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Math Help - (Linear Alg Question) Why is this matrix's eigenvector [1,1] * t for t belong to R ?

  1. #1
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    (Linear Alg Question) Why is this matrix's eigenvector [1,1] * t for t belong to R ?

    Hi, I'm new here, and couldn't find the 'Linear algebra' subforum, only noticed advanced algebra and this subforum. So I figured, since this is not a 2nd year subject I'd post here. My apologies if this post doesn't belong here.

    In class I came to understand that the eigenvector is the vector v, in the equation (A-lamda*I)*v = 0, where A is an nxn matrix, lamda belong to R, is the eigenvalue of this matrix.

    In an example we solved: for A =
    [1 3]
    [2 2]
    where lamda = -1 and 4
    For lamda = -1: its (A+I)*v=0 simplifies to:

    [1 3/2 | 0 ]
    [0 0 | 0 ] so converting this into vector form
    [v1] v2*[-3/2]
    [v2] = [ 1 ] So v2=t then the eigenvector is
    [-3] * t for t belong to R
    [2]
    I understand for the above step it was taken from the linear simultaneous equation system:
    1) v1+(3/2)v2=0
    2) 0(v1) + 0(v2) = 0
    Because of the 2nd row being all 0s, we have an infinite amount of solutions, which is why we had to add a "t belong to R" parameter to the eigenvector.

    BUT for lamda = 4:

    { [1 3] - 4* [1 0] } * v = 0
    { [2 2] [0 1] }

    it simplifies down to
    [1 0 | 0]
    [0 1 | 0]

    If I was to follow the same method of linear equation system as above, how is something like this converted into the eigenvector
    [1] * t for t belong to R?
    [1]
    since: v1+0(v2) = 0
    0(v1)+v2 = 0 shouldn't v =
    [0] instead of [1]
    [0] instead of [1] ? And that it is a unique solution without even the need for parameter t?
    Last edited by autumnwater; February 4th 2014 at 11:02 PM.
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  2. #2
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    Re: (Linear Alg Question) Why is this matrix's eigenvector [1,1] * t for t belong to

    Nevermind, found the mistake, it's actually
    [1 -1 |0]
    [0 0 |0]
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  3. #3
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    Re: (Linear Alg Question) Why is this matrix's eigenvector [1,1] * t for t belong to

    The advanced algebra sub-forum includes Linear Algebra (it's in the description). This sub-forum is for pre-university algebra.

    Solving for A - 4I = 0 leads to:

    [-3 3][x]....[0]
    [2 -2][y] = [0]

    which leads to x = y.

    So any non-zero scalar multiple of (1,1) is an eigenvector belonging to the eigenvalue 4.
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    Re: (Linear Alg Question) Why is this matrix's eigenvector [1,1] * t for t belong to

    [QUOTE=autumnwater;809869]Hi, I'm new here, and couldn't find the 'Linear algebra' subforum, only noticed advanced algebra and this subforum. So I figured, since this is not a 2nd year subject I'd post here. My apologies if this post doesn't belong here.

    In class I came to understand that the eigenvector is the vector v, in the equation (A-lamda*I)*v = 0, where A is an nxn matrix, lamda belong to R, is the eigenvalue of this matrix.[quote]
    Better to think "Av= lambda v". Of course that is the same as "(A- lambda I)v= 0" but is more fundamental.

    In an example we solved: for A =
    [1 3]
    [2 2]
    where lamda = -1 and 4
    An eigenvector for this matrix, with eigenvalue -1 must satisfy
    \begin{bmatrix}1 & 3\\ 2 & 2 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x+ 3y \\ 2x+ 2y\end{bmatrix}= \begin{bmatrix}-x \\ -y\end{bmatrix}
    which is equivalent to the two equations x+ 3y= -x and 2x+2y= -y. Those both reduce to 3y= -2x, y= (-2/3)x so that a eigenvector must be of the form <x, (-2/3)t>= 3t<3, -2>. A vector is an eigenvector of A corresponding to eigenvalue -1 if and only if it is a multiple of <3, -2>.

    For eigenvalue 4, the equivalent \begin{bmatrix}1 & 3\\ 2 & 2 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x+ 3y \\ 2x+ 2y\end{bmatrix}= \begin{bmatrix}4x \\4y\end{bmatrix}
    which gives the two equations x+ 3y= 4x and 2x+ 2y= 4y. Those both reduce to y= x so that an eigenvector corresponding to eigenvalue 4 is <x, x>= x<1, 1>. A vector is an eigenvector of A corresponding to eigenvalue 4 if and only if it is a multiple of <1, 1>.

    For lamda = -1: its (A+I)*v=0 simplifies to:

    [1 3/2 | 0 ]
    [0 0 | 0 ] so converting this into vector form
    [v1] v2*[-3/2]
    [v2] = [ 1 ] So v2=t then the eigenvector is
    [-3] * t for t belong to R
    [2]
    I understand for the above step it was taken from the linear simultaneous equation system:
    1) v1+(3/2)v2=0
    2) 0(v1) + 0(v2) = 0
    Because of the 2nd row being all 0s, we have an infinite amount of solutions, which is why we had to add a "t belong to R" parameter to the eigenvector.

    BUT for lamda = 4:

    { [1 3] - 4* [1 0] } * v = 0
    { [2 2] [0 1] }

    it simplifies down to
    [1 0 | 0]
    [0 1 | 0]

    If I was to follow the same method of linear equation system as above, how is something like this converted into the eigenvector
    [1] * t for t belong to R?
    [1]
    since: v1+0(v2) = 0
    0(v1)+v2 = 0 shouldn't v =
    [0] instead of [1]
    [0] instead of [1] ? And that it is a unique solution without even the need for parameter t?
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