Why is (I+A) invertible and what is its inverse?

Question

Let A and B be n x n (square) matrices where A^2 = 0 and B^2 = 0. Let I be the identity matrix.

(I+A) is invertible. Why? What is its inverse?

Does A^2 = 0 imply that the matrix A is nilpotent, making it non invertible?

Just trying to make sense of this situation.

What does $B$ have to do with anything? Also, $I^2-A^2 = (I+A)(I-A)$. — Randall, May 16 '24 at 00:51
$A^2 = 0$ does imply that the matrix $A$ is nilpotent. It also clearly has determinant $0$. It is non-invertible. — Henry, May 16 '24 at 00:52
If $A^2=0$, then $A$ does not have $-1$ as an eigenvalue. Therefore, $A+I$ does not have $0$ as an eigenvalue. — Arturo Magidin, May 16 '24 at 00:52

score 3 · Answer 1 · answered May 16 '24 at 01:15

Under some conditions on a real number $x,$ we have $$ \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - ... $$

Note that any power of a square matrix $A$ commutes with any other power. And

Under some conditions on a square matrix $A,$ we have $$ (I+A)^{-1} = I - A + A^2 - A^3 + A^4 - ... $$ The usual condition involves convergence of the infinite sum. In your case, it is not infinite, the sum "converges" and the equation is just true.

Why is (I+A) invertible and what is its inverse?

1 Answers1