Finding eigenvectors of a simple 2x2 matrix

[The thinker]

Homework Statement

The matrix is:
|1 2|
|3 4|

The Attempt at a Solution

I've worked out the eigenvalues to be $$\stackrel{\underline{5\pm\sqrt{33}}}{2}$$

But when I plug the first eigenvalue back in I get:

|1 - $$\stackrel{\underline{5+\sqrt{33}}}{2}$$......2 |
|3......4 - $$\stackrel{\underline{5+\sqrt{33}}}{2}$$ |

Multiplied by (x,y) = (0,0)

[Sorry I couldn't fit that on the same line as the matrix.]Which in turn gives two equations:

x*(1-$$\stackrel{\underline{5+\sqrt{33}}}{2}$$) + 2y = 0

and3x + y*(4-$$\stackrel{\underline{5+\sqrt{33}}}{2}$$) = 0Which I can't work out how to solve for the eigenvector.

Is the only solution x=y=0?

 

[Dick]

You solve them like you solve any two linear equations. Remember if (x,y) is an eigenvector then a*(x,y) is also an eigenvector. So you are going to have an undetermined parameter which you may as well may make x. Put say, x=1 and solve for y. You should get the same value of y from both equations, since one equation is really a multiple of the other.

 

[The thinker]

Thanks for the reply!

Thats exactly what I've been trying to do but X and/or Y always end up canceling and I just get some useless number.

 

[Dick]

Thanks for the reply!

Thats exactly what I've been trying to do but X and/or Y always end up canceling and I just get some useless number.

Are you trying to eliminate a variable from the two equations and solve for the other? That won't work. If you eliminate x then the resulting equation is 0*y=0. Is that your problem? Like I said before, that's happening because the two equations are really the same. They have an infinite number of solutions. Put x equal to any constant (like x=1) and then solve for y.

 

[The thinker]

AH!

Thank you!