# Vector Maximization

• Feb 24th 2010, 02:27 PM
AmberLamps
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 (Rofl)
• Feb 24th 2010, 03:30 PM
NonCommAlg
Quote:

Originally Posted by AmberLamps
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 (Rofl)

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 $x_k^TAx_k=\lambda_k x_k^T x_k=\lambda_k. \ \Box$
• Feb 24th 2010, 04:11 PM
AmberLamps
Thanks, that helps a lot. Don't suppose you know of any good textbooks for this topic?
• Feb 24th 2010, 05:47 PM
NonCommAlg
Quote:

Originally Posted by AmberLamps
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.