# Nilpotent Matrix Proof problem

1. Homework Statement
If B is any nilpotent matrix, prove that I-B is invertible and find a formula for (I-B)^-1 in terms of powers of B.

3. The Attempt at a Solution
If I make a matrix <<ab,cd>> then if 1/(ad-bc)$$\neq$$0 then the matrix has an inverse. Since I think all nilpotent matrices have a 0,0,0 leading diagonal with the other diagonal being not fully "0"s. Wouldn't it be impossible for nilpotent matrices to not have an inverse? I think I may have my wording jumbled.

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1. Homework Statement
If B is any nilpotent matrix, prove that I-B is invertible and find a formula for (I-B)^-1 in terms of powers of B.

3. The Attempt at a Solution
If I make a matrix <<ab,cd>> then if 1/(ad-bc)$$\neq$$0 then the matrix has an inverse. Since I think all nilpotent matrices have a 0,0,0 leading diagonal with the other diagonal being not fully "0"s. Wouldn't it be impossible for nilpotent matrices to not have an inverse? I think I may have my wording jumbled.
A nilpotent matrix cannot have an inverse. Say B^n = 0 where n is the smallest positive integer for which this is true. Now suppose it were invertible and let C be it's inverse. Then CB = I. But then 0 = CB^n = B^(n-1), a contradiction.

As to your original problem, you know B^n = 0 for some n.

Start small, suppose B^2 = 0 then notice (I - B)(I + B) = I. Now suppose B^3 = 0 what's an inverse for I - B in this case? Generalize this.

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Hmm. Why would you start out by assuming B^2 = I? Shouldn't it be 0 rather than I?

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that was a typo thanks, fixed, should be B^2 = 0

notice (I - B)(I + B) = I
I do not understand how this leads to me getting a formula for the inverse of (I-B).
The only equation I can think of that relates to this is (I-B)^-1(I-B)=I.... and I wouldn't have a clue how to change that if the powers of B were changing... any other ideas to get my brain into gear?

HallsofIvy
Homework Helper
notice (I - B)(I + B) = I.
I do not understand how this leads to me getting a formula for the inverse of (I-B).
The only equation I can think of that relates to this is (I-B)^-1(I-B)=I.... and I wouldn't have a clue how to change that if the powers of B were changing... any other ideas to get my brain into gear?
(I-B)(I+B)= I tells you that I+ B is the inverse of I-B; its product with I- B is I- that's all an inverse does.

If you don't get the idea yet, look at B3= 0. Then B3- I= -I or I-B3= I. But I- B3= (I- B)*what? Since their product is I, whatever that factor is is the inverse of I- B.

Now, can you find the general formula?

From that I'm assuming the general formula I need is I-B^k=(I-B)(I+B)?
I hope thats right, thanks alot you are very helpful

HallsofIvy