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 $\Gamma$?
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 $$(P^{-1}\Delta P)^5=P^{-1}\Delta^5 P=P^{-1}D P=A$$