# The VERY last part of finding an eigenvector

Printable View

• Dec 8th 2010, 12:08 PM
ironz
The VERY last part of finding an eigenvector
I can do everything, untill it gets to the point of actually putting the system into an eigenvector form, then im stuck. I won't put the actual numbers down, but say I have a system of equations like:

ax = by

How do I put this into eigenvector form?

Is it simply (a,b) or is it (1, b/a) or what?

I don't know what to do.

Thank you :)
• Dec 8th 2010, 12:37 PM
DrSteve
Simply choose any value for one of the variables, and then the other is determined. For example, choose y=a, then x=b. So the vector is (b,a).

Another possible choice would be y=1, then x=b/a (assuming a is not 0).

There are infinitely many choices.
• Dec 10th 2010, 04:55 AM
HallsofIvy
I am assuming you are writing your vectors as (x, y), not (y, x).

Do either of (a, b) or (1, b/a) satisfy ax= by? That is, is a(a)= b(b) or a(1)= b(b/a)?

From ax= by, you can solve for y: y= ax/b. Now choose x to be whatever you want and use that equation to find b. If you take x= 1, just because it is easy, you have y= a(1)/b= a/b. Your vector is (1, a/b) (not (1, b/a)).

If you take x= b, you have y= a(b)/b= a. Your vector is (b, a) (not (a, b)).

The set of all eigenvectors corresponding to a given eigenvalue is a subspace so any multiple of an eigenvector is also an eigenvector.