Assume that $A \in M_n(\mathbb{C})$ is an invertible matrix. Can we be sure that $A$ has a square root, i.e. like $T\in M_n(\mathbb{C})$ such that $T^2 = A$?
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See Under what conditions does a matrix $A$ have a square root? or $A = B^2$ for which matrix $A$? – Joe May 07 '22 at 12:58
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It is true and there are several prooves of this.
A possible proof is to use the fact that $\exp : M_{n}(\mathbb{C}) \rightarrow G L_{n}(\mathbb{C})$ is surjective (this result is not straightforward).
Then we can write $A = e^M = (e^{M/2}) (e^{M/2})$ and choose $T=e^{M/2}$.
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