For which positive integers \(n\) is it true that every invertible \(2\times2\) matrix \(A\) with real entries can be expressed as the \(n\)-th power of another \(2\times2\) matrix with real entries?
Solution
The answer is the odd positive integers. If \(n\) is even, then \(\begin{pmatrix}-1&0\\0&1\end{pmatrix}\) cannot be the \(n\)-th power of another \(2\times2\) matrix with real entries, because its determinant is not an \(n\)-th power of a real number.
Now assume \(n\) is odd. Thus every real number is an \(n\)-th power of a real number. The question of whether \(A\) is an \(n\)-th power is not affected by conjugation. Thus if \(A\) has distinct real eigenvalues, then without loss of generality we may assume that \(A\) is diagonal, in which we take the \(n\)-th roots of the diagonal entries to find another diagonal matrix \(B\) with \(B^n=A\).
If \(A\) has equal real eigenvalues, then by conjugation, we may assume
\[ A=\lambda\begin{pmatrix}1&c\\0&1\end{pmatrix} \]
where \(\lambda\in\RR^*\) and \(c\in\RR\). Then \(A=B^n\) where
\[ B=\lambda^{1/n}\begin{pmatrix}1&c/n\\0&1\end{pmatrix}. \]
Finally if the eigenvalues of \(A\) are not real, then the minimal polynomial of \(A\) is a quadratic polynomial \(f(x)\) with no real roots, so the \(\RR\)-subalgebra \(\RR[A]\) of \(M_2(\RR)\) generated by \(A\) is isomorphic to \(\RR[x]/(f(x))\cong\CC\). Since every element of \(\CC\) has an \(n\)-th root, the matrix \(A\) has an \(n\)-th root in \(\RR[A]\).