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Eigenvalues and determinants

  • #1

Homework Statement


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


Homework Equations


None


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 ~~
 

Answers and Replies

  • #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:
  • #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
 
  • #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)?
 
  • #5
How can i write our the characteristic polynomial if we're dealing with a general nxn matrix?
 
  • #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.
 
  • #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
 
  • #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?
 
  • #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?
 
  • #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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