# Do non-symmetric “strongly positive definite” matrices have unique positive definite square roots?

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!

• Did you mean for $B$ to be strongly positive definite as well? It’s not hard to construct a $2\times2$ matrix $A$ that satisfies your conditions but doesn’t have a unique square root. – amd Jul 16 '19 at 0:27
• Sure, thanks for your comment. I want B to be strongly positive definite too. – Lentes Jul 16 '19 at 15:00

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.

• Awesome, thanks a lot for your very complete answer and for linking the paper. – Lentes Jul 16 '19 at 19:41