If $A^2+A=0$,then $\lambda=1$ cannot be an eigenvalue of A. Prove the following statement:
If $A^2+A=0$,then $\lambda=1$ cannot be an eigenvalue of A.

I've been struggling on this question for a couple of hours and don't know how to approach it.
Let $\lambda$ be an eigenvalue of $A$ with eigenvector $v$.  What is $(A^2+A)v$?
I believe this may be a solution as well...

$A^2+A=0$

$A(A+I_n)=0$

$\therefore$ $A=[0]$  or  $A=-I_n$

Since $A=0$ and $A=-I_n$ are both diagonal matrices, the possible eigenvalues for $A$ are simply the entries of the diagonal.

$\therefore$ the possible eigenvalues of $A$ are:

$\lambda=0$, and $\lambda=-1$

Therefore, it is not possible for $\lambda=1$ to be an eigenvalue of $A$