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.
 A: Let $\lambda$ be an eigenvalue of $A$ with eigenvector $v$.  What is $(A^2+A)v$?
A: If there were an eigenvector $v$ of $A$ with eigenvalue $1$, we'd have
$$(A^2 + A)v = A A v + Av = Av + v = v + v = 2v$$
On the other hand,
$$(A^2 + A)v = 0v = 0$$
A: There is only one thing you can do with an eigenvalue $\lambda$ for a matrix $A$ without having to delve into the complicated or subtle: say that there exists an eigenvector $v$ for $\lambda$
And there is essentially only one thing you can know about $v$:
$$ Av = \lambda v$$
Thus, unless you want to start searching for more complicated ways to start this problem, there is pretty much only one thing you can do with the claim "1 is an eigenvalue": assert the existence of an eigenvector $v$, and multiply things by $v$ and see if you can make use of the equation $Av = v$.
$A^2 + A = 0$ is important in this problem. It's essentially the only "interesting" object you've been presented with. Thus, before we try inventing our own things to multiply by $v$, we should try multiplying this by $v$ to see what happens.


*

*$ (A^2 + A)v = 0v $

*$ A(Av) + Av = 0$

*$ Av + v = 0$

*$v + v = 0$

*$2v = 0$

*$v = 0$


Thus, we tried doing just about the only thing we can do without trying to be clever or complicated, and we were rewarded with a very useful piece of information: if $v$ is an eigenvector for $1$, then $v=0$.
This contradicts the fact that eigenvectors are nonzero. Thus, the claim "1 is an eigenvector" leads to a contradiction.
A: 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 $A$
