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I am following a proof of the next theorem, rephrased from Theorem 8.5 of the book "Introduction to the theory of differential inclusions" by Georgi V. Smirnov.

Let $\{\lambda_1,\dots,\lambda_n\}$ be eigenvalues of a constant matrix $A \in \mathbb{R}^{n\times n}$ and $X \in \mathbb{R}^{n\times n}$ be a variable in the set of symmetric matrices. Consider a linear system $XA + A^TX = 0$. It is equivalent to $Bx = 0$ where $x = (X_{11},X_{12},X_{22},\dots,X_{1n},\dots,X_{nn}) \in \mathbb{R}^{n(n+1)/2}$. Define \begin{align} C &= \{\text{all eigenvalues of $B$}\}, \\ D &= \{m_1\lambda_1 + \cdots + m_n\lambda_n \mid m_i \in \{0,1,2\},\,m_1 + \cdots + m_n = 2\}. \end{align} Then, C = D.

I understood the part $D \subset C$ but I am stuck in the part $C\subset D$. If $|D| = n(n+1)/2$, then it is obvious but in the case $|D| < n(n+1)/2$ the book says only "In the general case the result can be obtained taking the limit". Would you give me any hint or reference how to use the limit?

Edit

I hope this is a correct proof of the part $C \subset D$ that uses the limit.

Since $D\subset C$ and $|C|$ is at most $n(n+1)/2$, if $|D| = n(n+1)/2$, then $D = C$. Suppose $|D| < n(n+1)/2$. Consider a linear system $X(A + dA) + (A + dA)^TX = 0$ where $dA \in \mathbb{R}^{n\times n}$. It is equivalent to $(B + dB)x = 0$ and $\|dA\| \to 0 \Leftrightarrow \|dB\| \to 0$. Let $\{\lambda_1'(dA),\dots,\lambda_n'(dA)\}$ and $C = \{\rho_1,\dots,\rho_{n(n+1)/2}\}$ be eigenvalues of $A + dA$ and $B$, respectively. Define \begin{align} C' &= \{\rho_1'(dB),\dots,\rho_{n(n+1)/2}'(dB)\} = \{\text{all eigenvalues of $B + dB$}\}, \\ D' &= \{m_1\lambda_1'(dA) + \cdots + m_n\lambda_n'(dA) \mid m_i \in \{0,1,2\},\,m_1 + \cdots + m_n = 2\}. \end{align} Solutions of the characteristic equations $\det(A + dA - \lambda I) = 0$ and $\det(B + dB - \rho I) = 0$ are continuous functions with respect to parameters. So, for all $\varepsilon > 0$, there exists $\delta > 0$ such that $\max_i |\lambda_i'(dA) - \lambda_i| < \varepsilon$ and $\max_i|\rho_i'(dB) - \rho_i| < \varepsilon$ if $\|dA\| < \delta$ and $\|dB\|< \delta$; we assume that eigenvalues are properly listed. Also, there exists $\|dA\| < \delta$ such that all eigenvalues of $A + dA$ are distinct. Furthermore, there exists $\|dA\| < \delta$ satisfying $|D'| = n(n+1)/2$. Let $\rho \in C$. There exists $\rho' = m_1\lambda_1(dA) + \cdots + m_n\lambda_n(dA) \in C'$ such that \begin{equation} |\rho - (m_1\lambda_1 + \cdots + m_n\lambda_n)| \le |\rho - \rho'| + |\rho' - (m_1\lambda_1 + \cdots + m_n\lambda_n)| < (1 + 2n)\varepsilon. \end{equation} Since $\varepsilon$ is arbitrary, $\rho = m_1\lambda_1 + \cdots + m_n\lambda_n \in D$.

