According to https://en.wikipedia.org/wiki/Square_root_of_a_matrix#Matrices_with_distinct_eigenvalues, "An n×n matrix with $n$ distinct nonzero eigenvalues has $2^n$ square roots." And we can clearly see that they are given by decomposing $A$ into $SDS^{-1}$ where $D$ is diagonal and taking square roots of the eigenvalues; this generates $2^n$ pairwise distinct square roots. But how did we prove that these are the only matrices $B$ such that $B^2 = A$?
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$\begingroup$ Related questions here and elsewhere on this site. Search around and you'll likely find what you need. $\endgroup$– Lee MosherCommented May 1, 2022 at 14:35
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There needs to be some restriction on the field to make this true. In the wiki the field is implicitly $\mathbb{C}$ so let's work there.
Let $B^2=A$. Suppose the eigenvalues of $B$ are $\mu_1,\dots,\mu_n$. Then the eigenvalues of $A=B^2$ are $\mu_1^2,\dots,\mu_n^2$; we are told these are distinct, so that the $\mu_i$ are also distinct. So with respect to some basis $B$ is diagonal, and so is $A$. All is now clear.
answered May 1, 2022 at 14:34
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$\begingroup$ Do you have a counterexample where the field is not C? $\endgroup$– ions meCommented Sep 26, 2022 at 7:12
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$\begingroup$ And is the step in your proof that breaks down when in annother field that $\mu_i$ are distinct if $\mu_i^2$ are distinct ? $\endgroup$– ions meCommented Sep 26, 2022 at 7:14
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$\begingroup$ Trivially over $\mathbb{F}_2$ the $1\times 1$ matrix has only one square root. $\endgroup$ Commented Sep 26, 2022 at 15:26
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$\begingroup$ True, but the proof still sort of applies ... Just we identify the 2 square roots of [1] with each other. I.E. We identify the [1] = [-1] on F2. Is there a field that breaks this proof more? @ancient mathematician $\endgroup$– ions meCommented Sep 27, 2022 at 18:30
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$\begingroup$ The other sort of failure arises because square roots do not exist in the field: so if we are working over the rational field the matrix $[2]$ has no square roots. $\endgroup$ Commented Sep 28, 2022 at 6:51