Condition for $ \text{rank}(A) = \text{rank}(A^2) $ Let $ A $ be an $ n \times n $ complex matrix. Prove that $ \text{rank}(A) = \text{rank}(A^2) $ if and only if $ \lim_{\lambda \to 0} (A + \lambda I)^{-1}A $ exists.
I'm not exactly sure how to approach this sort of question. I've thought about putting $ A $ in Jordan form and working with Jordan blocks, but it seems messy.
 A: Your idea was correct. Essentially, you can put the matrix into lower triangular form with a nilpotent upper block, $N$, and an invertible lower block, $U$.  
Then $A=A^2$ if and only if $N=0$.
The question has now been reduced to proving that  $N=0$ if and only if $\lim_{\lambda \to 0} (N + \lambda I)^{-1}N$ exists. In one direction this is obvious!
Now $(N + \lambda I)^{-1}$ has elements $\frac {1}{\lambda }$ on the diagonal and then you can show that
 $ (N + \lambda I)^{-1}N$ will have elements which tend to infinity unless $N$ is  $0$. 
Hope that helps.
A: It is perhaps useful to notice that rank$(A)$=rank$(A^2)$ if and only if $\mathbb{C}^n=\ker(A)\oplus \text{Im}(A)$.
We can see that $ (A+\lambda I)(v)=\lambda v $ for all $v\in \ker(A)$, and thus $(A+\lambda I)^{-1}(v)=\frac{v}{\lambda}$.
So if $v\neq 0$ is in $\ker(A)\cap \text{Im}(A)$, we have that for $u\in A^{-1}[\{v\}]$, the limit does not exist.
We can write:
$$ (A+\lambda I)^{-1}A= I-\lambda(A+ \lambda I)^{-1} $$
If $ \mathbb{C}^n=\ker(A)\oplus \text{Im}(A) $, then:
$$\lim_{\lambda \to 0} (A+\lambda I)^{-1}A(v)=\begin{cases} 0 & v\in \ker(A) \\
v & v\in \text{Im}(A)  \end{cases} $$
And by linearity $\lim_{\lambda \to 0} (A + \lambda I)^{-1}A$ exists for all $v$. 
Edit: I see that the question already has an answer, but since it took some time to write it, I'll leave this answer.
