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Positive Definite Matrices

  • Thread starter angelz429
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  • #1
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[SOLVED] Positive Definite Matrices

1. Homework Statement

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. Homework Equations

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


3. The Attempt at a Solution

i'm not sure where to start
 

Answers and Replies

  • #2
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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.
 
  • #3
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hmm... i understand what you are saying, but I'm not sure how to apply it. My main problem is with proving part a.
 
  • #4
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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.
 
  • #5
Dick
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hmm... i understand what you are saying, but I'm not sure how to apply it. My main problem is with proving part a.
How would you find the eigenvalues of A-lambda*I? How would you find the eigenvalues of A for that matter?
 
  • #6
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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?
like any other eigenvalue i suppose do det((A-lambda*I)-alpha*I)=0 and solve for alpha.
 
  • #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?
 
  • #8
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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?
 
  • #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??
 
  • #10
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oh riiiight! Thanks!.... but how can we guarantee that alpha + lambda is > 0?
 
  • #11
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errr i mean that alpha - lambda is > 0
 
  • #12
Dick
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errr i mean that alpha - lambda is > 0
You'd better put some conditions on the eigenvalues of A. Look at what you are trying to prove.
 
  • #13
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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...
 
  • #14
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ahhh nevermind... i got it!!
 
  • #15
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Thanks!
 

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