Finding independent Eigen Vectors- Application

1. Dec 6, 2009

JoeSabs

This is the other test problem that's killing my group. Any help is appreciated! Here's the problem:

Seven masses are connected by springs to each other and to a rigid, immobile hexagonal frame, as shown. Each mass is free to move up and down, perpendicular to the plane of the frame, but they cannot move sideways. If a mass moves upward, the six springs attached to it try to pull it back down. But a mass is pulled upward by one spring (the spring connecting it to its neighbor) if its neighbor is pulled upward. Let x1, x2, x3,x4, x5, x6, and x7 be the displacements of the masses above the plane of the frame (that is, the x's are positive for upward displacements, negative for downward), and let b1, b2, b3,b4, b5, b6, and b7 be the forces exerted on the masses by the springs (again, positive=up, negative=down). Let x and b be the vector with components x1 thru x7 and b1 thru b7, respectively.

a. Construct a 7X7 Matrix A such that A*x=b.

I think I got this part:

-6 1 0 0 0 0 1 x1 b1 thru b7
1 -6 1 0 0 0 0 x2 like the x vector
0 1 -6 1 0 0 0 x3
0 0 1 -6 1 0 0 Times x4 Equals
0 0 0 1 -6 1 0 x5
0 0 0 0 1 -6 1 x6
1 0 0 0 0 1 -6 x7

Yes?

b. Find seven independent eigenvectors of A, and find the eigenvalue of each eigenvector. Hints: Don't start by finding the characteristic polynomial of A. That would require you to find the determinant of a 7X7 matrix, which will take you forever. Instead, try to guess what the eigenvectors are by looking at symmetrical patterns of the seven bodies moving up and down. For most of the eigenvectors, all of the components are 1, -1, and 0, and you just need to find a symmetrical arrangement of those numbers forming a hexagon with one number at the center. Warning: There are only seven independent eigenvectors. Because the hexagon has a 120 degree rotational symmetry, you might suppose that if you have an eigenvector, you can get two more eigenvectors by rotating the hexagon 120 degrees to the left and right. This procedure will indeed give you three eigenvectors, but they will not be independent. If two independent eigenvectors have the same eigenvalue (which sometimes happens with this matrix), then any linear combination of the eigenvectors will be another (but obviously not independent) eigenvector.

I have no idea how to do this part... Help??

c. After you have found the eigenvalues, find the determinant of A. Do not try to find the determinant directly, using Gaussian elimination or expansion by minors.

This is straightforward once I get the b. part down...

Thanks!!