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Matrix proof

  1. Apr 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Given a matrix B, if B = B2, is (B+I) invertible?

    2. The attempt at a solution

    det(B) = 0 or 1

    rref(rref(B) + I) is I, so (rref(B) + I) is invertible

    if det(B) = 1:
    let E1E2...En = B
    then E1E2...En(rref(B) + I) = B + E1E2...En

    I'm not sure if what I did is even useful =(
     
  2. jcsd
  3. Apr 24, 2009 #2

    CompuChip

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    What can you say about det(B + I)?
     
  4. Apr 25, 2009 #3
    If det(B) = 1, then B-1B = B-1B2, B = I
    so det(B+I) = det(2I) != 0, so B+I is invertible.

    I'm still stuck on if det(B) = 0..

    I'm quite sure that if B = B2, then B must be a diagonal matrix with entries being either 1 or 0, but I don't know how to prove it.
     
  5. Apr 26, 2009 #4

    HallsofIvy

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    If B= B2 then B2- B= B(B- I)= 0.
     
  6. Apr 26, 2009 #5

    matt grime

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    That follows from the fact you know its minimal poly divides X^2-X, hence you know all the possible eigenvalues (0 and 1), asndyou know the geometric multiplicity is 0 or 1.

    Alternatively remember that a definition of an eigenvalue is:


    t is an eigenvalue of X if and only if X-tI is not invertible.
     
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