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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?

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  • $\begingroup$ 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? $\endgroup$ – David Feb 28 '19 at 0:27
  • $\begingroup$ @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? $\endgroup$ – Jonathan Doe Feb 28 '19 at 0:32
  • $\begingroup$ Matrix multiplication is distributive over addition. It's not always commutative though $\endgroup$ – J. W. Tanner Feb 28 '19 at 0:35
  • $\begingroup$ 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. $\endgroup$ – Brian Tung Feb 28 '19 at 0:38
  • $\begingroup$ For the record, I agree with @BrianTung's elaboration of my comment $\endgroup$ – J. W. Tanner Mar 1 '19 at 16:56
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$$(I-A^2)(I-A^2)=I\times I - I \times A^2 - A^2 \times I + A^2 \times A^2 $$

$$=I - A^2 - A^2 + A^4 = I - 2A^2 + A^4,$$

so this is $I$ if $A^4=2A^2,$

so $(I-A^2)^{-1}=(I-A^2)$ in that case.

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  • $\begingroup$ This would work in any ring, which has a distributive law, not just matrices. Multiplicative inverse of a sum is not generally easy to work with algebraically $\endgroup$ – J. W. Tanner Feb 28 '19 at 2:48

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