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Thread: 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 $\displaystyle A=\begin{pmatrix}6 &13 &-8 \\
    2 &5 &-2 \\
    7 & 17 &-9\end{pmatrix}$, given $\displaystyle \lambda_1=-2$ and $\displaystyle \lambda_2=3$, then $\displaystyle v_1=\begin{pmatrix}1 \\0 \\1\end{pmatrix}$ and $\displaystyle v_2=\begin{pmatrix}1 \\1 \\2\end{pmatrix}$ behave as $\displaystyle Tv_i=\lambda_i$.

    You now know there exists a vector $\displaystyle x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$ such that $\displaystyle Tx=1x$. Thus $\displaystyle x$ is an eigenvector with eigenvalue $\displaystyle 1$. Since this does not match any of the previous eigenvalues, you know that this new $\displaystyle x$ is linearly independent of $\displaystyle v_1$ and $\displaystyle v_2$ (explain why). Then, you should be able to find the components of $\displaystyle x$ just as you found $\displaystyle v_2$.
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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 $\displaystyle v_1$ and $\displaystyle v_2$ are eigenvectors of T with corresponding eigenvalues $\displaystyle \lamba_1$ and $\displaystyle \lambda_2$ respectively, $\displaystyle \lambda_1\ne \lambda_2$, and that $\displaystyle av_1+ bv_2= 0$ for some scalars a and b. Applying T to both sides, $\displaystyle T(av_1+ bv_2)= aT(v_1)+ bT(v_2)= \lambda_1 av_1+ \lamba_2 bv_2= 0$. Multiplying the first equation by $\displaystyle \lambda_1$ gives $\displaystyle \lambda_1 av_1+ \lambda-1 bv_2= 0$ and subtracting those two equations, $\displaystyle (\lambda_2- \lambda_1) bv_2= 0$. Since $\displaystyle \lambda_2- \lambda_1$ is not zero, b= 0 and then similarly for a.
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