The problem seems to suggest that (Iₙ - A)⁴ = 0, but it turns out that no power of (Iₙ - A) is 0. So I am not sure how to use the first part of the problem. Is this your difficulty?
I have some difficulty proving the bolded part of the exercise:
If A is an nxn matrix establish the identity
I_{n}-A^{k+1}=(I_{n}-A)(I_{n}+A+A^{2}+...+A^{k}).
Deduce that if some power of A is the zero matrix then I_{n}-A is invertible.
Suppose now that
A=2 2 -1 -1
-1 0 0 0
-1 -1 1 0
0 1 -1 1
Compute the powers (I_{n}-A)^{i} for i=1,2,3,4 and, by considering
A=I_{4}-(I_{4}-A), prove that A is invertible and determine A^{-1}.
I would be grateful for any help you are able to provide...
The equation (Iₙ - A)⁴ = 0 is correct....(I computed it here Online Matrix Calculator) so I still have the same problem...
Hmm, I am not sure what the problem is with WolframAlpha, but you are right.
Have you done the first part where you were told to deduce the following fact:
If some power of B is the zero matrix then Iₙ - B is invertible (*)
(I replaced A with B)? If yes, then you put B := I - A for this particular A. You verified that B⁴ = 0; therefore, by (*) we have I - B = I - (I - A) = A is invertible. Moreover, by the first part of the problem,
(I - B)(I + B + B² + B³) = I - B⁴ = I,
so the inverse of I - B = A is I + B + B² + B³.