Is $\mbox{Rank}(A + A^2) \leq \mbox{Rank} (A)$?

Question

Here, $A$ is an $n \times n$ matrix. I am not able to find any counterexample but not able to prove this as well. The examples I have tried so far shows me that $\mbox{Rank} (A + A^2) = \mbox{Rank} (A)$. I don't know how to show the inequality.

Answer 1

$\DeclareMathOperator{\rank}{rank}$ $\DeclareMathOperator{\Img}{Im}$

To show the inequality, note that $\rank(M)=\dim(\Img(M))$. Since $\Img(A + A^2)$ is a subspace of $\Img(A)$, the dimension inequality $\dim(\Img(A + A^2))\leq\dim(A)$ holds.

Note that equality doesn't always hold. Consider $A=-I$ where $I$ is the identity. Then $A + A^2 = 0$. Assuming $n > 0$, we have $\rank(A+A^2)=0<n=\rank(A)$.