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!
Best Answer
Your condition "$A$ strongly PD" is equivalent to "$A+A^T$ is symmetric $>0$". According to,
Largest eigenvalues of matrix and its doubled symmetric part
every $\lambda\in spectrum(A)$ satisfies $Re(\lambda)>0$.
Thus $A$ admits $\log(A)$ as its principal logarithm and the principal square root $A^{1/2}$ is well defined; cf the first part of my post in
When is square root of transpose and transpose of square root of a matrix are equal?
Moreover, $A^{1/2}$ is the unique square root of $A$ whose all the eigenvalues have a positive real part. Thus, if $A$ admits a strong square root, then it is unique.
EDIT. The difficult part is to see if $A^{1/2}+{A^{1/2}}^T$ is $>0$.
That is true; cf. Corollary 8 in
https://www.sciencedirect.com/science/article/pii/S0024379500002433
cf. also
Square root of positive definite nonsymmetric matrix
where this question was studied.