Thread: 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+A²+...+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=I₄-(I₄-A), prove that A is invertible and determine A^-1.

I would be grateful for any help you are able to provide...

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?

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.

Thank you...you are genius!