Proving that if $A^n = 0$, then $I - A$ is invertible and $(I - A)^{-1} = I + A + \cdots + A^{n-1}$ [duplicate]

Solution 1:

Set $B:=I+A+\ldots+A^{n-1}$ and compute $(I-A)B$ by expanding the product: $$ (I-A)B=(I-A)(I+A+\ldots+A^{n-1})=I+A+\ldots+A^{n-1}-A(I+A+\ldots+A^{n-1}) $$ $$ =I+A+\ldots+A^{n-1}-A-A^2-\ldots-A^{n-1}-A^n=\;? $$

Observe the cancellations: you are left with $I$. This suffices to show that $I-A$ is invertible with inverse $B$. This is by the rank-nullity theorem: a square matrix is right invertible if and only if it is invertible. Or you can observe that $B$ and $I-A$ commute, so $(I-A)B=I$ yields also $B(I-A)=I$.

Solution 2:

How do you check a matrix $B$ is the inverse of matrix $A$? You multiply them together and see if you get the identity matrix. This is just the definition of inverse matrix.

You have a matrix $I-A$ and you want to check that its inverse is $1+A+A^2+...+A^{n-1}$ given that $A^n=0$.

So multiply together and after doing lots of cancellations etc you get $I - A^n = I$, which is what we were after!

Solution 3:

Well wenn $A^n=0$ then we konw that every eigenvalue of $A$ is zero, taking jordannormalform one sees that $I-A$ is invertible cause it is similar to a triangular matrix with every value 1 on the diagonal. Afterwards you make something like the geommetric series

$$\frac{1}{1-x}= 1+x+x^2+\dots$$ now we have matrices but it is essentially the same it give syou $$(I-A)^{-1}=1+ A +A^2 + \dots+ A^{n-1} + \underbrace{A^n}_{=0}+\dots$$ But as every $A^k$ for $k\geq n$ is zero it is finite.

As Git mentioned it is unneccessary to go to infty

We can even take the finite geometric series which says that $$\sum_{i=0}^{n-1} x^i = \frac{x^n -1}{x-1}=\frac{1-x^n}{1-x}$$

We have $$(I-A)^{-1}= (I-A^n) \cdot (I-A)^{-1} = \sum_{i=0}^{n-1} A^i = I + A + A^2 + \dots + A^{n-1}$$