# Finding eigenvectors.

1. Oct 10, 2009

### Dissonance in E

1. The problem statement, all variables and given/known data
I have matrix
A = 4 2
0 1

Whose eigenvalues I found to be 4 & 1

I need to find the eigenvectors for the same matrix
2. Relevant equations
(A-lambdaI)V=0

3. The attempt at a solution
Lambda = 4 gives
0 2 x V1 = 0
0 -3 V2

0v1 + 2v2 = 0
0v1 - 3v2 = 0

Eigenvector = 0 , 0
How does that give me any choice? Either theres some fundamental rule im missing or my maple skills need a workout as according to maple the vector is 1, 0
which would give 1+0 = 0 ? No?

Lambda = 1 gives

3 2 x V1 = 0
0 0 V2

3v1 + 2v2 = 0
0v1 + 0 v2 = 0

eigenvector = -2, 3
Ok I get that this works, why couldnt I say that the eigenvector is 2, -3?
wouldnt this give 3(2) + 2(-3) = 0
Whats the difference?

Clarifications greatly appreciated.

P.s: Sorry for the painful format, couldnt figure out how to draw matrices with latex.

Last edited: Oct 10, 2009
2. Oct 10, 2009

### Donaldos

How did you find those eigenvalues?

3. Oct 10, 2009

### Dissonance in E

Using det(a - lambdaI) = 0

4. Oct 10, 2009

### Donaldos

$$\lambda =4 \Rightarrow V_2=0$$ so $$V_1$$ can take any value, including 1.

$$\lambda =1 \Rightarrow 3V_1=-2V_2$$. Therefore the eigenvector you suggested is just as valid as the one you found using Maple.

Last edited: Oct 10, 2009
5. Oct 11, 2009

### HallsofIvy

Staff Emeritus
The eigenvalues of a trianguar (or diagonal) matrix are just the numbers on the main diagonal.

No, you only have 2v2= 0 and -3v2= 0 which are satisfied as long as v2= 0. v1 can be anything. An eigenvector for eigenvalue 4 is any vector of the form <x, 0>.

The definition of "eigenvalue" is that there exist a non-zero vector such that $Av= \lambda v$
Here you must have
$$\begin{bmatrix}4 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$$
$$= \begin{bmatrix}4x+ 2y \\ y\end{bmatrix}= \begin{bmatrix}4x \\ 4y\end{bmatrix}$$
which gives you the two equations 4x+ 2y= 4x and 4y= 4y. The first equation reduces to 2y= 0 which gives y= 0 and the second equation is automatically satisfied. Again, x can be anything.

Last edited: Oct 12, 2009