# Real and Zero eigenvalues

1. May 18, 2016

### rmiller70015

1. The problem statement, all variables and given/known data
$$\frac{d\vec{Y}}{dt} = \begin{bmatrix} 2 & 4 \\ 3 & 6 \end{bmatrix} \vec{Y}$$
Find the eigenvalues and eigenvectors
2. Relevant equations

3. 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?

2. May 18, 2016

### pasmith

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

3. May 18, 2016

### rmiller70015

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.

4. May 18, 2016

### Ray Vickson

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.

5. May 18, 2016

### rmiller70015

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

6. May 18, 2016

### Ray Vickson

I do not understand what you are trying to say. What is the correct $v_1$? What is the correct $v_2$? Instead of trying to describe these in words, just show the actual numerical entries.

7. May 18, 2016

### rmiller70015

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

Sory for putting so little work into this, I've got the answer and moved onto the next problem at this point.