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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.

matgaio
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    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 02:16
  • @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 02:18
  • 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 02:22
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    @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 02:25

3 Answers3

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

Martin Argerami
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  • 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 03:04
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    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 05:34
  • Just dropping by, how to choose that polynomial? – checkmath May 13 '12 at 05:27
  • @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: http://en.wikipedia.org/wiki/Lagrange_polynomial – Martin Argerami May 13 '12 at 05:51
  • does the degree of $p(B)$ always depend on $n$? In particular, for the case with $p(B)=\sqrt{B}$ can one find a polynomial whose degree is independent of $n$? – Condo Apr 16 '21 at 17:58
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    You need the degree of $p$ to be at least as big as the number of distinct eigenvalues of $B$. – Martin Argerami Apr 16 '21 at 19:07
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$AA^T$ 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 $O\sqrt\Lambda O^T=\sqrt{AA^T}$.

joriki
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  • Thank you very much. It's a really nice solution. – matgaio Apr 20 '12 at 02:50
  • @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 02:51
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Actually the result holds for any matrix that has a square root. If $\boldsymbol{B}$ is a real matrix that has no eigenvalues on the closed negative real axis, then there is a unique real-valued square root of $\boldsymbol{B}$ all of whose eigenvalues lie in the open right half-plane. This is the principal square root of $\boldsymbol{B}$. The principal matrix square root can be defined in terms of an integral as \begin{equation*} \sqrt{\boldsymbol{B}}=\frac{2}{\pi}\int_0^{\infty}(t^2\boldsymbol{I}+\boldsymbol{B})^{-1}\boldsymbol{B}\,dt. \end{equation*} By the Cayley-Hamilton theorem $(t^2\boldsymbol{I}+\boldsymbol{B})^{-1}$ is a polynomial in $\boldsymbol{B}$, and hence the entire term on the right is a polynomial in $\boldsymbol{B}$, i.e., $\sqrt{\boldsymbol{B}}$ is a polynomial in $\boldsymbol{B}$, say $p(\boldsymbol{B})$. Now if $\boldsymbol{A}\boldsymbol{B}=\boldsymbol{B}\boldsymbol{A}$, then $\boldsymbol{A}\boldsymbol{B}^n=\boldsymbol{B}^n\boldsymbol{A}$, so that $\boldsymbol{A}p(\boldsymbol{B})=p(\boldsymbol{B})\boldsymbol{A}$ which means that $\boldsymbol{A}\sqrt{\boldsymbol{B}}=\sqrt{\boldsymbol{B}}\boldsymbol{A}$.

It is easy to give an explicit expression for $p(\boldsymbol{B})$ in case $\boldsymbol{B}$ is diagonalizable. If $k$ is the number of distinct eigenvalues of $\boldsymbol{B}$, then \begin{equation*} \boldsymbol{B}=\sum_{i=1}^k \lambda_i \boldsymbol{P}_i, \end{equation*} where \begin{equation*} \boldsymbol{P}_i=\begin{cases} \prod_{\substack{j=1 \\ j\ne i}}^k \frac{\boldsymbol{B}-\lambda_j\boldsymbol{I}}{\lambda_i-\lambda_j},& \text{$k>1$} \\ \boldsymbol{I}, & \text{$k=1$}. \end{cases} \end{equation*} Assuming that there are no eigenvalues of $\boldsymbol{B}$ that lie in $(-\infty,0]$, the principal square root of $\boldsymbol{B}$ is given by \begin{equation*} \sqrt{\boldsymbol{B}}=\sum_{i=1}^k\sqrt{\lambda_i} \boldsymbol{P}_i. \end{equation*} For example, if $\boldsymbol{B}$ is a $3\times 3$ diagonalizable matrix with two distinct eigenvalue $\lambda_1\equiv \mu\ne \lambda_2=\lambda_3\equiv \lambda$, then by using the above formula, we get \begin{equation*} \sqrt{\boldsymbol{B}}=\sqrt{\mu}\left(\frac{\boldsymbol{B}-\lambda\boldsymbol{I}}{\mu-\lambda}\right)+\sqrt{\lambda}\left(\frac{\boldsymbol{B}-\mu\boldsymbol{I}}{\lambda-\mu}\right). \end{equation*} Note that no eigenvectors of $\boldsymbol{B}$ enter in the $\sqrt{\boldsymbol{B}}$ expression, and only eigenvalues of $\boldsymbol{B}$ are enough to find $\sqrt{\boldsymbol{B}}$ explicitly!

Jog
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