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Math Help - Vector Maximization

  1. #1
    Newbie AmberLamps's Avatar
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    Vector Maximization

    I have a host of questions that require me to maximise vectors/matrices. This is from an economic module so much of this material is new to me. Anyway there is a typical question below. Any examples that I can find in textbooks/online consider the problems of a positive definite matrix and I am unable to see how this changes the nature of the problem.

    Consider the problem of maximising  x^T Ax subject to  x^Tx=1 where x is a column vector of n variables and A is a negative-definite, symmetric n-by-n constant matrix. Show that the maximum value of the objective function is equal to the largest eigenvalue of A.

    Any help would be much appreciated
    Last edited by AmberLamps; February 25th 2010 at 11:28 AM. Reason: sp :p
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  2. #2
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    Quote Originally Posted by AmberLamps View Post
    I have a number of questions that require me to maximise vectors/matrices. This is from an economic module so much of this material is new to me. Anyway there is a typical question below. Any examples that I can find in textbooks/online consider the problems of a positive definite matrix and I am unable to see how this changes the nature of the problem.

    Consider the problem of maximising  x^T Ax subject to  x^Tx=1 where x is a column vector of n variables and A is a negative-definite, symmetric n-by-n constant matrix. Show that the maximum value of the objective function is equal to the largest eigenvalue of A.

    Any help would be much appreciated
    i don't think we need A to be negative-definite. we only need A to be (real) symmetric. then there exist an orthonormal basis of \mathbb{R}^n, say \{x_1, \cdots , x_n \}, which consists of eigenvectors of A.

    so we have Ax_i=\lambda_i x_i, for some \lambda_i \in \mathbb{R}. let \lambda_k= \max_i \lambda_i. now if x \in \mathbb{R}^n is any vector with x^Tx=1, then x=\sum_{i=1}^n \alpha_i x_i, for some \alpha_i \in \mathbb{R} and \sum_{i=1}^n \alpha_i^2=x^Tx=1. but then:

    x^TAx=\sum_{i=1}^n \alpha_i^2 \lambda_i \leq \lambda_k \sum_{i=1}^n \alpha_i^2=\lambda_k. it's also clear that x_k^TAx_k=\lambda_k x_k^T x_k=\lambda_k. \ \Box
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  3. #3
    Newbie AmberLamps's Avatar
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    Thanks, that helps a lot. Don't suppose you know of any good textbooks for this topic?
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  4. #4
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    Quote Originally Posted by AmberLamps View Post
    Thanks, that helps a lot. Don't suppose you know of any good textbooks for this topic?
    if this is a part of a course that you're taking, you should ask your instructor to give you some references. anyway, i guess you can find these stuff in non-elementary linear algebra textbooks.
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