Inverting I − A

#matrices#nilpotent

Statement

Let $A$ be a square matrix with $A^2=0$ . Show that $I-A$ is invertible and give its inverse.

Hint

Multiply $I-A$ by $I+A$ .

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Using $A^2=0$ ,
$(I-A)(I+A)=I-A^2=I\qquad\text{and}\qquad(I+A)(I-A)=I-A^2=I.$

So $I-A$ is invertible and $(I-A)^{-1}=I+A$ .