Eigenvalue Problems and Diagonalization

The best way to describe this procedure is with an example. First we will solve the following eigenvalue problem..

This eigenvalue problem represents a transformation that transforms coordinate (x,y) into coordinate (x',y').  In the case of an eigenvalue problem, (x',y') is just a scalar multiple of (x,y). The next step in the diagonalizing process is to write the two sets of equations, substituting the eigenvalues.

Notice that we wrote the set of two equations as one matrix equation.  You can verify this by multiplying both sides out.  If you find the eigenvectors, you will find that.

Making t=1, and normalizing each vector we find that...

Plugging these value into the above matrix equation and assigning each matrix a name we get..

So we can see that the diagonal matrix resulting is a matrix with its eigenvalues down the main diagonal and zeros everywhere else.  The matrix D is the matrix which describes in the (x',y') system the same deformation that M describes in the (x,y) system.  As you can see the matrix is a lot simpler.  This means that the eigenvectors of the transformation matrix M give the best basis to use for the transformation.  The matrix C is some sort of transformation, whether it be a rotation, reflection or some sort of skewing.  Note however that, if C is not an orthogonal matrix, then the transformation will lead to a non-orthogonal basis of eigenvectors.

Let us now do an example.  Imagine we have two masses, and between them, they are connected by a spring.  Then imagine that on the other side of each mass, there is a spring that is attached to the wall.  (We are ignoring gravity.)
Since, these are spring we use the potential energy of a spring in the equation of motion for each mass...

Note that we assumed a few things.  We are assuming both masses are the same, and all spring constants are the same.  Also note that "y" in this case represents the displacement of the third spring. The displacement of the second spring is the displacement of the first spring plus the negative of the displacement of the second spring.  Thus x-y.  For this situation, we will assume the un-damped oscillatory motion.

So now we have an expression to find the eigen-frequencies of the system..

The eigenvalues and eigenvectors describe the possible motions of the system.  In the first case where x=-y, the two masses move in opposite directions, first the spring goes out like this <--- ---> Then like this
---><---.  (Use your imagination, those are arrows!)  In the second case where x=y, the two masses move like --->---> then like <---<---.  Note that the eigenvectors produced were orthogonal,  this is because we had a symmetric matrix!  Let us now move on to the system where the masses and spring constants are not equal.

Suppose we had masses and spring constants of, from left to right, 2k,2m,6k,3m,3k.  This problem is done pretty much exactly as the previous one..

Note how we didn't need to actually solve for the constant vectors a and b since we know this will be a solution and it isn't needed to find the eigenvectors.  The behavior of the eigenvectors this time is the same as last time except in the case where the masses oscillate in opposite directions, one has 3/2 the magnitude of the other.  Note that since the matrix wasn't symmetric, the eigenvalues were not orthogonal! This could easily be fixed by a change of variables in the matrix or normalizing the existing vectors, but all we care about is the direction so there is no need.

As a final problem to this post, we will model a tri-atomic linear molecule with the two end masses of mass "m" and the middle molecule with mass "M."  We will model the forces holding them together as springs, so it will be very similar to the two previous problems.

So the first possible motion is all the masses moving to the right together, as in translation.  The second motion is when the middle mass stays stationary and the two side masses oscillate in opposite directions.  The third motion is both side masses oscillate in the same direction as the middle mass oscillates in the opposite direction with magnitude 2m/M.
1. m--->M-->m--->
2. m--->M<---m
3. m---><---Mm--->