Real square roots of operators with simple real spectrum ($B^2=A⟹{\lambda }_i(A)\ge 0$)

If a real linear operator $A$ on an $n$-dimensional vector space has $n$ distinct real eigenvalues, any real square root $B$ commutes with $A$ and preserves each one-dimensional eigenspace. There $B$ is multiplication by a real $b_i$, hence ${\lambda }_i=b_i^2\ge 0$. Conversely, if all the eigenvalues are nonnegative, choosing $b_i=\sqrt{{\lambda }_i}$ on an eigenbasis supplies a real square root. Without simple spectrum the necessity fails: a real quarter-turn squares to $-I$.