Eigenvectors

(in case my ascii treatment isn't obvious, M is a matrix)

M = (k 3)
____(0 2) , where k is not equal to 2.

By finding and solving the characteristic equation, I've found the eigenvalues to be k and 2. However, the problem comes in when I try to find a corresponding eigenvector to the eigenvalue k.

I get:

ka + 3b = ka
2b = kb

So I think b has to equal 0, but how do I find the possible values of k?

You are not required find the possible values of k. Atleast you can't find it with what is shown here.
The eigenvector for k is (a=1, b=0). eigenvector for 2 is (3/(2-k), 1)

The solution to that equation, a can be anynumber at all and b is 0. 1 is chosen because it is the simplest. eigenvectors are always used with a constant multiple to it.

Thanks, I get it now. I always try to solve the simultaneous equations rather than working out relationships between the constants.

>>3
I think an easy method is to make a variable 1 and try to solve the equations, you have one less variable so you can solve is straightforwardly. If the equations don't work out, make the variable 0 and try to solve the equations.

Oh by the way, for 2 by 2 matrices, there is only 1 equation and two variables (for 3 by 3, there are 2 eqns are 3 variables and so on). The two equations you got are actually the same.

>>5
Yes, I see. Oddly enough, I find this easy for 3x3 matrices.

Eigenvectors rule.