Question

Quick question on Pset 2, Problem 2.5.40. In the solutions, the alternative method is to write A = N - I, and take the inverse of N-I. Can anyone explain why the inverse of (N-I) is 1 + N + N^2 + N^3 + N^4?

Answer 1

Good question. It sure is not obvious (to me), but we can see they are taking advantage of a "telescoping sum"
Let's say
N =

0 1 0 0
0 0 2 0
0 0 0 3
0 0 0 0

>> N^2

0 0 2 0
0 0 0 6
0 0 0 0
0 0 0 0

>> N^3

0 0 0 6
0 0 0 0
0 0 0 0
0 0 0 0

>> N^4
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

Answer 2

Let A= (I - N) (or I + -1*N , which is the same thing)
Knowing about the "trick" of a telescoping sum, we do
\[ (I - N)(I +N +N^2 + N^3) \\=
(I-N)I + (I-N)N+ (I-N)N^2+ (I-N)N^3\\
= I -N +N -N^2 +N^2 -N^3 +N^3 -N^4
\]
notice the alternating signs on the N, N^2 and N^3 terms. we get
\[ (I - N)(I +N +N^2 + N^3) = I - N^4 \]
and for this problem N^4= 0 (with shape 4 x 4) and we have the result
\[ (I - N)(I +N +N^2 + N^3) = I \\
A (I +N +N^2 + N^3) = I\]
which shows that the expression \( (I +N +N^2 + N^3)\) is the inverse of A

Answer 3

Wow, thanks a ton! I would not have thought of that on my own.