if $A$ is a square matrix and $A^2 =0$, then $(I-A)$ is invertible

Question

I wanted to know if another way to solve this exercise is to first assume that $A$ is invertible on both the left and right sides, then it means that there exists a $B$ such that $AB=I$, therefore $A^2 B =A$, as $A^ 2 =0$ where $A= 0$.

Now if the matrix $IA=A$ then $A^t=0$.

$$IA = IA^t = A^t = 0$$

Since the symmetric matrix is null, we can consider the following

$$\begin{split} IA^t&=A^t\\ I(-A)&=A^t\\ I(-A)&=0\\ \end{split}$$

Therefore, since the null matrix is symmetric and at the same time antisymmetric, then it is not invertible.

Answer 1

Observe that $$(I-A)(I+A)=I-A^2=I.$$