I can't seem to find an eigenvector for this 2x2 matrix

[Hercuflea]

Homework Statement

I'm doing an ODE for homework and I can't find the eigenvector for this matrix (sorry, I don't know how to make matrices on here. Consider these as one matrix.):

[ 2 0] [v_{1}] = [0]
[ 1 1] [v_{2}] = [0]

Homework Equations

The only way he has taught us to find eigenvectors seems mostly like guesswork to me.

The Attempt at a Solution

2v_{1} + 0v_{2} = 0

v_{1} + v_{2} = 0

So 2v_{1} = 0 =====> v_{1} = 0

But v_{1} + v_{2} = 0, so v_{2} = 0 as well. But eigenvectors cannot consist of all zeros?

 

[jedishrfu]

wouldnt your eqn be more like

M * v = k * v where M is your matrix and k is some constant and v is your vector

2*v1 + 0*v2 = k * v1

v1 + v2 = k*v2

 

[Hercuflea]

Sorry maybe I didn't give the full background. (A-rI)v = 0

A =

[3 1]
[2 2]

one of the roots of the characteristic equation is r_{2} = 1

Since r_{2} = 1, (A-I)v=0

So

[ 2 0]
[ 1 1] v = 0

 

[Hercuflea]

!

Just found my mistake, I copied the matrix wrong!

This is embarrasseing...

 

[jedishrfu]

isn't that just Av = Iv which is Av = v ?

 

[Hercuflea]

Found my mistake, I accidentally subtracted 1 from ALL the elements in the original matrix A, when I should have only subtracted 1 from a11 and a22

 

[jedishrfu]

okay that makes sense.

 

[HallsofIvy]

Since you have found your error, expanding on jedishrfu's suggestion that you solve Av= v, that would be \begin{bmatrix}3 & 1 \\ 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x \\ y\end{bmatrix}.
which gives the equations 3x+ y= x and 2x+ 2y= y, both of which are equivalent to y= -2x. That is, any eigenvector, corresponding to eigenvalue 1, is of the form <x, y>= <x, -2x>= x<1, -2>.

For the other eigenvalue, 4, we have \begin{bmatrix}3 &amp; 1 \\ 2 &amp; 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= 4\begin{bmatrix}x \\ y\end{bmatrix}
which gives the equations 3x+ y= 4x and 2x+ 2y= 4y, both of which are equivalent to y= x. That is, any eigenvector, corresponding to eigenvalue 4, is of the form <x, y>=<x, x>= x<1, 1>.

Of course, solving (A- \lambda)v= 0 is equivalent to solving Av= \lambda v but the latter is more closely connected to the definitions of "eigenvalue" and "eigenvector".