3
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – amd Jul 16 '19 at 0:27
  • $\begingroup$ Sure, thanks for your comment. I want B to be strongly positive definite too. $\endgroup$ – Lentes Jul 16 '19 at 15:00
3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.