The problem is


$b = (1, -1)^\top, c = (1, 1)^\top, A \in \mathbb{R}^{2 \times 2}$, suppose the sum of reverse diagonal elements of $A$ is zero (i.e., $A\_{12} + A\_{21} = 0$), prove that the sum of reverse diagonal elements of $\sum\limits\_{r=0}^{n-1} A^r c b^\top A^{n-1-r} $ is zero for any $n \in \mathbb{N}^{+}$.


In fact, this is my conjecture and I have tested many examples in my computer. For diagonal case, it is easy to prove it, but for the general case I do not know how to do it. One idea come to my mind is to write $A$ as the sum of a diagonal matrix and an anti-diagonal matrix, then expand $A^r$ by binomial expansion, but unfortunately they do not commute.


Could someone give me some hints?


Thanks!



A problem of matrix polynomial expansion
Someone told me a simple method, I decide to post it here.


Note that for any $A \in \mathbb{R}^{2 \times 2}$, $A\_{12} + A\_{21} = 0$ if and only if


\begin{equation\*}
 A^\top = \sigma^{-1} A \sigma
\end{equation\*}
with
\begin{equation\*}
 \sigma = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\end{equation\*}


Let
\begin{equation\*}
 J = \sum\limits\_{r=0}^{n-1} A^r c b^\top A^{n-1-r} 
\end{equation\*}
then when $A\_{12} + A\_{21} = 0$ we have
\begin{align\*}
 J^\top & = \sum\limits\_{r=0}^{n-1} \left(A^\top\right)^{n-1-r} b c^\top \left(A^\top\right)^{r} \\
 & = \sigma^{-1} \sum\limits\_{r=0}^{n-1} A^{n-1-r} c b^\top A^{r} \sigma \\
 & = \sigma^{-1} J \sigma
\end{align\*}


Therefore
\begin{equation\*}
 J\_{12} + J\_{21} = 0
\end{equation\*}


