square root of a real matrix

Solution 1:

A real matrix $S$ can possess infinitely many real or nonreal square roots. For example, $$ S=I=\begin{pmatrix}1&t\\0&-1\end{pmatrix}^2 $$ for every $t\in\mathbb{C}$. Note that $S=I$ is real and positive definite.

Solution 2:

If $S$ is real, symmetric and positive definite, consider its eigenvalue / eigenvector decomposition $S = X \Lambda X^T$ where $\Lambda$ is diagonal and $X$ is orthogonal. Because $S$ is positive definite, $\Lambda_{ii} > 0$ for all $i$. The unique symmetric and positive definite square root of $S$ is given by $S^{1/2} = X \Lambda^{1/2} X^T$, where $\Lambda^{1/2}$ is the diagonal matrix with the $\sqrt{\Lambda_{ii}}$ on its diagonal. Indeed, $$ S^{1/2} S^{1/2} = X \Lambda^{1/2} X^T X \Lambda^{1/2} X^T = X \Lambda X^T = S, $$ because $X^T X = I$. So $S^{1/2}$ is indeed real.