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](https://mathoverflow.net/a/438232/497608) for the answer to this question. I am little confused at the reasoning given in the answer however, specifically in two places.


1. 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 $A$.
2. 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?



Questions regarding answer to complex symmetric square root of a complex symmetric invertible matrix
$\newcommand\la\lambda$


1. Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la\_j)=g^{(k)}(\la\_j)$ for each $j$ and all $k=0,\dots,m\_j-1$, where the $\la\_j$'s are the distinct eigenvalues of $A$ and the $m\_j$'s are their multiplicities. Then, by the [Leibniz rule](https://en.wikipedia.org/wiki/General_Leibniz_rule), $Q^{(k)}(\la\_j)=0$ for each $j$ and all $k=0,\dots,m\_j-1$.
2. So, for some polynomial $R(z)$,
\begin{equation}
 Q(z)=R(z)\prod\_j(z-\la\_j)^{m\_j}
=R(z)C(z), \tag{1}\label{1}
\end{equation}
where $C(z):=\prod\_j(z-\la\_j)^{m\_j}$ is the characteristic polynomial of $A$ (see details on the first equality in \eqref{1} below). So, the polynomial $Q(z)$ is indeed a multiple of the characteristic polynomial of $A$.


(The [previous answer](https://mathoverflow.net/a/438232/36721), referred to in your question, does not even mention indices.)




---


**Details on the first equality in \eqref{1}:**



> 
> **Lemma 1:** Let $q(z)$ be a polynomial such that $q^{(k)}(\la)=0$ for some $\la$ and all $k=0,\dots,m-1$, where $m$ is an integer $\ge1$. Then $(z-\la)^m$ is a divisor of the polynomial $q(z)$ -- that is,
> $q(z)=(z-\la)^m S(z)$ for some polynomial $S(z)$.
> 
> 
> 


*Proof of Lemma 1:* By shifting, without loss of generality $\la=0$.
Divide $q(z)$ by $z^m$ with a remainder $r(z)$ of degree $\le m-1$, so that $q(z)=z^m s(z)+r(z)$ for some polynomial $s(z)$. Then for all $k=0,\dots,m-1$ we have $0=q^{(k)}(0)=r^{(k)}(0)$. Because $r(z)$ is of degree $\le m-1$, it follows that $r(z)$ is the zero polynomial, so that $q(z)=z^m s(z)$, which completes the proof of Lemma 1. $\quad\Box$


Now, by part 1 of the answer, $Q^{(k)}(\la\_j)=0$ for each $j$ and all $k=0,\dots,m\_j-1$. So, for each $j$, by Lemma 1, the polynomial $(z-\la\_j)^{m\_j}$ is a divisor of the polynomial $Q(z)$. Also, the polynomials $(z-\la\_j)^{m\_j}$ are coprime for different values of $j$, since the $\la\_j$'s are distinct. So, $\prod\_j(z-\la\_j)^{m\_j}$ is a divisor of $Q(z)$; that is, the first equality in \eqref{1} holds.


It remains to provide


