I am trying to find out if for any symmetric (Not necessarily self-adjoint), invertible matrix $A$ over $\mathbb{C}$, there is a square root of the matrix that is also symmetric. I was able to figure out that for invertible matrices there always exists a square root since I can explicitly do it for a general Jordan block and go from there but I was hoping that it would necessarily be symmetric under this construction (which seems either not obvious or not true). Any thoughts?

  • $\begingroup$ You get problems if the ground field is not algebraically closed (or at least each eigenvalue of the matrix is a square). For example in $GL_2(\mathbb R)$ there is no square root of $\begin{pmatrix}0&1\\1&0\end{pmatrix}$. $\endgroup$ – Hagen von Eitzen Feb 26 '13 at 19:48
  • $\begingroup$ @HagenvonEitzen My ground field is $\mathbb{C}$, Ill edit the problem $\endgroup$ – Sean Ballentine Feb 26 '13 at 19:49
  • $\begingroup$ And you mean symmetric, not self-adjoint, right? $\endgroup$ – Julien Feb 26 '13 at 19:51
  • $\begingroup$ @julien Yes, I feel the question might be easier it was self-adjoint but I do not know that. $\endgroup$ – Sean Ballentine Feb 26 '13 at 19:52
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    $\begingroup$ @HagenvonEitzen Yes...a complex symmetric matrix which fails to be hermitian or skew-hermitian, or even normal...is just a mistake of Mother's Nature. $\endgroup$ – Julien Feb 26 '13 at 20:01

If $\|A-I\|<1$ you can always define a square root with the Taylor series of $\sqrt{1+u}$ at $0$: $$ \sqrt{A}=\sqrt{I+(A-I))}=\sum_{n\geq 0}\binom{1/2}{n}(A-I)^n. $$ If $A$ is moreover symmetric, this yields a symmetric square root.

More generally, if $A$ is invertible, $0$ is not in the spectrum of $A$, so there is a $\log$ on the spectrum. Since the latter is finite, this is obviously continuous. So the continuous functional calculus allows us to define $$ \sqrt{A}:=e^\frac{\log A}{2}. $$ By property of the continuous functional calculus, this is a square root of $A$.

Now note that $\log$ coincides with a polynomial $p$ on the spectrum (by Lagrange interpolation, for instance). Note also that $A^t$ and $A$ have the same spectrum. Therefore $$ \log(A^t)=p(A^t)=p(A)^t=(\log A)^t. $$ Taking the Taylor series of $\exp$, it is immediate to see that $\exp(B^t)=\exp(B)^t$. It follows that if $A$ is symmetric, then our $\sqrt{A}$ is symmetric.

Now if $A$ is not invertible, certainly there is no log of $A$ for otherwise $$ A=e^B\quad\quad\Rightarrow \quad 0=\mbox{det}A=e^{\mbox{Tr}B}>0. $$ I am still pondering the case of the square root.

  • $\begingroup$ I see how this gives you a square root but how do you check that the square root obtained is symmetric? (i.e. How does transpose commute with exp and log) $\endgroup$ – Sean Ballentine Feb 27 '13 at 16:04
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    $\begingroup$ @SeanBallentine In the first case $\|A-I\|<1$, it is clear by taking the transpose on each term of the series. In the second case, by the Taylor expansion series of $\exp$, again $\exp(A^t)=\exp(A)^t$. Now it is slightly more delicate to handle $\log(A^t)=\log(A)^t$. The holomorphic functional calculus furnishes $f(A)$ and $f(A^t)$ (note they have the same spectrum) by polynomial approximation of $f$. For each polynomial $p_n(A)^t=p_n(A^t)$, so passing to the limit yields $f(A)^t=f(A^t)$. There was a typo in my argument, I said complement instead of neighborhood. Edited. $\endgroup$ – Julien Feb 27 '13 at 17:00
  • $\begingroup$ When you think about exp and log in terms of their expansions this seems obvious. THANKS! $\endgroup$ – Sean Ballentine Feb 27 '13 at 17:06
  • $\begingroup$ @SeanBallentine Note that holomorphic functional calculus is really not needed here. I gave an argument using only continuous functional calculus. Which is basically Stone-Weierstrassand is fairly trivial on a finite spectrum. $\endgroup$ – Julien Feb 27 '13 at 17:55

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