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 $\lambda = 1$

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.
