Prove that for each $n$-dimensional column vector $\xi$, the equation $A'(A^2+A)X=A'\xi$ has a nonzero solution.

Question

Suppose $A$ and $B$ are $n\times n$ matrices which satisfy $$ A=\big(B-\frac{1}{110}E\big)'\big(B+\frac{1}{110}E\big). $$ Prove that for each $n$-dimensional column vector $\xi$, the equation $A'(A^2+A)X=A'\xi$ has a nonzero solution.

It suffices to prove that the column space of $A'$ is contained in the column space of $A'(A^2+A)$, and the column rank of $A'(A^2+A)$ must not be full rank, since $A'(A^2+A)X=0$ should admit a nonzero solution. But I don't know how to prove it rigorously.

Answer 1

The column space of $A'(A^2+A)$ is included in the column space of $A'$ so you need to show that they are equal, or equivalently that the two matrices have same rank.

This also means that $A$ cannot be full rank, which does not strike me as obvious from its definition. Maybe you forgot to mention what $B$ and $E$ are?