
invertible matrix
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...

Re: invertible matrix
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?

Re: invertible matrix
The equation (Iₙ  A)⁴ = 0 is correct....(I computed it here Online Matrix Calculator) so I still have the same problem...

Re: invertible matrix
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.

Re: invertible matrix
Thank you...you are genius!