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Quick eigenvector help

  1. May 14, 2008 #1

    rock.freak667

    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data

    The matrix,A,given by
    [tex]
    A = \left(
    \begin{array}{ccc}
    7 & -4 & 6\\
    2 & 2 & 2 \\
    -3 & 4 & -2 \
    \end{array}
    \right)
    [/tex]

    has eigenvalues 1,2,4 . Find a set of corresponding eigenvectors.

    Hence find the eigenvalues of B, where

    [tex]
    B = \left(
    \begin{array}{ccc}
    10 & -4 & 6\\
    2 & 5 & 2 \\
    -3 & 4 & 1 \
    \end{array}
    \right)
    [/tex]

    and state a corresponding set of eigenvectors.
    2. Relevant equations



    3. The attempt at a solution


    Well I easily found the eigenvectors
    [itex]
    \lambda=1[/itex] corresponds to
    [tex]
    \left(
    \begin{array}{c}
    -1\\
    0 \\
    1\
    \end{array}
    \right)
    [/tex]

    [itex]
    \lambda=2[/itex] corresponds to
    [tex]
    \left(
    \begin{array}{c}
    -4\\
    1 \\
    4\
    \end{array}
    \right)
    [/tex]

    [itex]
    \lambda=4[/itex] corresponds to
    [tex]
    \left(
    \begin{array}{c}
    2\\
    3 \\
    1\
    \end{array}
    \right)
    [/tex]


    Well for the one with B, just solve det(b-[itex]\lambda[/itex]I)=0 to get the e.values... but it says to state a set of e.vectors meaning that I am not supposed to work them out.
    The only thing I can really say about A and B is that in B all the elements in the main diagonal are the elements in the main diagonal of A with 3 added to them
     
  2. jcsd
  3. May 14, 2008 #2
    In other words, B = A + 3I. Can you find a simple way to get B's eigenvalues and eigenvectors without explicitly working through B? Use the definitions of eigenvectors and eigenvalues.
     
  4. May 14, 2008 #3

    rock.freak667

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    Homework Helper

    Well I could say that

    [itex]Det(B-\lamda I)=0[/itex] and put B=A+3I, then say that I can reduce A to a triangular matrix such that the diagonal elements(Eigenvalues of A)+(3-[itex]\lambda[/itex])=0 and solve from there. And get the e.values to be 7,4,5. But how would the e.vectors be altered?
     
    Last edited: May 14, 2008
  5. May 14, 2008 #4
    The determinant is just a way to extract the eigenvalues mechanically. The original definition should state that L is an eigenvalue of A iff Av = Lv for some vector v (an eigenvector of A), or equivalently, A - LI = 0. Thus, to find eigenvalues L, we are looking for solutions of the equation B - LI = 0, which is equivalent to A + (3 - L)I = 0. We already know the solutions of this equation, specifically we know that L - 3 = L' where L' is an eigenvalue of A. No messing around with matrix forms or determinants necessary. Similarly, you can use the original equation to see how the eigenvectors of A compare to the eigenvectors of B. Can you see the way from here? :)
     
    Last edited: May 14, 2008
  6. May 14, 2008 #5

    rock.freak667

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    Homework Helper

    I got the same answer but it all amounts to the same thing I guess.thanks

    But then the e.vectors would be unchanged?
     
  7. May 14, 2008 #6
    Indeed, you get the equation Av = (L - 3)v, where L is an eigenvalue of B and v is an eigenvector of B, implying that whenever v is an eigenvector for A, it is also an eigenvector for B, albeit with a different eigenvalue.
     
    Last edited: May 14, 2008
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