Double root eigenvalues?

Question

I'm solving for a particular solution to a system of differential equations that satisfies certain initial conditions, but when I solve for the eigenvalues I get a double root of 6.

...so I solved using the eigenvalue 6 for the eigenvector and got [1 2](vertical matrix)*e^(6t).


Are both the eigenvectors and functions the same? Cause then the constants cancel... so how do the general solutions change when you have double root eigenvalues?

Answer 1

When an eigenvalue is repeated there are a couple of scenarios. Suppose the eigenvalue is r and is repreated k times. In the first scenario, there are k independent eigenvectors and things proceed as usual. In the second, your solutions will not simply be of the form
A*e^(rt). You will have to consider solutions of the form A*te^(rt), B*t^2 e^(rt), etc. Until you get k independent solutions. I suspect that you are in the second scenario and will have to look at solutions of the form Ate^(6t) where A is a vertical matrix.

Answer 2

Not real clear but I know that for the identity matrix of size n, it has an eigenvalue of 1 that a repeated root n times. Any vector in the space is an eigenvector.

Answer 3

the issue is that there is a limiteless variety of them. really something orthogonal to the x = 10 eigenvector is an x = a million eigenvector. So the answer is honestly going to be a 2-dimensional subspace, that you'll represent as a span of two vectors. regrettably you may represent this span in a limiteless variety of diverse procedures! So your effect may be the finest option, it purely can be a diverse representation of an similar subspace than is given by the "professional" answer. As for the thanks to outline if 2 spans are an similar, I responded that query in this communicate board a short even as decrease back . . .

Answer 4

Whe you get double eigen vector the solutions do change. One of the solutions becomes t *e(6t)