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Thread: Eigenvectors!

  1. #1
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    Eigenvectors!

    (i) Determine all the eigenvalues of A
    (ii) For each eigenvalue of A, find the set of eigenvectors corresponding to

    A = ( 1 2 )
    ( 3 2 )

    I found eigenvalues of A to be 4 and -1.

    I also found the eigenvectors to be (2/3,1) for =4 and (-1,1) for =-1

    BUT the solution in the back of the book says (2,3) for =4 and (1,-1) for =1

    I'm soo confusedd! Can someone tell me what's wrongg? Am I calculating the eigenvectors wrong?

    Here's how i calculate eigenvector for =4

    A-4I = (-3 2)
    (3 -2)

    then (-3 2 |0)
    (3 -2 |0)

    and i end up with x1 -(2/3) x2 =0. so x2 = t, and x1 = (2/3)t. eigenvector = (2/3,1) but book says (2,3)

    and for =-1

    A-(-1)I = (2 3 )
    (2 3 )

    then (2 3 |0)
    (2 3 |0)

    and i end up with x1+x2=0. x2= t and x1 = -t. eigenvector = (-1,1) but book says (1,-1)
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  2. #2
    Senior Member roninpro's Avatar
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    You don't have a problem. Note that your eigenvectors are just multiples of the book's answer. This is fine!
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  3. #3
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    Smile

    Hi
    $\displaystyle \texttt{det}(A-I_{2}X)=\texttt{det}\begin{pmatrix}
    1-X & 2\\
    3 & 2-X
    \end{pmatrix}=(1-X)(2-X)-6=X^2-3X-4=(X-4)(X+1)$
    and thus your eigenvalues are 4 and -1.
    For 4
    $\displaystyle A\begin{pmatrix}
    x\\
    y
    \end{pmatrix}=4\begin{pmatrix}
    x\\
    y
    \end{pmatrix}\Leftrightarrow \begin{pmatrix}
    x+2y\\
    3x+2y
    \end{pmatrix}=$$\displaystyle \begin{pmatrix}
    4x\\
    4y
    \end{pmatrix}\Leftrightarrow \left\{\begin{matrix}
    -3x+2y=0\\
    3x-2y=0
    \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
    x=\frac{2}{3}t\\
    y=t,t\in \mathbb{R}
    \end{matrix}\right.$
    and thus the eigenvectors are of the form,
    $\displaystyle x=x_1(\frac{2}{3},1),x_1\in \mathbb{R}\setminus \left \{ 0 \right \}$
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  4. #4
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    The set of all eigenvectors of a matrix A, corresponding to a given eigenvalue, $\displaystyle \lambda$, form a subspace. In particular, any multiple of an eigenvector is also an eigenvector, corresponding to the same eigenvalue: If $\displaystyle Av= \lambda v$ and "r" is any scalar, then $\displaystyle A(rv)= r(Av)= r(\lambda v)= (r\lambda)v= \lambda (rv)$.
    Last edited by HallsofIvy; Jul 11th 2010 at 08:52 AM.
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