It is well known that if $A$ is a symmetric positive definite matrix, then it has a unique square root which is positive definite. My question is whether this result extends to a strongly positive definite nonsymmetric matrix.
More precisely, let $A$ be a real nonsymmetric $n\times n$ matrix, which satisfies the following strong positive definite condition: there exists $a>0$ such that for each $x\in\mathbb R^n$, the estimate $$ \langle Ax, x\rangle\geq a|x|^2 $$ holds. Is it true then that there exists a unique (edit: strongly positive definite) matrix $B$ such that $B^2=A$? I would be very interested in knowing the answer to this result, and reference to a proof. Thanks!