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

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?

• if $A$ is invertible the logarithm of $A$ exists, and a square root is $e^{\tfrac{1}{2}\log A}$; this is symmetric because $(e^B)^\top=e^{B^\top}$ and $(\log A)^\top=\log A^\top$. Jun 8 at 13:59

$$\newcommand\la\lambda$$

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.)

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

• Thank you for your help in answering these questions. I now fully understand the issue raised in part 1. I have some questions on part 2 though and I don't know if I should "ask a new question" or just write each of them in their own comment in this section? I would like to include my own thoughts and I don't know if the allowable space here is conducive to that? On the other hand, maybe it is better to keep them all together? What do people think? Jun 9 at 10:18
• @jcb2535 : I suggest you just tell me what still seems unclear about part 2. If doing this requires a substantial write-up, perhaps you can create a link to it. Jun 9 at 15:26
• I will write a short version of the questions here and see if it suffices. 1. Why is $n_j \geq m_j$? How can we know this for certain? 2. We obtain that $\lambda_j$ are the roots of $Q(z)$ but why do we need to show that the derivatives of $Q(t)$ equal 0 at each $\lambda_j$ for this? Is showing that $Q(\lambda_j) = 0$ for each $\lambda_j$ not enough? 3. Where does $R(z)$ come from? Since we know the roots and multiplicities of $Q(z)$, shouldn't $R(z)$ not be a constant instead of a polynomial? Jun 9 at 15:58
• @jcb2535 : I have added the requested details. Jun 9 at 17:39
• Previous comment continued: Anyhow, I do not want to repeat that anymore. Instead, I have tried to simplify the proof, at the same time providing more details. If anything is still unclear, just let me know what is the first logical transition that is unclear to you. Try not to ask me why I did or did not do something (sometimes, this is hard or impossible to explain). Instead, ask me only why a logical conclusion is true (if you have doubts about it). Jun 12 at 17:56

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.

• Thanks very much! Yes it is a very good book for this topic. The explanations are clear and they give some examples which really helps with understanding. Jun 9 at 13:23