If $A^k = 0$, $k>=1$, prove that $(I-A) ^{-1} = I +A + A^2 +\cdots+ A^ {k-1}$

Question

I was solving questions related to inverse matrices and I got this question. How do I attempt this question?

Answer 1

Multiply both sides by $(I-A)$ and expand the brackets.

Answer 2

Hint. Simply compute the following products: $$(I-A)\sum_{i=0}^{k-1}A^i\textrm{ and }\left(\sum_{i=0}^{k-1}A^i\right)(I-A).$$

If you want to get some insight on the formula, let us proceed formally:

• In the ring $R[[X]]$ of formal series over a ring $R$, the inverse of $1-X$ is given by $\displaystyle\sum_{k=0}^{+\infty}X^k$. This result can be thought as a analog of the geometric summation formula.

• If $A^k=0$, then the above sum is finite and one can substitute $A$ to $X$.