# Why does an invertible complex symmetric matrix always have a complex symmetric square root?

Chapter XI Theorem 3 from here implicitly states that an invertible complex symmetric matrix always has a complex symmetric square root.

It's clear that a square root exists, by appealing to the Jordan Normal Form and the fact that the matrix is invertible. But why should this also be symmetric?

Note also that not all square roots of invertible symmetric matrices are symmetric: For instance, one square root of $$\begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$$ is $$\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$$, which is not complex symmetric.

What I've tried: By adapting the proof of the Spectral Theorem to complex symmetric matrices, one gets that "most" symmetric matrices can be diagonalised using an orthogonal matrix. For those matrices, the square root claim is true. What do I mean by most (as opposed to all) such matrices can be diagonalised? Call a vector $$v$$ a null vector if $$v^T v = 0$$. The adapted Spectral Theorem states that as long as a complex symmetric matrix has no null eigenvectors, it must be diagonalisable by an orthogonal matrix. The adapted Spectral Theorem is in fact false for matrices which have null eigenvectors. I don't know how to prove the square root claim for those matrices.

Higham, in Functions of Matrices, Theorem 1.12, shows that the Jordan form definition is equivalent to a definition based on Hermite interpolation. That shows that the square root of a matrix $$A$$ (if based on a branch of square root analytic at the eigenvalues of $$A$$) is a polynomial in $$A$$. Therefore, if $$A$$ is symmetric so is its square root.