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I have been working on a research problem and have encountered a situation in which two matrices $A, B \in \mathbb{R}^{n \times n}$ are such that \begin{equation} A^k \mathbf{1} = B^k \mathbf{1} \quad \forall 0 \leq k \leq n \end{equation} where $\mathbf{1}$ is the vector of ones. Can I now say anything about the relationship between $A$ and $B$? Obviously, no explicit relationship need exist between the two. If $\mathbf{1}$ were an eigenvector of both, then clearly the equations above hold. Suppose $\mathbf{1}$ weren't an eigenvector of at least one of $A,B$. Is it still possible for the relationships above to hold?

tgk1729
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  • How about $$A=\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix},~B=\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}.$$ You have $A\mathbf 1=\begin{pmatrix} 1 \ 1 \end{pmatrix}=B\mathbf 1$ and $B^2=A$. – Hirshy Jul 01 '16 at 12:11
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    Yes, of course. As I mentioned in my problem above, if $\mathbf{1}$ were an eigenvector of both, then the relationship above clearly holds. My question was that if at least one of them did not have $\mathbf{1}$ as an eigenvector, can something still be said about the two? Is it still possible for the relationship to exist? – tgk1729 Jul 01 '16 at 12:16
  • True, I missed that sentence in your question, sorry about that. – Hirshy Jul 01 '16 at 12:19

1 Answers1

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For your first question, this situation is similar to the following thread:

$A,B,C \in M_{n} (\mathbb C)$ and $g(X)\in \mathbb C[x]$ such that $AC=CB$- prove that $A^jC=CB^j$ and $g(A)C=Cg(B)$

Let $A,B\in M_n(F)$ for some algebraically closed field $F$. Suppose $v\in F^n$ is a nonzero vector such that $A^kv=B^kv$ for $k=1,2,\ldots,n$. Let $m_B(x)$ denotes the minimal polynomial of $B$. Then $m_B(A)v=0$. Since $v$ is nonzero, $m_B(A)$ must be singular. In turn, $A$ and $B$ must share an eigenvalue over $F$.

Anyway, in your case, $A,B$ are real. Therefore the two matrices share at least one real eigenvalue or at least one conjugate pair of complex eigenvalues.

Now, for your second question, the answer is yes. E.g. consider $$ A=\pmatrix{0&1&0&-1\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0}, \ B=\pmatrix{0&-1&0&1\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0}, \ v=\pmatrix{0\\ 1\\ 0\\ 1}. $$ Then $Av=Bv=(0,0,1,0)^\top$. Since $A^2=B^2=0$, we also have $A^kv=B^kv$ for $k=1,2,3,4$. Note that $v$ is not an eigenvector of the two matrices, because $A,B$ are nilpotent but $Av=Bv\ne0$. Now, by a change of basis, we may transform $v$ into $\mathbf 1$ and we may take the transformed $A$ and $B$ as examples.

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