Questions regarding answer to complex symmetric square root of a complex symmetric invertible matrix

Question

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.

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?

Answer 1

$\newcommand\la\lambda$

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, $Q^{(k)}(\la_j)=0$ 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, referred to in your question, does not even mention indices.)

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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

Answer 2

This technique of computing matrix functions via interpolating polynomials is described in more detail in Chapter 1 of this book.

Higham, Nicholas J., Functions of matrices. Theory and computation, Philadelphia, PA: Society for Industrial and Applied Mathematics (SIAM) (ISBN 978-0-898716-46-7/hbk; 978-0-89871-777-8/ebook). xx, 425 p. (2008). ZBL1167.15001.

I suggest you check it out if you are interested in knowing more; the same chapter has also more insight on the different square roots that can be defined for a matrix.