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Eigenvalues and determinants

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Let A be an nxn matrix, and suppose A has n real eigenvalues lambda_1, .....lambda_n repeated according to multiplicities. Prove that det A = lambda_1....lambda_n


    2. Relevant equations
    None


    3. The attempt at a solution
    Could someone explain what is meant by 'repeated according to multiplicities'? and give me a hint as to how to start this proof? thank u ~~
     
  2. jcsd
  3. May 25, 2009 #2

    Dick

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    The eigenvalues are the roots of the characteristic polynomial p(lambda)=det(I*lambda-A), the multiplicity of the eigenvalue is the multiplicity of the root. E.g. (lambda-1)^2 has a root 1 of multiplicity 2. Roots of polynomials correspond to linear factors of the polynomial. Think about how to find p(0).
     
    Last edited: May 25, 2009
  4. May 25, 2009 #3
    to find p(0), i wud just sub 0 in the place of every variable and solve. I will be left with a constant, if there is a one
    Is this wat u r asking? but how does this relate to the question? =S
     
  5. May 25, 2009 #4

    Dick

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    I'm asking you to write out the characteristic polynomial in terms of lambda and lambda_1...lambda_n. Then think about what constant you get if lambda=0. How is it related to lambda_1...lambda_n? And how is that constant related to det(A)?
     
  6. May 25, 2009 #5
    How can i write our the characteristic polynomial if we're dealing with a general nxn matrix?
     
  7. May 25, 2009 #6

    Dick

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    You know the eigenvalues. E.g. if lambda_1 is a eigenvalue, then it's a root of the characteristic polynomial p(lambda). That means p(lambda) has a factor (lambda-lambda_1), right? Remember what you know about how roots of polynomials are related to factors of polynomials.
     
  8. May 25, 2009 #7
    the factors of polynomials are the roots of the polynomials

    i think...
    det(A)=(lambda-lambda_1)(lambda-lambda_2)....(lambda-lambda_n) so the eigenvalues are lambda_1.....lambda_n
     
  9. May 25, 2009 #8

    Dick

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    Ok, but that's not quite det(A), it's det(lambda*I-A). det(A) doesn't have a lambda in it. So if lambda=0 on both sides what do you get?
     
  10. May 25, 2009 #9
    ohhhhhh i think i got it now

    det(A-lambda*I) =(lambda-lambda_1)(lambda-lambda_2)....(lambda-lambda_n)
    if lambda =0, then we have
    det(A) =(lambda_1)(lambda_2)....(lambda_n)

    but, can we just set lambda = 0 like that?
     
  11. May 25, 2009 #10

    Dick

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    Yes, you can. But you have to pay attention to the minus signs. And the characteristic polynomial is det(lambda*I-A). det(A) and det(-A) are different. But that's pretty close.
     
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