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If A and B are invertible square matrices, there exists

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data

    If A and B are invertible matrices over an algebraically closed field [itex] k [/itex], show there exists [itex] \lambda \in k [/itex] such that [itex] det(\lambda A + B) = 0 [/itex].


    3. The attempt at a solution

    Can anyone agree with the following short proof? I tried looking online for a confirmation, but I wasn't exactly sure how to search exactly for this problem.

    Suppose [itex] det(\lambda A + B) \neq 0 ~~\forall \lambda \in k [/itex]

    Since [itex] det(A^{-1}) \neq 0 [/itex], then

    [itex] det(A^{-1}(\lambda A + B)) = det(\lambda I + A^{-1}B) = det(A^{-1}B - (-\lambda)I) \neq 0 ~~\forall \lambda \in k [/itex]


    So [itex] -\lambda [/itex] is not an eigenvalue for [itex] A^{-1}B [/itex] for all [itex] -\lambda \in k [/itex]. i.e. [itex] \lambda [/itex] is not an eigenvalue of [itex] A^{-1}B [/itex] for all [itex] \lambda \in k [/itex]. Since [itex] k [/itex] is algebraically closed this cannot be the case.

    Is this okay? I guess it is not necessarily true when [itex]k[/itex] is not algebraically closed?
     
    Last edited: Apr 2, 2014
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  3. Apr 2, 2014 #2

    HallsofIvy

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    ??? What are you asked to show about [itex]det(\lambda A+ B)[/itex]? You have a subject with no verb!

     
  4. Apr 2, 2014 #3
    My bad...early morning. Edited.
     
  5. Apr 2, 2014 #4

    Dick

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    You bet it can fail. Take A=[[0,1],[1,0]] and B=[[2,0],[0,1]]. What goes wrong if k is the field of rationals?
     
  6. Apr 2, 2014 #5

    D H

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    I didn't see that last bit about what happens when k is not algebraically closed. It can also fail with the reals. Just construct your matrices so the eigenvalues of A-1B are complex.
     
  7. Apr 2, 2014 #6

    Dick

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    Sure. But that's because the reals aren't algebraically closed. The det gives you a real polynomial. If it has no real roots, you are out of luck.
     
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