A problem of matrix polynomial expansion

Question

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!

Answer 1

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 $J^r=2^{-r} (a+c)^r\left( \begin{array}{cc} 1 & 2 r (a+c)^{-1} \\ 0 & 1 \\ \end{array} \right)$.

Answer 2

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*}