# Does certain matrix commutes with square root of another one?

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

$$\sqrt{AA^T}=O\sqrt{\Lambda}O^T,$$

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:

$$A\sqrt{AA^T}=\sqrt{AA^T}A$$

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

Any help will be appreciated.

• If we have $A^T=-A$ then $AA^T=-A^2$ so we will have always $A$ commute with $AA^T$ !! Commented Apr 20, 2012 at 2:16
• @Abdelmajid: Yes, that's already stated in the question. The question is whether $A$ also commutes with the square root of $AA^T$. Commented Apr 20, 2012 at 2:18
• what do you mean by $O$ ? presumably is the transition matrix, if it's so i think you should write $AA^T=O\Lambda O^{-1}$. Commented Apr 20, 2012 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 $AA^T$. Commented Apr 20, 2012 at 2:25

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 $$\sqrt{B}=S\sqrt{D}S^{-1}=Sp(D)S^{-1}=p(SDS^{-1})=p(B).$$

• 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. Commented Apr 20, 2012 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. Commented Apr 20, 2012 at 5:34
• Just dropping by, how to choose that polynomial? Commented May 13, 2012 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: en.wikipedia.org/wiki/Lagrange_polynomial Commented May 13, 2012 at 5:51
• You need the degree of $p$ to be at least as big as the number of distinct eigenvalues of $B$. Commented Apr 16, 2021 at 19:07

$AA^T$ 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 $O\sqrt\Lambda O^T=\sqrt{AA^T}$.

• Thank you very much. It's a really nice solution. Commented Apr 20, 2012 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. Commented Apr 20, 2012 at 2:51