Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can't figure out if this is true:

Suppose $A^T=-A$ and that the symmetric matrix $AA^T$ is a positive definite (so diagonalizable, lets say $AA^T=O\Lambda O^T$, with all eigenvalues positive). Therefore, we can define a square root by


where $\sqrt{\Lambda}=\textrm{diag}(\sqrt{\lambda_1},\ldots\sqrt{\lambda_n})$, $\lambda_i$ eigenvalues of $AA^T$. One can see that $A$ commutes with $AA^T$, once $A^T=-A$. But I can't see if it is true that:


In other words, does $A$ commute with $AA^T$ implies $A$ commutes with $\sqrt{AA^T}$?

Any help will be appreciated.

share|cite|improve this question
If we have $A^T=-A$ then $AA^T=-A^2$ so we will have always $A$ commute with $AA^T$ !! – Abdelmajid Khadari Apr 20 '12 at 2:16
@Abdelmajid: Yes, that's already stated in the question. The question is whether $A$ also commutes with the square root of $AA^T$. – joriki Apr 20 '12 at 2:18
what do you mean by $O$ ? presumably is the transition matrix, if it's so i think you should write $AA^T=O\Lambda O^{-1}$. – Abdelmajid Khadari Apr 20 '12 at 2:22
@Abdelmajid: Since $AA^T$ is symmetric, $O$ can be chosen orthogonal, and then $O^{-1}=O^T$. It's quite usual to make use of that; the columns of $O$ can then be viewed as an orthonormal system of eigenvectors for $AA^T$. – joriki Apr 20 '12 at 2:25
up vote 12 down vote accepted

In more generality, if $B$ is positive semidefinite and $AB=BA$, then $A\sqrt B=\sqrt B\,A$. The key observation is that there exists a polynomial $p\in\mathbb{R}[x]$ such that $\sqrt B=p(B)$. Then we have $$ AB^2=(AB)B=(BA)B=B(AB)=B(BA)=B^2A; $$ similarly we deduce that $AB^n=B^nA$ for any $n\in\mathbb{N}$, and so $Ap(B)=p(B)A$ for any polynomial.

The existence of the required polynomial is shown as follows: as $B$ is positive semidefinite, it is diagonalizable, so $B=SDS^{-1}$ with $D$ diagonal. Now choose a polynomial $p$ such that $p(d_{jj})=\sqrt{d_{jj}}$. Then $$ \sqrt{B}=S\sqrt{D}S^{-1}=Sp(D)S^{-1}=p(SDS^{-1})=p(B). $$

share|cite|improve this answer
Thanks a lot. Your solution answer another question I had. It was my first trying to prove that $AB=BA$ then $A$ commutes with the square root, but you've found a really nice way of proving much more. Paeticularly, I like the way you recover the square root of $B$ based in constructing a polynomial. Thanks again. – matgaio Apr 20 '12 at 3:04
You are welcome. Of course, as you see, this works for any other function too, i.e. you can define $f(B)$ for any $f$ using the same trick. – Martin Argerami Apr 20 '12 at 5:34
Just dropping by, how to choose that polynomial? – checkmath May 13 '12 at 5:27
@chessmath: you have $n$ points $d_{11},\ldots,d_{nn}$ and you want a polynomial with prescribed values at those points. The canonical way of doing it is Lagrange Polynomial: – Martin Argerami May 13 '12 at 5:51

$AA^T$ and $\sqrt{AA^T}$ are diagonalized by the same matrix $O$, and they have the same pattern of equal or distinct eigenvalues. We can also form $O^TAO$, which commutes with $O^TAA^TO=\Lambda$. The matrices that commute with a given diagonal matrix $\Lambda$ are all matrices that have non-zero entries $a_{ij}$ only where $\lambda_i=\lambda_j$. Since this condition is the same for $\Lambda$ and $\sqrt\Lambda$, the same matrices commute with $\Lambda$ and $\sqrt\Lambda$. Since $O^TAO$ commutes with $\Lambda$, it also commutes with $\sqrt\Lambda$, and thus $A$ commutes with $O\sqrt\Lambda O^T=\sqrt{AA^T}$.

share|cite|improve this answer
Thank you very much. It's a really nice solution. – matgaio Apr 20 '12 at 2:50
@matgaio: You're welcome. But Martin's is also good, a bit more abstract but also more general, and it's useful to be familiar with that sort of approach to matrices, too. – joriki Apr 20 '12 at 2:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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