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Math Help - Equation for eigenline

  1. #1
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    Equation for eigenline

    Hi

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

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

    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.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    You simply then constuct a basis vector that fits that equation: (x, x3/2). For example, (1, 3/2).
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  3. #3
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    Hi

    Thanks for the reply but I must be being dense, could you elaborate please.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    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!
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  5. #5
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    Hi

    Thanks, I've got it.
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  6. #6
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    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!
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  7. #7
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    Hi

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

    Thanks
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  8. #8
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    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.
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  9. #9
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    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).
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  10. #10
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    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...
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