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Positive Definite Matrices

  1. Apr 25, 2008 #1
    [SOLVED] Positive Definite Matrices

    1. The problem statement, all variables and given/known data

    a) If A is Symmetric show that A-λI is positive definite if and only if all eigenvalues of A are >λ, and A-λI is negative definite if and only if all eigenvalues of A are <λ.

    b) Use this result to show that all the eigenvalues of
    [ 5 2 -1 0]
    [ 2 5 0 1]
    [-1 0 5 -2]
    [ 0 1 -2 5]
    are between zero and eight.



    2. Relevant equations

    If all eigenvalues of A>0, A is positive definite.


    3. The attempt at a solution

    i'm not sure where to start
     
  2. jcsd
  3. Apr 25, 2008 #2
    here's a hint: since A is symmetric, you can get a set of eigenvectors of A which are an orthonormal basis for the vector space.
     
  4. Apr 25, 2008 #3
    hmm... i understand what you are saying, but I'm not sure how to apply it. My main problem is with proving part a.
     
  5. Apr 25, 2008 #4
    to show that a matrix M is positive definite, it is sufficient to show that v^T M v > 0 for any vector v in an orthonormal basis. since you can get such a basis with eigenvectors of A, you just need to show that for any eigenvector v of A, v^T (A-λI) v > 0. remember that you know how A acts on eigenvectors.
     
  6. Apr 25, 2008 #5

    Dick

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    How would you find the eigenvalues of A-lambda*I? How would you find the eigenvalues of A for that matter?
     
  7. Apr 25, 2008 #6
    like any other eigenvalue i suppose do det((A-lambda*I)-alpha*I)=0 and solve for alpha.
     
  8. Apr 25, 2008 #7

    Dick

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    Right. So when you do that alpha is an eigenvalue of A-lambda*I. Do you see that makes (lambda+alpha) an eigenvalue of A? So what's the relation between the eigenvalues of A and those of A-lambda*I?
     
  9. Apr 25, 2008 #8
    ok, so det(A-lambda*I-alpha*I) implies det (A-alpha*lamba*I) therefore alpha is an eigenvlaue od A-lambda*I and alpha*lambda is an eigenvalue of A.

    But what does this say about the relationship between alpha and lambda? I'm supposed to see that if alpha > lambda A is positive definite & is alpha < lambda A is negative definite.

    If alpha > lambda, alpha*lambda > lambda
    If alpha < lambda, alpha*lambda can be < lambda

    That's all I can see. How does this show that all alpha*lambda> 0 or alpha*lambda<0?
     
  10. Apr 25, 2008 #9

    Dick

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    det(A-lambda*I-alpha*I)=det(A-(lambda+alpha)*I). '+', not '*'! If alpha+lambda is an eigenvalue of A when alpha is an eigenvalue A-lambda*I, I would say that the eigenvalues of A-lambda*I are obtained by subtracting lambda from the eigenvalues of A. Wouldn't you agree??
     
  11. Apr 25, 2008 #10
    oh riiiight! Thanks!.... but how can we guarantee that alpha + lambda is > 0?
     
  12. Apr 25, 2008 #11
    errr i mean that alpha - lambda is > 0
     
  13. Apr 25, 2008 #12

    Dick

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    You'd better put some conditions on the eigenvalues of A. Look at what you are trying to prove.
     
  14. Apr 25, 2008 #13
    Alright, so I understand the first part... now how do I use this to show that all the eigenvalues of B are between zero & eight? I just checked that B is positive definite...
     
  15. Apr 25, 2008 #14
    ahhh nevermind... i got it!!
     
  16. Apr 25, 2008 #15
    Thanks!
     
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