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FP2 Eigenvectors with an unknown element in 2x2 matrix. Watch

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    Hello,

    I am currently in need of assistance on the first part of the question below.

    I have found both eigenvalues to be:

    Lambda = k

    Lambda = 2

    How would I go about finding the eigenvectors? Would they have to be in terms of k?
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    (Original post by Wunderbarr)
    Hello,

    I am currently in need of assistance on the first part of the question below.

    I have found both eigenvalues to be:

    Lambda = k

    Lambda = 2

    How would I go about finding the eigenvectors? Would they have to be in terms of k?
    Well, one eigenvector is easy enough.

    It's \mathbf{M}\begin{pmatrix}x\\y \end{pmatrix} = 2\begin{pmatrix}x\\y \end{pmatrix}

    And the other will be in terms of k, as you rightly said:

    \mathbf{M}\begin{pmatrix}x\\y \end{pmatrix} = k\begin{pmatrix}x\\y \end{pmatrix}
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    am I too late today?
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    (Original post by Zacken)
    x
    (Original post by TeeEm)
    x
    Could either of you or someone else post some sort of help/guidance for finding the eigenvector for either eigenvalue?

    Assuming I don't know anything, as my method of multiplying the matrix M by the 2x1 matrix (x, y) isn't currently working for me
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    (Original post by Wunderbarr)
    Could either of you or someone else post some sort of help/guidance for finding the eigenvector for either eigenvalue?

    Assuming I don't know anything, as my method of multiplying the matrix M by the 2x1 matrix (x, y) isn't currently working for me
    I am not an expert but I can only find one eigenvector for lamda = 2
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    (Original post by Wunderbarr)
    Could either of you or someone else post some sort of help/guidance for finding the eigenvector for either eigenvalue?

    Assuming I don't know anything, as my method of multiplying the matrix M by the 2x1 matrix (x, y) isn't currently working for me
    Multiply out the matrix with the vector. Then you have two equations in x and y, so you can find y in terms of x.
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    (Original post by Wunderbarr)
    Could either of you or someone else post some sort of help/guidance for finding the eigenvector for either eigenvalue?

    Assuming I don't know anything, as my method of multiplying the matrix M by the 2x1 matrix (x, y) isn't currently working for me
    Let's work with the first one:

    kx + 3y = 2x and 2y = 2y \Rightarrow y=y, so we're good here; y can take any value, pick one.
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    (Original post by Zacken)
    Let's work with the first one:

    kx + 3y = 2x and 2y = 2k \Rightarrow y=k. Plug this back into the first equation: kx + 3k = 2x \Rightarrow x(k-2) = -3k \Rightarrow x = -\frac{3k}{k-2}.


    So our eigenvector is any multiple of \displaystyle \begin{pmatrix} -\frac{3k}{k-2} \\ k \end{pmatrix} = k \begin{pmatrix} -\frac{3}{k-2} \\ 1 \end{pmatrix} assuming I haven't made a silly algebraic mistake somewhere.
    So where did the 2y = 2k come from? :3
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    (Original post by Wunderbarr)
    So where did the 2y = 2k come from? :3
    Do you remember your matrix multiplication rules?

    \displaystyle \mathbf{M}\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} kx + 3y \\ 0x + 2y \end{pmatrix}
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    (Original post by Zacken)
    Do you remember your matrix multiplication rules?

    \displaystyle \mathbf{M}\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} kx + 3y \\ 0x + 2y \end{pmatrix}
    Yesss I do, except I don't seem to be getting it anywhere.

    Anyway I seem to have figured something out and got (1, 0) for the other eigenvector.
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    (Original post by Wunderbarr)
    Yesss I do, except I don't seem to be getting it anywhere.

    Anyway I seem to have figured something out and got (1, 0) for the other eigenvector.
    Correctomundo. (I think we actually have x = \lambda for any \lambda \in \mathbb{R}, but yours is as good as any, somebody will jump in and correct me if I'm wrong).
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    (Original post by Wunderbarr)
    So where did the 2y = 2k come from? :3
    Right, sorry about that. The woes of grogginess. That was completely wrong.
 
 
 
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