# Formula for the inverse problem

danielI
If $$A^2 = 0$$, show that $$I - A$$ is invertible.

So we know that $$\det(A^2) = (\det A)^2 = \det 0 \Leftrightarrow \det A = 0$$

We should now show that $$\det(I-A) \not= 0$$.

But I'm not sure how to do that. Could someone kick me in the right directon?

Muzza
There's no need for determinants here. There is a very simply formula for the inverse of I - A (when A^2 = 0). Any further hints will practically give away the entire solution, but (I - A)^(-1) "looks" a lot like I - A...

Homework Helper
forget matrices for one second, think taylor series

what is (1-x)^{-1} as a series expansion when |x|<1?

This can be suitably altered to tell us what the inverse of 1-A is for any A up to some convergence questions. Since A^2=0 these convergence questions vanish entirely as they do if instead of A^2=0 we had A^r=0

obviously since you have been given r=2 i could be using something, talyor series, you've never heard of, and instead you are supposed to use something that, up to making a sign change, is what Muzza said (difference of two squares, anyone?)

Homework Helper
and you don't have to be clever. try showing I-A has no kernel. this suffices at least in finite dimensions.

danielI

$$(I-A)^{-1}=(I+A)\text{, since }A^n=A^2\cdot A^{n-2} = 0\cdot A^{n-2}=0$$

$$(I - A)(I + A) = II + IA - AI - AA = I + A - A - 0 = I$$
$$(I + A)(I - A) = II - IA + AI - AA = I - A + A - 0 = I$$

Now when I've solved this problem I would like to see your solutions (if it's different). It's allways good to learn more ways to solve a problem

mathwonk: exactly what do you mean by 'kernel'?

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