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

  • Thread starter Hercuflea
  • Start date
588
49
1. 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[itex]_{1}[/itex]] = [0]
[ 1 1] [v[itex]_{2}[/itex]] = [0]

2. Homework Equations

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

3. The Attempt at a Solution

2v[itex]_{1}[/itex] + 0v[itex]_{2}[/itex] = 0

v[itex]_{1}[/itex] + v[itex]_{2}[/itex] = 0

So 2v[itex]_{1}[/itex] = 0 =====> v[itex]_{1}[/itex] = 0

But v[itex]_{1}[/itex] + v[itex]_{2}[/itex] = 0, so v[itex]_{2}[/itex] = 0 as well. But eigenvectors cannot consist of all zeros?
 
10,956
4,456
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
 
588
49
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[itex]_{2}[/itex] = 1

Since r[itex]_{2}[/itex] = 1, (A-I)v=0

So

[ 2 0]
[ 1 1] v = 0
 
588
49
!!!!

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

This is embarrasseing...
 
10,956
4,456
isn't that just Av = Iv which is Av = v ?
 
588
49
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
 

HallsofIvy

Science Advisor
Homework Helper
41,715
876
Since you have found your error, expanding on jedishrfu's suggestion that you solve Av= v, that would be [tex]\begin{bmatrix}3 & 1 \\ 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x \\ y\end{bmatrix}[/tex].
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 [tex]\begin{bmatrix}3 & 1 \\ 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= 4\begin{bmatrix}x \\ y\end{bmatrix}[/tex]
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 [itex](A- \lambda)v= 0[/itex] is equivalent to solving [itex]Av= \lambda v[/itex] but the latter is more closely connected to the definitions of "eigenvalue" and "eigenvector".
 

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