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

## Homework Statement

I'm doing an ODE for homework and I cant 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?

## Answers and Replies

jedishrfu
Mentor
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

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

!!!!

Just found my mistake, I copied the matrix wrong!!!

This is embarrasseing...

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

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
Mentor
okay that makes sense.

HallsofIvy
Science Advisor
Homework Helper
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 $$\begin{bmatrix}3 & 1 \\ 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= 4\begin{bmatrix}x \\ y\end{bmatrix}$$
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".