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Eigenvalues and eigenvectors Watch

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    When you find the eigenvalues and corresponding eigenvectors of a 3x3 matrix, when is there an invariant plane rather than just an invariant line?
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    Is this for MEI?
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    (Original post by Louisb19)
    Is this for MEI?
    I'm doing AQA actually. FP4.
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    (Original post by B_9710)
    When you find the eigenvalues and corresponding eigenvectors of a 3x3 matrix, when is there an invariant plane rather than just an invariant line?
    when you find the eigenvectors of a 3 x 3 matrix, for a particular value you expect the equations to be dependent and you end up with two equations from which you can find an eigenvector (direction of invariant line through O)

    Sometimes you get all 3 equations to be identical ...
    then that is a plane through the origin
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    There is an eigenplane when there is no unique solution to the 3 systems of equations.

    Basically, say you have x'=ax+by+cz, y' is also ax+by+cz, etc.

    So you can write you have 2 eigenvectors, let's say the equation is x+2y+3z=3, one eigenvector would be (1,1,0) and another one would be (2,2,-1).

    They will go through the origin when the constant is 0.
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    So the determinant of the [M-λ] matrix would be 0 for that particular value of λ?
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    (Original post by B_9710)
    So the determinant of the [M-λ] matrix would be 0 for that particular value of λ?

    No, the determinant is always 0 of the [M-λ] matrix! .

    Anyway, the determinant being 0 doesn't imply there is no solution, just that there is no unique solution.
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    (Original post by Argylesocksrox)
    There is an eigenplane when there is no solution to the 3 systems of equations.

    Basically, say you have x'=ax+by+cz, y' is also ax+by+cz, etc.

    So you can write you have 2 eigenvectors, let's say the equation is x+2y+3z=3, one eigenvector would be (1,1,0) and another one would be (2,2,-1).

    They will go through the origin when the constant is 0.
    Did you mean no unique solution?
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    (Original post by Argylesocksrox)
    No, the determinant is always 0 of the [M-λ] matrix! .

    Anyway, the determinant being 0 doesn't imply there is no solution, just that there is no unique solution.
    Of course
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    (Original post by B_9710)
    Did you mean no unique solution?

    Yeah, sorry.
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    So basically when all 3 equations formed are the same, you will be able to find 2 linearly independent eigenvectors and so will have an invariant plane?
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    (Original post by B_9710)
    So basically when all 3 equations formed are the same, you will be able to find 2 linearly independent eigenvectors and so will have an invariant plane?
    Yeah, that's right.
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    Thanks to everyone that helped.
 
 
 
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