Show that if $(A+2I)^2=0$, then $A+\lambda I$ is invertible for $\lambda \ne 2$. 
Show that if $(A+2I)^2=0$, then $A+\lambda I$ is invertible for $\lambda \ne 2$.

I tried to solve this by treating $(A+\lambda I)v=0$ as linear equation system, and proving that $v$ must be $0$ (trivial solution) therefore $A+\lambda I$ echelon form is I and it's invertible..
Would love to hear another solutions, and tips to my own proof:
$$A+\lambda I=A+2I+(\lambda-2) I.$$
$$(A+\lambda I)v=0 \Rightarrow (A+2I)v+(\lambda-2)Iv=0.$$
$$(A+2I)(A+2I)v+(A+2I)(\lambda-2)Iv=0 \Rightarrow (\lambda-2)(A+2I)Iv=0 \Rightarrow (A+2I)Iv=0$$ (multiply by $A+2I$ and we know $\lambda \ne 2$).
$$(A+2I)v+(\lambda-2)Iv=0 \wedge (A+2I)Iv=0\Rightarrow (λ−2)Iv=0 \Rightarrow v=0$$
Therefore if v is solution for $(A+\lambda I)v=0$ it must be 0.
(We have also shown that for $\lambda = 2$, $v$ can be $(A+2I)$ which shows $A+2I$ isn't invertible)
BTW If A is such that $(A+2I)^2=0$, prove that $A+I$ is invertible. solves the basic case for $\lambda = 1$
 A: Below is a more general statement.  I have provided three solutions to your question.  The first two use the proposition, and the last one is as you requested. 

Proposition. Let $n$ be positive integer and $A$ an $n$-by-$n$ matrix over a field $\mathbb{K}$.  Write $I$ for the $n$-by-$n$ identity matrix.  For $\mu\in K$, the matrix $A-\mu\, I$ is not invertible if and only if $\mu$ is an eigenvalue of $A$ (or equivalently, $\det(A-\mu\, I)=0$, or $\ker(A-\mu\,I)\neq  \{0\}$). 

Proof.  For each $\mu\in \mathbb{K}$, let $V_\mu\in \mathbb{K}^n$ denote the kernel (i.e., the nullspace) of $A-\mu\,I$.  If $A-\mu\,I$ is invertible, then for any $v\in V_\mu$, we have $(A-\mu\,I)\,v=0$ and so $$v=(A-\mu\,I)^{-1}(0)=0\,,$$ implying that $V_\mu=\{0\}$.  Conversely, if $V_\mu=\{0\}$, then $A-\mu\,I$ is an injective linear map from $\mathbb{K}^n$ to itself.  It is well known that any injective linear map on a finite-dimensional vector space is also surjective, whence bijective and so invertible.  (This well known result is not true for infinite-dimensional vector spaces, by the way.)  Therefore, $A-\mu\,I$ is invertible. 

First Solution.
In your case, the only eigenvalue of $A$ is $-2$.  Thus, $A-\mu\,I$ is invertible if and only if $\mu\neq -2$, which is equivalent to saying that $A+\lambda\, I$ is invertible if and only if $\lambda\neq 2$.  

Second Solution.
We shall prove that $\ker(A+\lambda\,I)=\{0\}$ when $\lambda\neq 2$.  Suppose $v\in \ker(A+\lambda\,I)$.  Then, $(A+\lambda\,I)\,v=0$, so $Av=-\lambda\,v$ and $A^2v=A(Av)=A(-\lambda\,v)=-\lambda\,(Av)=-\lambda\,(-\lambda v)=\lambda^2\,v$.  Since $(A+2\,I)^2=0$, we also have $(A+2\,I)^2\,v=0$.  Therefore, $$\lambda^2\,v-4\,\lambda\,v+4\,v=A^2v+4\,Av\,+4\,v=0\,.$$
Ergo, $(\lambda-2)^2\,v=0$.  Since $\lambda\neq 2$, $v=0$, which implies $\ker(A+\lambda \,I)=\{0\}$.

Third Solution.
Since $\lambda \neq 2$, we have
$$\frac{(x+2)^2-(x+4-\lambda)(x+\lambda)}{(\lambda-2)^2}=1\,,$$
where $x$ is a dummy variable.  Therefore, $$\frac{(A+2\,I)^2-\big(A+(4-\lambda)\,I\big)\,(A+\lambda \, I)}{(\lambda-2)^2}=I\,.$$
As $(A+2\,I)^2=0$, we get
$$\left(-\frac{1}{(\lambda-2)^2}\,\big(A+(4-\lambda)\,I\big)\right)\,(A+\lambda\,I)=I\,.$$  Thence, $A+\lambda\,I$ is invertible and
$$(A+\lambda\,I)^{-1}=-\frac{1}{(\lambda-2)^2}\,\big(A+(4-\lambda)\,I\big)\,.$$
A: Have you heard about eigenvalues? YOur equation for $A$ shows that $-2$ is the only eigenvalue of A(because any eigenvalue would satisfy $(x+2)^2=0$, the equation satisfied by $A$), so no other $\lambda$ could be an eigenvalue for $A$, qed.
