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Identity Matrix and Inverses

  1. Mar 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Suppose [tex]P \in L(V)[/tex] and P^2 = P. Prove that (I+P) is invertible.

    2. Relevant equations
    3. The attempt at a solution
    Am I right to assume that since P^2 = P, P = I?
     
  2. jcsd
  3. Mar 17, 2008 #2

    Dick

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    Nope. If P=[[1,0],[0,0]], P^2=P. Is that all you wanted to know? Try solving (I+P).(I+aP)=I for a real number a. If you can solve it, you've found an inverse.
     
  4. Mar 18, 2008 #3

    Hurkyl

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    Don't give out answers. :tongue:
     
    Last edited: Mar 18, 2008
  5. Mar 18, 2008 #4

    Hurkyl

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    That doesn't even work for real numbers: x^2 = x has two solutions. (What are they? How would you find them?)

    I assume you wanted to cancel a P, but that's not always a legal operation. What operation are you doing when you cancel something? When are you allowed to do it? How is that operation different when you're working with matrices instead of real numbers?
     
  6. Mar 18, 2008 #5

    Dick

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    Ok, I'll admit, that's a little close to giving the answer for comfort. I thought about stopping after the '?'. But that seemed a little too close to just teasing the OP.
     
  7. Mar 18, 2008 #6

    Hurkyl

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    The problem, IMHO, is that there is nothing left of the original problem -- you've simply presented him with an entirely different (and much easier) problem, and absolutely no motivation why someone would ever think of the new problem.
     
  8. Mar 18, 2008 #7

    Dick

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    I, in fact, solved it by ansatz. I guessed a form for the solution which seemed to have enough parameters to be solvable given powers of P reduce to linear functions of P. That's all. If you have a better motivation, go for it. Suppose I could have explained that, though.
     
  9. Mar 18, 2008 #8
    Ok. here are my thoughts now. Using the fact that P^2 = P, we can say that P is a diagonal matrix. The entries along this diagonal are either 0 or 1 (solving the equation x^2 = x). Therefore we can conclude that M=(I+P) is also a diagonal matrix with entries of either 1 or 2 along the diagonal. By a proposition (5.16 in Axler's Linear Algebra Done Right), T (upper triangular) is invertible iff all the entries on the diagonal are nonzero. Therefore my M is invertible. Further, we can say that the inverse of this matrix is also a diagonal matrix with entries 1/x(i,i) where x(i,i) are the entries in M.

    Am I making any incorrect assumptions?
     
  10. Mar 18, 2008 #9

    Hurkyl

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    Why?

    In fact, I have a counterexample:
    [tex]\left(
    \begin{array}{cc}
    1/2 & 1/2 \\
    1/2 & 1/2
    \end{array}
    \right)[/tex]
     
  11. Mar 18, 2008 #10

    Hurkyl

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    The original poster is on one right track, I think. The problem is easy if the matrix is diagonal, and if he can prove the problem can be reduced to the diagonal case, he's done. I don't remember how easy that is, though.

    I'm a little disappointed that my favorite trick of expanding 1/(1+P) as a power series doesn't quite work; it requires you to do a deformation to arrive at the correct answer. Though this would give one method of motivating your approach.

    Another method of motivating your approach would be to invoke the knowledge that matrices have minimal polynomials, and there is an easy way to extract the inverse of your matrix from its minimal polynomial.


    Of course, you could simply guess at the form of the answer, as you did, but I feel like simply writing the form of the answer doesn't really help the OP learn.
     
  12. Mar 18, 2008 #11

    Dick

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    I don't THINK P is necessarily diagonalizable. It might be but I don't see what compels it to be. I did the power expansion first as well and realized it was useless. As you might expect since P could have an eigenvalue of 1 so the power series doesn't necessarily converge. I then decided that there might be a more general form for a possible inverse. And guessed it. Hopefully the OP will learn from this exchange as well.
     
  13. Mar 18, 2008 #12

    Hurkyl

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    P and 1-P are complementary projection matrices.

    I'm hoping so. :smile:
     
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