Symmetric Square Root of Symmetric Invertible Matrix 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?
 A: 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.
