They can be used to find OBB orientation, for rigid body physics and many more cool things BUT : What the heck are they anyway? I followed a linear algebra course at college but they where not covered in that course... I read about it on the internet but I dont understand how its working. How can we compute them from a 3x3, 4x4, NxN matrix? What exactly do they represent? Is there a fast way to compute them in c++? Thanks

I google that already information already, but I do not figure how to something usefull with it, for example : what
does mean "to solve for the eigenvalues find the roots of the polynomial"?

I would like an example.

Not a single bit of eigen stuff was covered in my course, it was not even mentionned in my textbook...

quote:Original post by etaylor27
I'm very surprised to hear you say that this wasn't introduced in a linear algebra course at all.

It's not supposed to be covered in the first course, in Quebec.

Here's the full derivation:

MX = lX

X is an unknown column matrix and l is an unknown scalar. How do you solve this?

MX - lX = 0
MX - (lI)*X = 0 [try this identity yourself if you're not convinced]
(M - lI) * X = 0 [distributivity]

By "0", I mean a column-matrix, like X, whose elements are all 0. Let M' = M - lI. We have, then:

M' * X = 0

How can we solve this? This is like an equation system (in the case where X = [x y] transposed):

ax + by = 0
cx + dy = 0

This system always has a trivial solution of x = 0 and y = 0 (i.e.: X = 0). In "most" cases, this is the only solution. However, X = 0 is not the solution we're looking for; it's the trivial one. We're looking for the other solutions, those that are not null. If you try to solve for y, you'll get, at some point:

(ad - bc) * y = 0

This involves that either y = 0 or (ad - bc) = 0. y = 0 is the trivial solution we already know, so we consider (ad - bc). Surely, you've recognized here the determinant of a 2x2 matrix. It's not a conincidence; the equation system has a non-zero solution only if the determinant of M' is 0 (remember, we had M' * X = 0). In fact, if its determinant is 0, it has an infinity of solutions, since we can always multiply X by a constant and maintain the equality.

Now, we want det(M') = 0. Since M' = M - lI, then we want det(M - lI) = 0. Finding the determinant of M' is pretty straightforward, although it is tricky for large matrices. The determinant will be a polynomial in l. Finding its roots allows you to find the values of l that satisfies det(M') = 0. These are the eigenvalues.

Once you have l (there may be multiple values), you can plug it back in (M - lI) * X = 0, and solve for X, like a regular equation system. There are as many eigenvector "families" as there are eigenvalues (I use the term family because, as mentionned above, if X is an eigenvector, then kX is also one).

That's it. If you don't understand everything, try to do it yourself. If you've had the first linear algebra course in a cegep, you should be able to do all of the intermediate steps on your own.

Cédric

[edited by - cedricl on May 6, 2003 10:31:33 PM]

Geometrically, the eigenvalue represents the degree of inflation (scale) in a matrix. A matrix with eigenvalue 1 will transform any mesh into another mesh with the same volume; i e it''ll only contain translation and rotation.

quote:Original post by Anonymous Poster
Geometrically, the eigenvalue represents the degree of inflation (scale) in a matrix. A matrix with eigenvalue 1 will transform any mesh into another mesh with the same volume; i e it''ll only contain translation and rotation.

What if there is more than one eigenvalue?

Cédric

quote:Original post by Shnoutz
Is there a fast way to compute them in c++?

I use the lapack libraries http://www.netlib.org/lapack/