Beefy Boxes and Bandwidth Generously Provided by pair Networks
Don't ask to ask, just ask

Re^2: Help with Matrix math!

by pc88mxer (Vicar)
on Dec 17, 2007 at 05:04 UTC ( #657368=note: print w/replies, xml ) Need Help??

in reply to Re: Help with Matrix math!
in thread Help with Matrix math!

In this case you have to take into account the constraint on the vector x. The quadratic form xiDijxj is unbounded on Rn, so a derivative test (even with an Hessian test) is only going to find local extreme points, and there is no guarantee that those points will lie on the surface of interest (i.e. x'x = 1.) The difference is that for a point to be a local maximum when x is constrained, the derivative only needs to be zero when evaluated in the tangent space of the constraining surface at that point, not zero in every direction.

The standard way to include the surface constraint is to use Lagrangian multipliers.

Replies are listed 'Best First'.
Re^3: Help with Matrix math!
by randyk (Parson) on Dec 18, 2007 at 01:41 UTC
    I was thinking of the problem differently - find the extremal points of xDx, which leads to the eigenvalue equation Dx = 0. This equation doesn't determine xx completely, so one is free to impose, for example, xx = 1 as a normalization condition. But you're right that if xx = 1 is intended as a true constraint, then a method like Lagrange multipliers should be used.

Log In?

What's my password?
Create A New User
Node Status?
node history
Node Type: note [id://657368]
and the web crawler heard nothing...

How do I use this? | Other CB clients
Other Users?
Others having an uproarious good time at the Monastery: (4)
As of 2016-10-24 05:54 GMT
Find Nodes?
    Voting Booth?
    How many different varieties (color, size, etc) of socks do you have in your sock drawer?

    Results (303 votes). Check out past polls.