How prove this $A=\left(B-\frac{1}{110}I\right)^T\left(B+\frac{1}{110}I\right)$

Question

let $A_{n\times n},B_{n\times n} $ real matrix ,and such $$A=\left(B-\dfrac{1}{110}I\right)^T\left(B+\dfrac{1}{110}I\right)$$ show that:

For any n-dimensional column vector $\xi\neq 0$,then the equation exsit $$A^T(A^2+A)X=A^T\xi$$ nonzero root.

My try: since $$A=\left(B-\dfrac{1}{110}I\right)^T\left(B+\dfrac{1}{110}I\right)=B^TB-\dfrac{1}{12100}E+\dfrac{B^T}{110}-\dfrac{B}{110}$$ so $$A+E=B^TB+\dfrac{12099}{12100}E+\dfrac{1}{110}(B^T-B)$$ then I can't,Thank you

There must be some other constraints. If we let $B=I$ then $A=\frac{12099}{12100}I$ and $A^T\xi=0\implies\xi=0$. — robjohn, Dec 23 '14 at 14:28
Why do you ask for nonzero roots? It could very well happen that $X=0$ is not a root. — Mizar, Dec 23 '14 at 14:38
@chinamath: evidently, I misread your intent. You mentioned roots rather than solutions, and I thought you were looking for $A^T\xi=0$ (the roots of $A^T\xi$). Instead, you are looking for solutions $X$ to $A^T(A^2+A)X=A^T\xi$ given a $\xi\ne0$. — robjohn, Dec 23 '14 at 15:53

Daniel · Accepted Answer · 2014-12-23T21:19:06.920

Let's prove that $\ker(A^t(A^2+A))=\ker(A)$. Thus, $rank(A^t(A^2+A))=rank(A)=rank(A^t)$. Since $\Im(A^t(A^2+A))\subset\Im(A^t)$ then $\Im(A^t(A^2+A))=\Im(A^t)$ and the result follows.

First, $A=(B-\frac{1}{110}I)^t(B+\frac{1}{110}I)=B^tB+\frac{1}{110}(B^t-B)-\frac{1}{110^2}I$.

Notice that if $v\in \mathbb{R^n}$ then $v^t(B^t-B)v=0$, since $v^tB^tv=v^tBv$.

Assume $v^tv=1$. Therefore, $$v^tAv=v^t(B^tB+\frac{1}{110}(B^t-B)-\frac{1}{110^2}I)v=v^tB^tBv-\frac{1}{110^2}\geq -\frac{1}{110^2}>-1.$$

Thus, $-1$ can not be an eigenvalue of $A$, otherwise would exist $v\in \mathbb{R^n}$ such that $v^tv=1$ and $v^tAv=-1$.

Now, if $A^tA(A+I)w=0$ then $w^t(A+I)^tA^tA(A+I)w=0$ and $A(A+I)w=0$. Thus, $A(Aw)=-(Aw)$. Since $-1$ is not an eigenvalue of $A$ then $Aw=0$. Therefore $w\in\ker(A)$. So $\ker(A^t(A^2+A))\subset\ker(A)$. it is obvious that $\ker(A)\subset\ker(A^t(A^2+A))$. So $\ker(A^t(A^2+A))=\ker(A)$.

How prove this $A=\left(B-\frac{1}{110}I\right)^T\left(B+\frac{1}{110}I\right)$

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