A problem of matrix polynomial expansion 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: Let $A=\pmatrix{a&-b\\ b&c},\,B=\pmatrix{1&-1\\ 1&-1}$ and $R=\pmatrix{0&1\\ 1&0}$. It is straightforward to verify that
\begin{cases}
RB=-BR=B,\\
\operatorname{tr}(BR)=\operatorname{tr}(RB)=0,\\
\operatorname{tr}(ABR)+\operatorname{tr}(BAR)
=\operatorname{tr}(ABR)+\operatorname{tr}(ARB)
=\operatorname{tr}(-AB)+\operatorname{tr}(AB)=0,\\
\operatorname{tr}(ABAR)
=\operatorname{tr}\pmatrix{-(a-b)(b+c)&(a-b)^2\\ -(b+c)^2&(a-b)(b+c)}=0.\\
\end{cases}
By Cayley-Hamilton theorem, the powers of $A$ are linear combinations of $A$ and $I$. Let $A^k=pA+qI$ and $A^{n-1-k}=rA+sI$. Then
\begin{align}
&\operatorname{tr}(A^kB A^{n-1-k}R+A^{n-1-k}B A^kR)\\
&=\operatorname{tr}\left((pA + qI) B (rA + sI) R + (rA + sI) B (pA + qI) R\right)\\
&=2pr\operatorname{tr}(ABAR) + (ps+qr)\operatorname{tr}(ABR + BAR) + 2qs\operatorname{tr}(BR)\\
&=0.\tag{1}
\end{align}
In particular, when $n-1=2k$, we have $\operatorname{tr}(2A^kBA^kR)=0$ and hence
$$
\operatorname{tr}\left(A^{(n-1)/2}BA^{(n-1)/2}R\right)=0.\tag{2}
$$
It follows from $(1)$ and $(2)$ that $\sum_{k=0}^{n-1}A^kBA^{n-1-k}R$ has zero trace. Since the trace of $\sum_{k=0}^{n-1}A^kBA^{n-1-k}R$ is precisely the sum of the two anti-diagonal elements of $\sum_{k=0}^{n-1}A^kBA^{n-1-k}$, the conclusion follows.
A: 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*}
