If A is such that $(A+2I)^2=0$, prove that $A+I$ is invertible.

Question

I'm stuck on the second question of this matrix exercise:

Consider a matrix $A$ $\in$ $\mathcal{M}_n(K)$ such that $(A+2I)^2=0$, where $I$ is the identity matrix.

1. Prove that $A$ is invertible and describe the $A^{-1}$ matrix as a function of $A$.

2. Prove that $(A+I)$ is invertible.

Solution for 1.:

$(A+2I)^2=0$ $\Rightarrow$

$\Rightarrow A^2 + 4AI +4I^2 = 0 \Rightarrow$

$\Rightarrow A(A+4I) = -4I \Rightarrow$

$A(-\frac {A+4I}{4}) = I$, therefore A is invertible, and $A^{-1} = -\frac 14A-I.$

I've tried pretty much every replacement I could think of, but no matter what, I couldn't come to the form $(A+I)B=I$.

Many thanks in advance.

Answer 1

\begin{align}(A+2I)^2=0&\iff (A+I+I)^2=0\\&\iff(A+I)^2+2(A+I)+I=0\\&\iff(A+I)\bigl((A+I)+2I\bigr)=-I\\&\iff(A+I)\bigl(-(A+I)-2I\bigr)=I.\end{align}

Answer 2

From $A^2 + 4AI +4I^2 = 0 $ one gets $-\frac 1 4A^2 = (A + I)$. Since $A$ is invertible , so is $A+I$

Answer 3

$A+I$ is not invertible implies there exists $x\neq 0$ with $A(x)=-x$, this implies that $(A^2+4A+4I)(x)=x-4x+4x=x=0$ contradiction.