I've been chewing on the following problem for some time and I just don't have any more ideas how to tackle it. I have matrices $A_1,...,A_k \in \mathbb R^{n\times n}$ and I'm observing that the spectrum of the product is the same whatever order of multiplication I choose, that is, for an arbitrary permutation $\pi$ on ${1,...,k}$, I have $$\sigma(A_1\cdot ...\cdot A_k) = \sigma(A_{\pi(1)}\cdot ...\cdot A_{\pi(k)}).$$ Of course, the form of the matrices is important since the above equality doesn't hold in general. All are of the form: $$ A_j = \left[ \array{0 & 1 & 0 & \cdots & 0 \\ 0 & \ddots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & 1 & 0 \\ 0 & \cdots & \cdots & 0 & 1 \\ a_j & \cdots & \cdots & \cdots & a_j } \right],$$ with possibly all different $a_j \in\mathbb{R},\ j=1,...,k$. I find this quite peculiar and I don't see the reason, although all tested examples confirm the statement... I would be very curious and grateful if someone has an idea how to solve that... Thanks ahead for your comments!
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1$\begingroup$ shouldn't $n$'s in the formula for $\sigma$ actually be $k$'s? $\endgroup$– Dima PasechnikCommented Oct 1, 2014 at 12:41
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$\begingroup$ Of course! Sorry! edited... $\endgroup$– LeoloCommented Oct 1, 2014 at 12:48
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1$\begingroup$ As Denis Serre pointed out, the claim is false when $(n,k)=(2,4)$. It is also false in some cases where $n>k$, such as $(5,4)$. However, when $n=k<9$, one can check that the characteristic polynomial is completely symmetric. I do not know whether that is still true for $n=k\geq 9$. $\endgroup$– Neil StricklandCommented Oct 1, 2014 at 16:51
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3$\begingroup$ This reminds me a lot of Fiedler companion matrices. If you haven't yet, you should definitely check out A note on companion matrices, M. Fiedler, 2003. It does not prove your conjecture (which seems to be false, according to the answers below), but it provides a family of matrices $\hat{A}_j$, not too different from your $A_j$'s, which have the property you are looking for (the spectrum of their product does not depend on the order). $\endgroup$– Federico PoloniCommented Oct 1, 2014 at 18:22
3 Answers
It seems that the claim is false when $n=2$ and $k=4$, because $${\rm Tr} (A_1A_2A_3A_4)=a_1a_3+a_2a_4+\sigma_3({\bf a})+\sigma_4({\bf a})$$ is not symmetric in ${\bf a}=(a_1,a_2,a_3,a_4)$.
Edit: just let me add a classical result. Let $C_1,\ldots,C_r$ be the companion matrices of unitary polynomials $P_1,\ldots,P_r$ of same degree $n$. Suppose that $\lambda$ is a common root to all the $P_j's$. Then $\lambda^r$ is an eigenvalue of the product $C_1\cdots C_r$.
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$\begingroup$ Thanks Dennis, this proves my conjecture wrong, of course. $\endgroup$– LeoloCommented Oct 10, 2014 at 10:59
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$\begingroup$ @Leolo. You know that France is a poor country, if not underdevelopped: here, Denis has only one N. $\endgroup$ Commented Oct 10, 2014 at 12:18
As suggested in Neil Strickland's comment, the claim seems to hold indeed for $n=k$. Inspection shows that the characteristic polynomial of the product can be written simply as $$\det(x-A_1\cdot ...\cdot A_n)=x^n-\sum_{i=1}^n (x-1)^{i-1} x^{n-i}s_i,$$ where $s_i$ is the $i$-th symmetric function of $a_1,…,a_n$. The proof of this should not be hard using some kind of induction.
E.g. for $n=4$, in expanded form it is $$x^4-(s_1+s_2+s_3+s_4)x^3+(s_2+2s_3+3s_4)x^2-(s_3+3s_4)x+s_4.$$
Write $diag(\lambda_1,\dots,\lambda_n)$ for the diagonal matrix with entries $\lambda_1,\dots,\lambda_n$. The matrix in question as $J(a_j)=diag(1,\dots,1,a_j)P$ for a fixed matrix $P$. For a given matrix $A$, each coefficient of the characteristic polynomial $\det(x-diag(1,\dots,a)A)$ is a polynomial in $a$ of degree $\le 1$. Now each coefficient of the characteristic polynomial $$ \det(x-J(a_1)\cdots J(a_k)) $$ is a polynomial in the $a_j$ which is invariant under cyclic permutations and is of degree $\le 1$ in each variable $a_j$. Each such polynomial must be invariant under the whole permutation group.
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$\begingroup$ The matrix $A_j$ in the question has the entire bottom row equal to $a_j$, whereas your solution seems to only have one entry in the bottom row equal to $a_j$, and the rest $0$, right? $\endgroup$ Commented Oct 1, 2014 at 15:12
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1$\begingroup$ The expression $a_1a_3+a_2a_4$ from Denis's answer provides a counterexample to your final claim. (I take it by "degree $1$" you mean in each of the $a_j$'s separately, if you treat the others as constants.) $\endgroup$ Commented Oct 1, 2014 at 16:06