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Math Help - Transformations/eigenvectors

  1. #1
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    Transformations/eigenvectors

    I have the following question, sorry in advance for my poor mathematical notation, having some serious computer issues!
    A a 3x3 matrix:
    6 13 -8
    2 5 -2
    7 17 -9

    I was given an eigenvector: v=(1 0 1)^T where its 3x1. I had to find the corresponding eigenvalue which is -2. Then had to show 3 is an eigenvalue and its corresponding eigenvector is (1/2 1/2 1)^T

    I am now stuck at the following question:

    The matrix A defines a linear transormationT: from real numbers of dimension 3 to the real numbers of dimension 3 by T(x)=Ax. It is known that T fixes a non-zero vector x, T(x)=x. Use this information to determine another eigenvalue of A

    Any help would be gratefully received!
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  2. #2
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    Eigenvalues are of course scaling factors by which a transformation may multiply a corresponding eigenvector. So in your case, you know that

    if A=\begin{pmatrix}6   &13  &-8              \\<br />
 2   &5    &-2 \\<br />
 7 &  17  &-9\end{pmatrix}, given \lambda_1=-2 and \lambda_2=3, then v_1=\begin{pmatrix}1 \\0 \\1\end{pmatrix} and v_2=\begin{pmatrix}1 \\1 \\2\end{pmatrix} behave as Tv_i=\lambda_i.

    You now know there exists a vector x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} such that Tx=1x. Thus x is an eigenvector with eigenvalue 1. Since this does not match any of the previous eigenvalues, you know that this new x is linearly independent of v_1 and v_2 (explain why). Then, you should be able to find the components of x just as you found v_2.
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  3. #3
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    Thats a great help, thank you. Why is it that we know for sure that this new x is linearly independent of v1 and v2?
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  4. #4
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    Eigenvectors corresponding to distinct eigenvectors are always independent.

    Suppose v_1 and v_2 are eigenvectors of T with corresponding eigenvalues \lamba_1 and \lambda_2 respectively, \lambda_1\ne \lambda_2, and that av_1+ bv_2= 0 for some scalars a and b. Applying T to both sides, T(av_1+ bv_2)= aT(v_1)+ bT(v_2)= \lambda_1 av_1+ \lamba_2 bv_2= 0. Multiplying the first equation by \lambda_1 gives \lambda_1 av_1+ \lambda-1 bv_2= 0 and subtracting those two equations, (\lambda_2- \lambda_1) bv_2= 0. Since \lambda_2- \lambda_1 is not zero, b= 0 and then similarly for a.
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