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Linear Algebra / Proof

  • Thread starter Wildcat
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  • #1
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Homework Statement



Let A be any nxn symmetric positive definite matrix. Show that (x‡0,xεRn)
x^TAx/x^Tx = the smallest eigenvalue of A.



Homework Equations





The Attempt at a Solution



Our hint was to first prove this for a diagonal matrix
For x^TAx/x^Tx I get L1x1² + L2x2² +...+Lnxn²/x1² + x2² +...+ xn² (I'm using L as lambda, the diagonal entries)
I know this is ≥1 since x1² + x2² +...+xn²/x1² + x2² +...+ xn² = 1 ≤ L1x1² + L2x2² +...+Lnxn²/x1² + x2² +...+ xn²
For the eigenvalues of A, If I choose x1, x2,.. to be 1 then xn = L1+L2+..L(n-1)/-Ln
I'm stuck here, help!
 

Answers and Replies

  • #2
lanedance
Homework Helper
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how about noting a symmetric matrix has an orthogonal set of eignevectors, so have you tried expanding x in terms of the eigenvector basis?

also, as its poistive definite, you know it has eigenvalues > 0.
 
  • #3
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how about noting a symmetric matrix has an orthogonal set of eignevectors, so have you tried expanding x in terms of the eigenvector basis?

also, as its poistive definite, you know it has eigenvalues > 0.
I'm not sure how to expand x in terms of the eigenvector basis. Would that mean that each of the xn^2 terms are =1?
 
  • #4
lanedance
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so if ui are the eigenvectors you should be able to write the vector x as
[tex] x = a_1. u_1 + \ .. \ + a_i. u_i + \ .. [/tex]

where, ai are scalars, to find them examine the dot product of x defined as above with a single eignevector, noting that the eignevectors are orthogonal
 
  • #5
1,444
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Something is wrong with your statement:

x^TAx/x^Tx = the smallest eigenvalue of A.

If x is an eigenvector to the eigenvalue a then x^TAx/x^Tx=a. So, you probably want to show that:

[tex]x^TAx/x^Tx\geq a_{min}[/tex]
 
  • #6
lanedance
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Homework Statement



Let A be any nxn symmetric positive definite matrix. Show that (x‡0,xεRn)
x^TAx/x^Tx = the smallest eigenvalue of A.
however i'm not quite convinced this is true... take x = (1,0,0,...) and A a diagonal matrix with diagonal elements (2,1,1,..) then
x^TAx/x^Tx = 2 and the smallest eigenvalue is 1?
 
  • #7
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however i'm not quite convinced this is true... take x = (1,0,0,...) and A a diagonal matrix with diagonal elements (2,1,1,..) then
x^TAx/x^Tx = 2 and the smallest eigenvalue is 1?
I left an important piece of info out of my original problem. It should say
Show that (x‡0,xεRn)
min x^TAx/x^Tx = the smallest eigenvalue of A.
min is in front of x^TAx/x^Tx with (x‡0,xεRn) under it.
 
  • #8
lanedance
Homework Helper
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ok that makes more sense, try the eigenvalue expansion of a generic vector to show arkajad's inequality & minimise or use the eigenvector with smallest eigenvalue to demostrate equality
 
  • #9
1,444
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Or use the http://en.wikipedia.org/wiki/Symmetric_matrices" [Broken] of symmetric matrices and check that your expression has the same value for the diagonal form of your matrix. The rest will be then very easy.
 
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  • #10
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Or use the http://en.wikipedia.org/wiki/Symmetric_matrices" [Broken] of symmetric matrices and check that your expression has the same value for the diagonal form of your matrix. The rest will be then very easy.
I like very easy but I'm afraid I'm not on your level :( We have a theorem to use after we show that the original statement is true for a diagonal matrix. The Theorem states that since A is symmetric we can find an orthogonal matrix P and a diagonal matrix D such that P^TAP=D. Is this the diagonalization property? I think once I show the original statement is true for a diagonal matrix, I can apply the theorem to show its true for any nxn symmetric matrix.

I'm going to ask this because I'm just not sure what the min in front of the expression means? I've looked through my notes and can't find anything, it's probably something I should already know which is why its not in my notes. This is part of my problem, I'm still not sure about the eigenvalue expansion of a generic vector??
 
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  • #11
lanedance
Homework Helper
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yeah so knowing that P^T = P^(-1) you can use P^T P = I

(x^TAx)/(x^Tx) = (x^T I A I x)/(x^TI x) = (x^T P^T P A P^T Px)/(x^T P^T P x)

now consider the vector u = P x, and with P A P^T = D, then you get
(x^TAx)/(x^Tx) = (u^TDu)/(u^Tu)
 
  • #12
1,444
4
And then

[tex]u^TDu=\sum_i \lambda_i |u_i|^2\geq\sum_i\lambda_{min} |u_i|^2=\lambda_{min}\,u^Tu[/tex]
 

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