flyingwith
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    The question is not clearly stated. If both $A$ and $X$ are given, then $XA + A^TX = 0$ is either true of (more likely) false, but it is not a linear system. Apparently $X$ is not really given but considered as a collection of indeterminates; given the imposed symmetry only some of the entries of $X$ can be chosen independently. I gather from the expression for $x$ that maybe the "weakly above diagonal" part of $X$ is chosen for this (and so $B$ has size $\binom{n+1}2$, which was not at all clear to me on first reading). – Marc van Leeuwen Jun 04 '17 at 04:05
  • @MarcvanLeeuwen There is nothing unclear about this question. $XA+A^T X=0$ is a linear system as $X$ is a collection of $n^2$ indeterminants. If the system $Bx=0$ with $x\in\mathbb{R}^{n^2}$ is equivalent to $XA+A^T X=0$, then the matrix $B$ will be given in terms of entries of $A$ as a $n^2\times n^2$ matrix. – Sungjin Kim Jun 04 '17 at 17:09
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    Saying "[let] $X \in \mathbb{R}^{n\times n}$ be symmetric" means you assume it is one, symmetric, matrix (though it is not stated which one), not a whole family; this is just like "$A \in \mathbb{R}^{n\times n}$" means that $A$ is given. Also, because of the symmetry imposed on $X$, there is no way there can be $n^2$ independent unknowns. I can more or less guess what is meant, I was just commented that this is not the right way to state the question. – Marc van Leeuwen Jun 05 '17 at 19:27
  • @MarcvanLeeuwen I get your point. Actually, I solved this without assuming that $X$ is symmetric. I think the assumption is needless. – Sungjin Kim Jun 05 '17 at 20:15
  • @i707107 The assumption is necessary. Otherwise, $|D| < |C|$ is possible. – flyingwith Jun 06 '17 at 06:33
  • Can you show me the example with $|D|<|C|$? – Sungjin Kim Jun 06 '17 at 06:37
  • @i707107 I was wrong. At my first thought, if $X$ is not symmetric, then $B \in \mathbb{R}^{n^2}$, so there could be at most $n^2$ distinct eigenvalues of $B$, while $|D|$ is at most $n(n+1)/2$, so $|D| < |C|$ is possible. However, I checked from the book that the assumption of symmetry is not necessary because it uses a quadratic form that can be always possible to make the matrix symmetric. – flyingwith Jun 06 '17 at 07:24
  • @MarcvanLeeuwen I modified the question. I hope that it is clear now. – flyingwith Jun 06 '17 at 07:26
  • @flyingwith Your proof of $C\subset D$ looks okay to me. That is probably what the author meant by "taking the limit". In my answer, I did not use $X$ is symmetric. This means that whenever there is a "nonzero" matrix $X$ satisfying $X(A-\lambda I) +A^T X= 0$, there are eigenvalues $\lambda_i, \lambda_j$ of $A$ such that $\lambda-\lambda_i = \lambda_j$. Of course, the converse is also true. That is what I meant by "X being symmetric is not necessary". – Sungjin Kim Jun 06 '17 at 20:32
  • @i707107 Thanks a lot for your all replies. You also let me know a very interesting point that symmetry of X is not necessary. One day I may study abstract algebra and understand your proof. – flyingwith Jun 07 '17 at 08:51

1 Answers1

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Let $A$ be a $n\times n$ matrix. Define $M^A$ the module over $\mathbb{R}[x]$ as $\mathbb{R}^n$ equipped with $x\cdot v = A v$ for any $v\in\mathbb{R}^n$.

By the structure of finitely generated module over principal ideal domain, we have $$ M^A = \oplus \mathbb{R}[x]/(f_i^{r_{i s_i}}) \ \ (*) $$ with some irreducible polynomials $f_i$ and integers $r_{i s_i}\geq 1$ that satisfy $\sum_{i=1}^r\sum_{j=1}^{s_i} r_{i j}=n$.

Now, if $\lambda$ is an eigenvalue of $B$, then there is a nonzero matrix $X$ such that $$ X(A-\lambda I) + A^T X = 0 \ \ (1) $$ where $I$ is $n\times n$ identity matrix.

The solution set for the equation $(1)$ can be described as a group of module homomorphisms $$ \mathrm{Hom}_{\mathbb{R}[x]} (M^{\lambda I - A}, M^{A^T}). \ \ (2) $$

It is well-known that $A$ and $A^T$ is similar. Thus, $(2)$ is isomorphic to

$$ \mathrm{Hom}_{\mathbb{R}[x]} (M^{\lambda I - A}, M^{A }). \ \ (3) $$ From the structure $(*)$, the group of homomorphism $(3)$ has a nontrivial element if and only if $$ \lambda - \lambda_i = \lambda_j $$ where $\lambda_i, \lambda_j$ are eigenvalues of $A$. Therefore, we have $C=D$.

Sungjin Kim
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  • Thanks for your answer. Unfortunately, I cannot understand this proof because I don't have knowledge of abstract algebra. Nevertheless, it is interesting to see that there's a proof that is quite different to the one that I studied. That proof shows $D \subset C$ by constructing $X \neq 0$ that makes $X(A - \lambda I) + A^TX = 0$. For the part $C \subset D$, it only says we can prove it by taking the limit but I cannot catch the point how to use limit. I just guess continuity of determinant might be used. – flyingwith Jun 05 '17 at 06:55
  • @flyingwith I'd like to see the proof of $D\subset C$. Can you tell me which textbook it was? – Sungjin Kim Jun 05 '17 at 20:36
  • I put the title of the book to the question. – flyingwith Jun 06 '17 at 07:11