# 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?

• Hint: you do not have to actually find any inverse in this problem. If I say "$C$ is the inverse of $B$", what does that actually mean? – David Feb 28 '19 at 0:27
• @David Oh right! That means that their product equals the identity matrix, so (I - A^2)(I - A^2). I wouldn't be able to distribute that, would I? – Jonathan Doe Feb 28 '19 at 0:32
• Matrix multiplication is distributive over addition. It's not always commutative though – J. W. Tanner Feb 28 '19 at 0:35
• With respect to J.W. Tanner's comment: Although in general, matrices $A$ and $B$ do not commute, the identity matrix will commute with any other matrix, and of course any matrix commutes with itself. – Brian Tung Feb 28 '19 at 0:38
• For the record, I agree with @BrianTung's elaboration of my comment – J. W. Tanner Mar 1 '19 at 16:56

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