# Equation for eigenline

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• May 14th 2009, 01:27 AM
bobred
Equation for eigenline
Hi

Im having trouble with finding the equation of an eigenline. I can see that

$\left(\begin{array}{cc} 1 & 2\\
3 & 2\end{array}\right)\left(\begin{array}{c}
x\\
y\end{array}\right)=4\left(\begin{array}{c}
x\\
y\end{array}\right)
$

gives the following equations

$x+2y=4x$
$3x+2y=4y$

Which is

$-3x+2y=0$
$3x-2y=0$

But I'm not sure how to reduce these equations to find the equation of the eigenline

$3x-2y=0$

It's probably simple but I just can't see it.

Thanks in advance.
• May 14th 2009, 01:46 AM
Swlabr
You simply then constuct a basis vector that fits that equation: $(x, x3/2)$. For example, $(1, 3/2)$.
• May 14th 2009, 03:36 AM
bobred
Hi

Thanks for the reply but I must be being dense, could you elaborate please.
• May 14th 2009, 04:30 AM
Swlabr
The work you have done has given you conditions for the $x$ and $y$. So a vector is an eigenvector if and only if it it obeys those restrictions and is non-zero. That is to say, it is of the form $(x, x3/2)$ where $x \in F \setminus \{0\}$. Clearly, $(1, 3/2)$ fulfills those conditions.

The bit which says "v an eigenvector if and only if it it obeys those restrictions" holds because:
$\Rightarrow$ as $(x,y)$ was an arbitrary eigenvector (for the eigenvalue 4)
$\Leftarrow$ any vector that adheres to those conditions must also be an eigenvector for that eigenvalue as you just showed that it fulfils the conditions to be an eigenvector ( $v \lambda = M v$).

I apologise if that still doesn't make sense! Feel free to ask again. I still haven't got to grips with Linear algebra, so I can totally understand what you mean about feeling dense!
• May 14th 2009, 05:03 AM
bobred
Hi

Thanks, I've got it.
• May 14th 2009, 01:06 PM
HallsofIvy
You do understand, don't you, that the y= (3/2)X that you got from swlabr is the same as 2y= 3x or 3x-2y= 0, exactly the equation, you posted originally?

The "eigenline" is given by 3x- 2y= 0!
• May 15th 2009, 04:47 AM
bobred
Hi

Yes I do. The problem was that for one stupid moment I could'nt see how the eigen line equation was found.

Thanks
• January 16th 2013, 11:38 AM
Paulos
Re: Equation for eigenline
Hi, I have come across this same example, and I am the same I don't get it. The text says "these equations both reduce to 3x-2y=0. Thus the eigenlines corresponding to the eigenvalue k =4 has the equation y=(3/2)x". I understand that 3x-2y=0 is the same as y=(3/2)x but cannot seem to get my head around how the equations reduce to 3x-2y=0. I'm sure this is just basic manipulation but I can't see it.
• January 16th 2013, 12:41 PM
HallsofIvy
Re: Equation for eigenline
Are you referring to the equations x+ 2y= 4x and 3x+ 2y= 4y?

Subtract 4x from both sides of x+ 2y= 4x to get -3x+ 2y= 0. Subtract 4y from both sides of 3x+ 2y= 4y to get 3x- 2y= 0. Those are clearly the same (one is the other multiplied by -1).
• January 18th 2013, 04:43 AM
Paulos
Re: Equation for eigenline
Thanks for that, I have tried this on an assignment question, could someone please check? I'm not after the answer, just where I'm going wrong.
A=5 7
-2 -4
I believe the eigen values to be k=3 and -2

This gives me the eigen equations for k=3 5x+7y = 3x and -2x-4y =3y

These I believe cancel down to y=(2/7)x

Which then gives me an eigen vector of 7
2

I have done similar with the second eigen value, I know these are wrong because a subsequent part of the question asks for the pdp-1 which I have carried out and does no bring me back to the original matrix.

I believe the error is with the cancel down which is the bit I cannot grasp...