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Questing about eigenvector order

  1. Mar 24, 2014 #1

    939

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  2. jcsd
  3. Mar 25, 2014 #2

    DrClaude

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    Staff: Mentor

    The equation he obtains is
    $$
    \left( \begin{array}{cc} 8 & 1 \\ -8 & -1 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = 0
    $$
    ##(x,y) = (1, -8)## is a solution, but not ##(x,y) = (-8, 1)##. Try it for yourself: substitue the possible solutions in the above equation, and see what works.

    By the way, there is an infinite number of possible solutions: any vector ##a (1, -8)## is also a solution (with ##a## a scalar). Which one to choose is arbitrary. In the video, he could have chosen ##y=1## instead, and found ##(x,y) = (-1/8, 1)## (you can check for yourself that solution also).
     
  4. Mar 25, 2014 #3

    HallsofIvy

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    You talk here, and in a similar question in the homework section as if the "eigenvector" were just two numbers in arbitrary order. It is not- it is a vector which, in this case, can be represented[/b] as an ordered pair of numbers.

    You should, yourself, have done the multiplication
    [tex]\begin{pmatrix}8 & 1 \\ -8 & -1\end{pmatrix}\begin{pmatrix}-8 \\ 1\end{pmatrix}= \begin{pmatrix}-64+ 1 \\ 64- 1\end{pmatrix}= \begin{pmatrix}-63 \\ 63\end{pmatrix}[/tex]
    which is NOT
    [tex]0\begin{pmatrix}-8 \\ 1 \end{pmatrix}[/tex]
     
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