TITLE: Prove that if matrix $A$ is nilpotent, then $I+A$ is invertible.
QUESTION [8 upvotes]: So my friend and I are working on this and here is what we have so far.
We want to show that $\exists \, B$ s.t. $(I+A)B = I$. We considered the fact that $I - A^k = I$ for some positive $k$. Now, if $B = (I-A+A^2-A^3+ \cdots -A^{k-1})$, then $(I+A)B = I-A^k = I$. My question is: in matrix $B$, why is the sign for $A^{k-1}$ negative? Couldn't it be positive, in which case we'd get $(I+A)B = I + A^k$?
Thank you.

REPLY [6 votes]: By incrementing if necessary by $1$, we can make sure that $k$ is *odd. Then we have the familiar identity (for odd $k$) $1+x^k=(1+x)(1-x+x^2-x^3+\cdots +x^{k-1})$.