Question

Done

Answer 1

\[A ^{k}=O _{(n,n)} \] for some natural integer k. Show that \[I _{n} + A \] is invertible.

Answer 2

You want play off the geometric formula:\[(1+x)(1-x+x^2-x^3+\ldots+(-1)^{n-1}x^{n-1})=1+(-1)^{n-1}x^{n}\]If you replace x with A, and 1 with the identity matrix I, you get:\[(I_n+A)(I_n-A+A^2-A^3+\ldots+(-1)^{n-1}A^{n-1})=I_n+(-1)^{n-1}A^{n}\]for any natural number n. Take n=k, that power that makes A^k the zero matrix. Then the right hand side becomes just the identity matrix, which shows (I+A) is invertible.

Answer 3

Ohhh that is cool! Thank you, that question was infuriating me!