If $A^2=0$, then $I−A$ is invertible

Question

If $A^2=0$, then show that $I−A$ is invertible.

I am getting nowhere that leads me to the hint: $I+A$.

score 11 · Answer 1 · answered Mar 08 '15 at 21:42

11

$(I-A)(I+A)=I^2-A^2=I$

What does that tell you?

answered Mar 08 '15 at 21:42

KittyL

can we do that in a constructive proof?... like is it legal? – misosoup Mar 08 '15 at 23:12
1

@misosoup Yes. In general a "constructive proof" is one where you explicitly create or demonstrate the existence of something which implies what it is you want to prove. In this case, the only thing "constructed" was $(I-A)^{-1}$ which happens to have an easy representation. Further more since $(I-A)^{-1}$ exists, it proves the claim. (Requiring more explanation on your part when writing the proof: what is the inverse of $(I-A)$ and why do we know it is in fact the inverse?) – JMoravitz Mar 08 '15 at 23:50

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