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If A is nilpotent square matrix then I+A is invertibl

  1. Oct 25, 2007 #1
    Theorem: If A is a nilpotent square matrix (that is for some natural number k>0, A^k =0) then (I + A) is an invertible matrix.

    Pf: Let B denote the inverse which will constructed directly. Let n be the smallest integer so that 2^n>k.


    then, B(1+A)=(1+A)B=1-A^(2^n)=1 - (A^(k))(A^(2n-k)) = 1
    now if

    I wonder if this is a sufficient proof? I've never taken a linear algebra course so I really don't know!
    Last edited: Oct 25, 2007
  2. jcsd
  3. Oct 25, 2007 #2


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    You've skipped far too many steps.


    These two equalities are not sufficiently obvious to be written without justification.
  4. Oct 25, 2007 #3
    if you write it out it's easy 2 see. i could prove the equation by induction.

    the idea is after you multiply (1+A) with (1-A) you get 1-A^2 - this is our base case.
    Now for inductive step; we have B(1+A)/(1+A^(2^n))= 1 -A^(2^n), solving for B(1+A) we get B(1+A)=1-A^[2^(n+1)].

    HA! I made a computational error, but this doesn't matter, as the last term in B is then the identity!
  5. Oct 25, 2007 #4
    Can someone redo this but with big pictures and a james earl jones voice over cuz its a question id like to know how to do, not sure my pea sized brain gets your proof tho :P
  6. Oct 25, 2007 #5
    B as written above (actually B=(1-A)(1+A^(2^1))(1+A^(2^2))....(1+A^[2^(n-1)])) is an inverse of I + A.

    if you multiply it out you get I.
  7. Oct 25, 2007 #6

    matt grime

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    The point was not necessarily that we couldn't see it was true (or not), but that if you're attempting a proof like this and want people to check your work it is a good idea to justify all of your assertions.

    What is (1+x)^-1?
  8. Oct 25, 2007 #7
    the inverse of the real (or complex) # 1+x. So we know A cannot be the negative identity because A^k = (plus/minus) Identity, for all k>0. So 1+A is not the zero matrix.

    Yeah, it's probably a good idea I start trying to be more meticulous with the algebra being as that I'm new to it.

    I've recently been using Latex to write out solved problems from rudin's Principles and have become rather 'fast' with my proofs, making the reader fill in what I feel are obvious gaps. I've come to hate prefacing an argument with "because the definition is equivalent to *blank*" if the equivalence is a well known theorem proved in the text BEFORE the problem set or if i've proven it above. But yeah, again, because I'm new to the subject...
  9. Oct 25, 2007 #8

    matt grime

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    I meant what is the talyor series of 1/(1+x). I should have been clearer.
  10. Oct 25, 2007 #9


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    The fact that multiplying out (1+A)B gives you an iterative difference-of-squares thing is certainly one thing that should have been stated.

    But B(1+A) is a different story; remember that matrix algebra is noncommutative! It's not enough to simply multiply out (1+A)B to get I, you also have to multiply out B(1+A). Again, because matrix algebra is noncommutative, you cannot simply divide by a matrix; you have to multiply (either on the left or on the right) by its inverse. Oh, and have you even shown that each of the 1+A^(2^m) are invertible?

    (p.s. you already gave n a specific purpose in your opening post, so you shouldn't use it here for a new purpose)
  11. Oct 25, 2007 #10
    But B(1+A) is a different story; remember that matrix algebra is noncommutative

    Whoa! I blanked on this one, lol - yes you are completely right Hurkyl I have no way of showing that B(1+A) gives us the same difference of squares thing.

    Matt Grime:

    boooyaa! Let n = k-1.

    B=Sum of (-1)^i A^i as i runs through {0,1,2....,n}

    B(1+A) = B+BA =1 - A+,-...+/- A^n + A -A^2 +A^3 -,+...-/+A^n +/- A^k

    We see that this telescopes to leave us with

    B(1+A) =1 +/- A^k =1 +/- 0=1

    for (1+A)B we can argue similarly because BA=AB (this is due to the fact that B is the sum of matrices each of which A commutes with, like the identity and integer powers of A).

    Thanks Matt Grime!
  12. Oct 25, 2007 #11


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    let B = -A for simplicity, and then just factor I = I^k - B^k =(I-B)(I+B+....+B^k-1).
  13. Oct 25, 2007 #12

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