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Linear Algebra Determinants Proof

  1. Feb 12, 2015 #1

    B18

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    1. The problem statement, all variables and given/known data
    Prove that if A is an n x n matrix with the property A3=A, then det(A)=-1, det(A)=0, or det(A)=1

    2. Relevant equations


    3. The attempt at a solution
    At first I started with the property A3=A
    I then applied the determinant to both sides.
    From this point I don't really see any properties that I can use to solve this equation for the det(A).
    I understand how the solutions work. Im just unsure on how to achieve the proof of those solutions to the property.
    Am I missing something simple here?
     
    Last edited: Feb 12, 2015
  2. jcsd
  3. Feb 12, 2015 #2

    Dick

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    Did you apply the property that det(AB)=det(A)det(B)? A^3=AAA.
     
  4. Feb 12, 2015 #3

    B18

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    Hi Dick, sorry. I did apply this property. I I'm just trying to get past this step.
     
  5. Feb 12, 2015 #4

    Dick

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    Then show me what you got after you applied the property.
     
  6. Feb 12, 2015 #5

    B18

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    det(A)*det(A)*det(A)=det(a)
    (det(A))^3=det(A)
     
  7. Feb 12, 2015 #6

    Dick

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    det(A)*det(A)*det(A) is not equal to 3det(A). It's equal to ##det(A)^3##. Let x=det(A). Then your equation is ##x^3=x##. Can you solve that equation for x?
     
  8. Feb 12, 2015 #7

    B18

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    (det(A))3=det(A)
    (det(A))3-det(A)=0
    det(A)(det(A)2-1)=0
    Thus:
    det(A)=0
    and det(A)2=1
    det(A)=+/-1
     
  9. Feb 12, 2015 #8

    Dick

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    Much better.
     
  10. Feb 12, 2015 #9

    B18

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    Is there a property that allows us to manipulate determinants this way? Or is it applicable because a determinate is a real number. Just need to build my proof format now. Really appreciate the help.
     
  11. Feb 12, 2015 #10

    Dick

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    As you said, a determinant is just a number. You treat them just like numbers. That's why I said let x=det(A). ##x^3=x## looks less mysterious.
     
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