Why is the square root of a complex symmetric matrix also complex symmetric

Question

I am looking at trying to show that a complex symmetric matrix always has a complex symmetric square root. Showing a square root exists is fairly easy if the matrix is also invertible by using the Jordan Canonical Form.

I have seen on here that showing that the square root of a matrix A is a (Hermite) polynomial in A proves that if A is symmetric then so is its square root. My question is, why is this true?

The reference for this would be Function of Matrices, Defn 1.2 (Matrix Function using Jordan Canonical Form) and Defn 1.4 (Matrix Function using Hermite Interpolation) and Theorem 1.12 (which shows that the two definitions given are equivalent).

Answer 1

If your complex symmetric matrix $A$ is not invertible, it might not have a square root at all, e.g. $$ \pmatrix{i & 1\cr 1 & -i\cr}$$ If $A$ is invertible, let $\lambda_j$ be the eigenvalues of $A$ and $m_j$ their multiplicities. Let $P(z)$ be a polynomial such that $P(z)$ and the first $m_j-1$ of its derivatives agree with some branch of $\sqrt{z}$ and the first $m_j-1$ of its derivatives at each $\lambda_j$. Then the polynomial $P(z)^2 - z$ and the first $m_j - 1$ of its derivatives are $0$ at $\lambda_j$, implying that $P(z)^2 - z$ is divisible by the characteristic polynomial of $A$, and so the Cayley-Hamilton theorem implies $P(A)^2 = A$, i.e. $P(A)$ is a square root of $A$.

Of course if $A$ is symmetric, $P(A)$ is symmetric.