# Real and Zero eigenvalues

## Homework Statement

$$\frac{d\vec{Y}}{dt} = \begin{bmatrix} 2 & 4 \\ 3 & 6 \end{bmatrix} \vec{Y}$$
Find the eigenvalues and eigenvectors

## The Attempt at a Solution

I found the eigenvectors to be
$$\vec{v_1} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} , \vec{v_2} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$$

I found a widget on Wolfram Alpha that says the second eigenvector should be:
$$\begin{bmatrix} 2 \\ 3 \end{bmatrix}$$

I am more inclined to believe wolfram alpha is correct, but can someone show me why?

pasmith
Homework Helper
How are we to determine whether, and if so how, you have made an error if you don't post your working?

It was mostly by looking at the matrix and trying to make them the same and now that you say that when I do the algebra on it to post it, I see what I did wrong. I just tried to look at 3 and -2 to find the solution when I should have said 3x-2y=0 and 3x=2y then the vector becomes clear.

Ray Vickson
Homework Helper
Dearly Missed
It was mostly by looking at the matrix and trying to make them the same and now that you say that when I do the algebra on it to post it, I see what I did wrong. I just tried to look at 3 and -2 to find the solution when I should have said 3x-2y=0 and 3x=2y then the vector becomes clear.

Neither of your ##v_1## or ##v_2## are eigenvectors. If ##A## is your matrix we have ##Av_1 = (8,12)^T##, which cannot be a multiple of ##v_1##: the first component of ##v_1## is larger than the second component, while the opposite is true for ##Av_1##. Similarly, ##v_2## cannot be an eigenvector of ##A## because the components of ##v_2## have opposite signs, while those of ##Av_2## have the same sign.

Also: you did not show the eigenvalues.

rmiller70015
Eigens are 0,8. The problem was v2 I ommited a negative sign in the first vector on accident.

Ray Vickson
$$\vec{v_1}= \begin{bmatrix} -2 \\ 1 \end{bmatrix} , \vec{v_2}= \begin{bmatrix} 2 \\ 3 \end{bmatrix}$$