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The VERY last part of finding an eigenvector Watch

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    I can do everything, untill I get to the point of actually putting the system of equations into that eigenvector "form". I won't use my actual numbers, but say I have solved everything and got:

    ax = by
    ax = by

    Does this mean that for everyone 1 'y', I get (a/b) lots of x's, and so my vector is (a/b, 1). Or is it the other way round so for everyone 1 'x', I get (b/a) y's, so my vector is (1,b/a). Or is it simply (a,b)?

    Thank you for your help?
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    It might be clearer if you did use your numbers. x is an eigenvector if Ax=cx, where A is a matrix, x a vector and c a scalar.

    Is your 'by' a scalar multiple of x? If so, and if a is your matrix, then you x is an eigenvector.
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    (Original post by Hopple)
    It might be clearer if you did use your numbers. x is an eigenvector if Ax=cx, where A is a matrix, x a vector and c a scalar.

    Is your 'by' a scalar multiple of x? If so, and if a is your matrix, then you x is an eigenvector.
    If I use numbers, this is what has happened:

    I have the matrix:

    10 -6
    -6 5

    I worked out the eigenvalues to be 14 and 1 and then to work out the eigenvector when lambda = 14, I get

    (A - lambda I). v = 0
    v = (x,y)

    I get -4x - 6y = 0
    and -6x - 9y = 0

    So by rearranging and simplifying, I get

    -2x = 3y
    -2x = 3y

    Now I need to put this into the vector form thing, but I don't get what you have to do.
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    That shows your equations are consistent. Pick a number for x - let's pick 3 (I picked 3 to make the numbers nice, but any number will do), so y=-2. An eigenvector is then (3,-2). Picking different values for x will give you a scalar multiple of that eigenvector.
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    (Original post by claret_n_blue)
    If I use numbers, this is what has happened:

    I have the matrix:

    10 -6
    -6 5

    I worked out the eigenvalues to be 14 and 1 and then to work out the eigenvector when lambda = 14, I get

    (A - lambda I). v = 0
    v = (x,y)

    I get -4x - 6y = 0
    and -6x - 9y = 0

    So by rearranging and simplifying, I get

    -2x = 3y
    -2x = 3y

    Now I need to put this into the vector form thing, but I don't get what you have to do.
    This shows that any eigenvector of A with eigenvalue 14 has the form \begin{bmatrix} x \\ -\frac{2x}{3} \end{bmatrix} where x \neq 0
 
 
 
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