2017-02-07
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    Yes it does work !! – 2017-02-07
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    @WillJagy : thanks, I will edit my confusing notation. – 2017-02-07
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    About your side question: can you answer it when $n=1$? – 2017-02-07

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It does not extend to any power and diagonalisable matrix: since the eigenvalues of $\Gamma^k$ are the $k$-th powers of the eigenvalues of $\Gamma$, you need either the exponent to be odd or the eigenvalues to be $\ge 0$.

The proof is: there is a diagonal matrix $D$ such that $PAP^{-1}=D$. Consider $\Delta$ a (easily constructed) diagonal matrix such that $\Delta^5=D$. Then, observe that $$(P^{-1}\Delta P)^5=P^{-1}\Delta^5 P=P^{-1}D P=A$$

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    probably better to edit out the first word, "No." – 2017-02-07
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    @WillJagy Fair enough. – 2017-02-07
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    I see: I looked at the question again, first he asks the thing I noticed, in yellow, but then he really does add on a secondary question as far as generalization, and you are saying no to that. Makes sense. Better if the OP asks one question at a time – 2017-02-07
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    Thank you for your answer. I'm not sure how to derive the diagonal matrix $\Delta$ (even though you're saying it is easily constructed). Is it the fifth root of the diagonal elements of $D$? This is why the eigenvalues must be $\geq 0$. Is that correct? – 2017-02-07
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    @Elix Indeed they are. – 2017-02-07