First-year matrix problem: How do you show that a sum of an identity matrix and another matrix is equal to the sum's inverse? The following is the problem at hand:
$A^4 = 2A^2.$ Prove that $(I-A^2) = (I-A^2)^{-1}$
My attempts at a solution:
$I = A^{-1} * A,$ therefore we can start with $(A^{-1}A - A^2) = (A^{-1}A - (1/2)A^4) = (A^{-1}A - A^2) = (A^{-1}A - (1/2)(A^{-4})^{-1}) = ???$
I'm having problems with the fundamentals of this question. Sure, I can play around with the individual matrices on the left side, but that doesn't help to handle the inverse of a sum of matrices, like how the transpose of a sum is equal to the sum of the transpose of its individual matrix terms. How do you approach this without delving into more complex matrix operations?

$$(I-A^2)(I-A^2)=I\times I - I \times A^2 - A^2 \times I + A^2 \times A^2 $$
$$=I - A^2 - A^2 + A^4 = I - 2A^2 + A^4,$$
so this is $I$ if $A^4=2A^2,$ 
so $(I-A^2)^{-1}=(I-A^2)$ in that case.
