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: This is a bit of a brute force approach, but it's effective. Note that the sum of the reverse diagonal elements of a $2\times 2$ matrix $M$ equals ${\rm tr}\,\sigma M$ with
$$\sigma=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$
For the most general form of the matrix
$$A=\begin{pmatrix}a&b\\ -b&c\end{pmatrix},\;\;\text{and for}\;\;D=\mathbf c\mathbf b^{\rm T}=\begin{pmatrix}1&-1\\1&-1\end{pmatrix},$$
I calculate
$$J(r,n)={\rm tr}\,\sigma A^r DA^{n-1-r}=$$
$$=\frac{2^{-n-1} (a+c-z)^{-r} (a+c+z)^{-r}}{(a-2 b-c) \left(a c+b^2\right)} \left[\left(z (a+c)-(a-c)^2+4 b^2\right) (a+c+z)^n (a+c-z)^{2 r}-\left(z (a+c)+(a-c)^2-4 b^2\right) (a+c-z)^n (a+c+z)^{2 r}\right],$$
with the definition $z=\sqrt{(a-c)^2-4 b^2}$.
Then I evaluate for $n\geq 1$ the sum
$$\sum_{r=0}^{n-1}J(r,n)=\frac{2^{-n-1} (a+c) \left((a-c)^2-4 b^2-z^2\right) \left((a+c-z)^n-(a+c+z)^n\right)}{z (a-2 b-c) \left(a c+b^2\right)}.$$
Substitution of the definition of $z$ finally gives the desired result
$$\sum_{r=0}^{n-1}J(r,n)=0.$$


Details of the calculation: I may assume $b\neq 0$ (otherwise $A$ is diagonal and the identity follows trivially). Then the matrix $A$ is diagonalizable when $b\neq \tfrac{1}{2}|a-c|$, in the form
$A=U\Lambda U^{-1}$ with $$U=\left(
\begin{array}{cc}
 z-a+c & -z-a+c \\
 2 b & 2 b \\
\end{array}
\right),\;\;\Lambda={\rm diag}\,\left(\tfrac{1}{2} \left(-z+a+c\right),\tfrac{1}{2} \left(z+a+c\right)\right)$$
With this decomposition we can readily evaluate $A^r=U\Lambda^r U^{-1}$.
If $b=\tfrac{1}{2}(a-c)\neq 0$ we instead use the Jordan decomposition $A=VJV^{-1}$ with
$$V=\left(
\begin{array}{cc}
 -1 & -\frac{2}{a-c} \\
 1 & 0 \\
\end{array}
\right),\;\;J=\left(
\begin{array}{cc}
 \frac{a+c}{2} & 1 \\
 0 & \frac{a+c}{2} \\
\end{array}
\right).$$
Then $A^r=VJ^r V^{-1}$, with $J^r=2^{-r} (a+c)^r\left(
\begin{array}{cc}
 1 & 2 r (a+c)^{-1} \\
 0 & 1 \\
\end{array}
\right)$.

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*}
