# Positive Definite Matrices

1. Apr 25, 2008

### angelz429

[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. Apr 25, 2008

### eok20

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. Apr 25, 2008

### angelz429

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. Apr 25, 2008

### eok20

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. Apr 25, 2008

### Dick

How would you find the eigenvalues of A-lambda*I? How would you find the eigenvalues of A for that matter?

6. Apr 25, 2008

### angelz429

like any other eigenvalue i suppose do det((A-lambda*I)-alpha*I)=0 and solve for alpha.

7. Apr 25, 2008

### Dick

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. Apr 25, 2008

### angelz429

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. Apr 25, 2008

### Dick

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. Apr 25, 2008

### angelz429

oh riiiight! Thanks!.... but how can we guarantee that alpha + lambda is > 0?

11. Apr 25, 2008

### angelz429

errr i mean that alpha - lambda is > 0

12. Apr 25, 2008

### Dick

You'd better put some conditions on the eigenvalues of A. Look at what you are trying to prove.

13. Apr 25, 2008

### angelz429

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. Apr 25, 2008

### angelz429

ahhh nevermind... i got it!!

15. Apr 25, 2008

Thanks!