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Why is this always the case A=[itex]\begin{array}{cc}1 & 6

  1. Nov 4, 2011 #1
    Why is this always the case...

    1 & 6 \\
    4 & 3 \\



    1-7 & 6 \\
    4 & 3-7 \\

    -6 & 6 \\
    4 & -4 \\

    Why is [itex]A_{11}[/itex] and [itex]A_{12}[/itex]
    always a multiple of
    [itex]A_{21}[/itex] and [itex]A_{22}[/itex]?

    Is this a feature of Eigenvalues or is this done on purpose to make solving the eigenvectors easier?
  2. jcsd
  3. Nov 4, 2011 #2


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    Staff Emeritus
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    Gold Member

    Re: Eigenvalues/Eigenvectors

    You seem to mean [itex](A-\lambda I)_{11}[/itex], [itex](A-\lambda I)_{12}[/itex] and so on.

    [itex]Ax=\lambda x[/itex] implies [itex](A-\lambda I)x=0[/itex]. So if [itex]A-\lambda I[/itex] is invertible, x=0. An eigenvector of A with eigenvalue [itex]\lambda[/itex] is by definition a non-zero x that satisfies [itex]Ax=\lambda x[/itex]. So A has an eigenvector with eigenvalue [itex]\lambda[/itex] if and only if [itex]A-\lambda I[/itex] is not invertible. For a matrix to be not invertible, its rows must be linearly dependent.
  4. Nov 4, 2011 #3
    Re: Eigenvalues/Eigenvectors

    Clear as day now, thank you
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