SOLUTION: Suppose that A is a square matrix and A4 = 0 (the zero matrix). Show that (10 points) (I − A) −1 = I + A + A^2 + A^3 .

Algebra ->  Finance -> SOLUTION: Suppose that A is a square matrix and A4 = 0 (the zero matrix). Show that (10 points) (I − A) −1 = I + A + A^2 + A^3 .      Log On


   



Question 1164621: Suppose that A is a square matrix and A4 = 0 (the zero matrix). Show that (10 points)
(I − A)
−1 = I + A + A^2 + A^3
.

Answer by ikleyn(52824) About Me  (Show Source):
You can put this solution on YOUR website!
.
Suppose that A is a square matrix and A%5E4 = 0 (the zero matrix). Show that
%28I+-+A%29%5E%28-1%29 = I + A + A%5E2 + A%5E3.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let  B  be the matrix  I + A + A%5E2 + A%5E3.



We need to show that  (I-A)*B = I  and  B*((I-A) = I.



For the first equality, we have

    (I-A)*B = %28I-A%29%2A%28I+%2B+A+%2B+A%5E2+%2B+A%5E3%29 = I%2A%28I+%2B+A+%2B+A%5E2+%2B+A%5E3%29+-+A%2A%28I+%2B+A+%2B+A%5E2+%2B+A%5E3%29 = 

                                        = %28I+%2B+A+%2B+A%5E2+%2B+A%5E3%29 - %28A+%2B+A%5E2+%2B+A%5E3+%2BA%5E4%29 = I+-+A%5E4 = I.


    Here we used the fact that  A%5E4 = 0,  which is given.

    Thus, the property  (I-A)*B = I  is proven.



For the second equality, we have

    B*(I-A) = %28I+%2B+A+%2B+A%5E2+%2B+A%5E3%29%2A%28I-A%29 = %28I+%2B+A+%2B+A%5E2+%2B+A%5E3%29%2AI - %28I+%2B+A+%2B+A%5E2+%2B+A%5E3%29%2AA = 

                                        = %28I+%2B+A+%2B+A%5E2+%2B+A%5E3%29 - %28A+%2B+A%5E2+%2B+A%5E3+%2BA%5E4%29 = I+-+A%5E4 = I.


    Again, here we used the fact that  A%5E4 = 0,  which is given.

    Thus, the property  B*(I-A) = I  is proven.

At this point, the problem is solved completely.