Basic confusion with eigenvectors

[estro]

Given the matrix:
$$\left( \begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array} \right)$$

I'll find eigenspace for the eigenvalue of t=0, So I have to solve:

$$\left( \begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array} \right) {(x,y)}^t=0$$

Then I do $$R_2->R_2-2R_1$$ and get x+2y=0 => x=-2y => get the eigenvector (-1,2).

But wolframalpha tells me the eigenvector for this eigenvalue should be (1,2).

Where is my sin?

 

[I like Serena]

Hi estro!

You've got the wrong definition of an eigenvector.
Check for instance: http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors

 

[estro]

I'n my first pose I did a mistake, I wrote eigenvalue but meant eigenvector.
Fixed it.

 

[I like Serena]

I'n my first pose I did a mistake, I wrote eigenvalue but meant eigenvector.
Fixed it.

Still, an eigenvector of a matrix A is a vector v that satisfies:
$$A v = \lambda v$$
for some number lambda, which is the eigenvalue.

To solve it you need to set up the so called characteristic quadratic equation and solve that...

 

[estro]

Thank you for your response, but I know what is eigenvalue, eigenvector and characteristic polynomial, I left out the technical details and only showed how i calculate the eigen space for eigenvalue 0. [this matrix has 2 eigenvalues 5 and 0]

 

[I like Serena]

Thank you for your response, but I know what is eigenvalue, eigenvector and characteristic polynomial, I left out the technical details and only showed how i calculate the eigen space for eigenvalue 0. [this matrix has 2 eigenvalues 5 and 0]

Sorry! My bad! I guess I didn't read carefully enough.

In that case, your solution is the wrong way around: it should be for instance (2, -1).

The solution you mentioned from WolframAlpha is the eigenvector belonging to eigenvalue 5.

 

[Ray Vickson]

Your sin is an incorrect solution: from x = -2y you get (-2,1) (by putting y = 1) or (2,-1) (by putting y = -1). There is no way to get (-1,2) or (1,-2).

RGV

 

[estro]

Thanks!