Proving Vector Space Identity: I-T Bijectivity

AI Thread Summary
To prove that I - T is bijective when T^2 = 0, it is sufficient to show that (I - T)(I + T) = I. This demonstrates that I - T has an inverse, specifically I + T, confirming its bijectivity. The calculation is correct, and the identity holds, indicating that I - T is indeed invertible. The discussion highlights the importance of understanding linear maps and their properties in vector spaces. Thus, I - T is bijective under the given conditions.
Pearce_09
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Hello,
I am having trouble with particular algebra question. I don't know where to start and it would be greatly appreciated if someone could point me in the right direction.

Here is the questoin:

Let V be a vector space, where T is a linear map of V
prove if T^2 = 0 then I - T is bijective where I is the identity matrix

I tried (I-T)(I+T) = I - T^2 which equals I, but i am not sure where to go from here or if this even correct.
thanks for the time and help
regards,
adam
 
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You are done!
You just showed that I-T is invertible/bijective by showing that (I-T)(I+T) = (I+T)(I-T) = I. Which means, by definition, (I-T)^{-1} = (I+T)
 
thx for the help incredible
 

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