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Math Help - Eigenvectors/Eigenvalues help!

  1. #1
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    Eigenvectors/Eigenvalues help!

    Hi everyone

    I have a piece of maths coursework of an engineering course which the final part of which is providing some difficulties, so was wondering if someone here would be kind enough to help please.

    I am aware of what eigenvalues/vectors are, but i am not certain how to apply the things i know to this question - nor does anyone on my course or any other doctor i have asked!

    Anyway, the question (or the very first part!).

    Show that:

     <br /> <br />
\left[ \begin{array}{cccc} cos(x) & -sin(x) & 0 \\ sin(x)& cos(x) & 0 \\ 0 & 0 & 1 \end{array} \right]<br /> <br />

    Has the an eigenvector


     <br /> <br />
\left[ \begin{array}{cccc} 1 \\ i \\ 0 \end{array} \right]<br /> <br />

    Write the answer in exponential form.

    Id appreciate any help on the matter

    thanks in advance!
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  2. #2
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    " Write the answer in exponential form" means "Use the fact that \cos x+i\sin x = e^{ix}".

    Multiply the matrix by the given eigenvector to get \begin{bmatrix} \cos x& -\sin x& 0 \\ \sin x& \cos x& 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1\\i\\0\end{bmatrix}= \begin{bmatrix}\cos x-i\sin x\\ \sin x + i\cos x \\0\end{bmatrix}= \begin{bmatrix}e^{-ix}\\ ie^{-ix} \\0\end{bmatrix} = e^{-ix}\begin{bmatrix}1\\i\\0\end{bmatrix} .

    Now you should be able to see what the eigenvalue is.
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  3. #3
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    Hi

    Thanks for the fast reply. I understand the first part of your reply, but i really dont understand how to get an eigenvalue from this.

    Sorry for appearing to not know a thing here really, I am really keen to understand all this though as its essential for my engineering degree.

    Thanks
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  4. #4
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    Quote Originally Posted by farso View Post
    I really don't understand how to get an eigenvalue from this.
    The definition of an eigenvalue of a matrix A is that it is a number \lambda such that for some nonzero vector v, the equation Av = \lambda v holds.

    In this problem, if we call the matrix A, and v is the vector \mathbf{v}=\begin{bmatrix}1\\i\\0\end{bmatrix}, then the calculation in my previous comment shows that A\mathbf{v} = e^{-ix}\mathbf{v}. That shows that v is an eigenvector for the eigenvalue \lambda = e^{ix}.
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  5. #5
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    Thanks very much, it tok me a fair bit of work to get my head around iSinx + Cos x in exponential form, but finally got it.
    Perhaps you can help me with another part of the same question?

    If the first matrix i said was Rz
    and
    <br />
Ry = \left[ \begin{array}{cccc} cos(x) & 0 & sin x \\ 0 & 1 & 0 \\ -sin x & 0 & cos x \end{array} \right]<br />
    which are both used to represent movment of a robot arm starting at.
    <br />
\left[ \begin{array}{cccc} 1 \\ 0 \\ 0 \end{array} \right]<br />
    and its movment is applied pi/2 to z axis followed by the y axis
    show that it ends in position
    <br />
\left[ \begin{array}{cccc} 0 \\ 1 \\ 0 \end{array} \right]<br />

    Thanks for your help!
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  6. #6
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    Quote Originally Posted by Opalg View Post

    Multiply the matrix by the given eigenvector to get \begin{bmatrix} \cos x& -\sin x& 0 \\ \sin x& \cos x& 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1\\i\\0\end{bmatrix}= \begin{bmatrix}\cos x-i\sin x\\ \sin x + i\cos x \\0\end{bmatrix}= \begin{bmatrix}e^{-ix}\\ ie^{-ix} \\0\end{bmatrix} = e^{-ix}\begin{bmatrix}1\\i\\0\end{bmatrix} .
    doesn't sinx + icosx = ie^ix ???
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  7. #7
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    Quote Originally Posted by geo3 View Post
    doesn't sinx + icosx = ie^ix ???
    No, ie^{ix} = i(\cos x+i\sin x) = i\cos x - \sin x.
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