I am looking at showing that a complex symmetric invertible matrix always has a complex symmetric square root and I refer to this Q&A for the answer to this question. I am little confused at the reasoning given in the answer however, specifically in two places.

Firstly, how can we show that the polynomial $Q(z) := P(z)^2 -z$ and its $m_j-1$ derivatives are all $0$ at each $\lambda_j$? I have done out a the first few examples which all work, but how do we know this in general? Note that $\lambda_j$ are the eigenvalues of some $n \times n$ complex matrix $A$ and $m_j$ are the indices of $\lambda_j$ in the Jordan Canonical Form of $A$.

Once we have established the result in (1) above, the answer to the OP says that this implies that $Q(z)$ is a multiple of the characteristic polynomial of $A$, denoted here by $C_A(z)$.

My understanding for (2) is this: For $Q(z)$ and $C_A(z)$ to be a multiplicative constant, they both must have the same roots and these roots must have the same multiplicities. Both $C_A(z)$ and $Q(z)$ have $\lambda_j$ as their roots, this is obvious. Initially, I thought that the derivatives of $Q(z)$ up to order $m_j-1$ being $0$ implied something about the multiplicities of each of the roots but now I don't think that is true because the multiplicity and the index aren't necessarily the same.

Since then I have hit a bit of a wall. Does anyone have any insight into these two problems?