-4+2y=0 and y=2x comes from Eigenvectors

1. Apr 1, 2009

Ry122

I don't understand where -4+2y=0 and y=2x comes from
Is it obtained after finding the determinant or are the equations reconstructed that the matrix was created from?
http://users.on.net/~rohanlal/eigen1.jpg [Broken]

I also don't understand what's happening here. Where did the x(1,2) come from and how is e1=(1,2)
derived from this?
http://users.on.net/~rohanlal/eigen2.jpg [Broken]

Last edited by a moderator: May 4, 2017
2. Apr 1, 2009

lanedance

Re: eigenvectors

hi Ry22

if A is you matrix first solve the characteristic equation given by
$$det(\textbf{A} - \lambda.\textbf{I})$$
this give you the eigenvalues

Then to solve for the eignevectors corresponding to the ith eginevalue lambda, solve
$$\textbf{A}.\textbf{x} = \lambda_i.\textbf{x}$$

it is not clear whether your original matrix is the one given, of which one of the eigenvalues is zero, or this is actually the matrix B related to an eigenvalue given by
$$\textbf{B}.\textbf{x} = (\textbf{A}- \lambda_i.\textbf{I}).\textbf{x}= \textbf{0}$$

probably the 2nd case where the eigenvalue is known and included in the matrix - do you have any more info?

3. Apr 2, 2009

Ry122

Re: eigenvectors

the matrix does include the eigenvalues and i understand how they're obtained. im just not sure about how the eigenvectors are obtained

4. Apr 2, 2009

lanedance

Re: eigenvectors

defintion of eignevectors for eignevalue lambda i, are that they mapped into a scalar multiple (= lamda i) of themselves.
$$\textbf{A}.\textbf{x} = \lambda_i.\textbf{x}$$

re-writing the equation gives:
$$\textbf{B}.\textbf{x} = (\textbf{A}- \lambda_i.\textbf{I}).\textbf{x}= \textbf{0}$$

In your 2d case x = (x,y) where x and y are unknown, so an equation from B.x to get x in terms of y. (note both equations give you the same relation)

note the eigenvector is not unique, if u is an eigenvector, then so is k.u for any scalar k. So you only determine the eigenvector upto a constant.

Last edited: Apr 2, 2009
5. Apr 2, 2009

HallsofIvy

Staff Emeritus
Re: eigenvectors

Do the matrix multiplication:
$$\begin{bmatrix}-4 & 2 \\ -2 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-4x+ 2y \\-2x+ y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$
and, since two matrices are equal if and only if corresponding terms are equal, that gives -4x+ 2y= 0, -2x+ y= 0. Both of those equations reduce to the same thing, y= 2x. Now, by scalar multiplication of a matrix, it is true that
$$x\begin{bmatrix}1 \\ 2\end{bmatrix}= \begin{bmatrix}x \\ 2x\end{bmatrix}$$

Actualy, we could have just chosen x= 1, or any other non-zero value, to get an eigen vector.

6. Jun 21, 2009

Ry122

Re: eigenvectors

where did the x come from in the matrix on the RHS?
$$x\begin{bmatrix}1 \\ 2\end{bmatrix}= \begin{bmatrix}x \\ 2x\end{bmatrix}$$

7. Jun 21, 2009

HallsofIvy

Staff Emeritus
Re: eigenvectors

From the "x" on the LEFT side, of course! I just multiplied the x into the vector.

8. Jun 21, 2009

Ry122

Re: eigenvectors

well where did the 1 come from for the vector?

9. Jun 21, 2009

lanedance

Re: eigenvectors

which 1?

the key to understanding this I think is to recognise each eignvalue has in fact a subspace associated with it & the eigenvectors for a given eigenvalue is a set of vectors that spans that subspace.

In this case the subspace is a line, so you can pick any vector that spans the line given by -4x + 2y = 0, or equivalently y=2x. This is what I meant by determining the vector upto a constant.

Hey Halls, on a seprate issue, any new latex seems to display as black for me & I can;t read it any more - any ideas on how I can fix it?

10. Jun 22, 2009

Ry122

Re: eigenvectors

so i have y=2x
and a matrix is derived from it.
it consists of x and 2x. i can understand where 2x came from because y=2x.
but why do we make the other element of this matrix x?

11. Jun 22, 2009

lanedance

Re: eigenvectors

I think Halls was using an easy way to show the line y = 2x is parallel to the vector [1,2] as follows

i can't read tex at the moment so here goes, these should be column vectors to match above
[x,y]
=[x,2x] from the equation of the line through the origin y=2x
=x[1,2]

12. Jun 22, 2009

Feldoh

Re: eigenvectors

It's simple matrix multiplication, a scalar times a matrix is just the product of each term in the matrix and the scalar (in this case x).

13. Jun 22, 2009

Fredrik

Staff Emeritus
Re: eigenvectors

We don't. It's been x the whole time. You started by calling the first component x and the second y. Then you found out that they satisfy the relationship y=2x. That means that you can write your two components as x and 2x. If you prefer, you can write them as y/2 and y instead.

14. Jun 22, 2009

HallsofIvy

Staff Emeritus
Re: eigenvectors

You are asking the right person because the same thing happened to me! I asked the administrators and they said they had put in an update of the LaTex and suggested that I was using an old version of "Internet explorer". It turned out I was (I was using version 6 and the latest is version 8). I downloaded the new version and now the LaTex works fine.

15. Jun 22, 2009

lanedance

Re: eigenvectors

thanks - will try the update