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Find a matrix that diagonalizes the matrix A

  1. Dec 5, 2006 #1
    This could be a dumb question..but here goes:

    Im trying to find a matrix that diagonalizes the matrix A. If I find my eigenvalues and then proceed to get my eigenvectors...after row reducing I get 1, 1, 0, 0.(2X2 matrix, two 1's on top).what is the eigenvector?
     
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  3. Dec 5, 2006 #2

    Hootenanny

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    In this case you have a single eigenvector of value (1,0) and I say you would have made a mistake somewhere in the question since n eigenvectors are required to diagonalise an nxn matrix.
     
  4. Dec 5, 2006 #3
    I got two eigenvectors, i just didnt know what to do with:[tex]\left(\begin{array}{cc}1&1\\0&0\end{array}\right)[/tex]
     
  5. Dec 5, 2006 #4

    Hootenanny

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    Apologies, I misread your post. To find your eigenvector you need the following to be true;

    [tex]\left(\begin{array}{cc}1&1\\0&0\end{array}\right)\cdot \left(\begin{array}{c}x\\y\end{array}\right) = 0[/tex]

    Do you know why?

    P.S. I assume you have obtained the above matrix by applying the Characteristic equation, namely [itex]\det(A-\lambda\cdot I) = 0[/itex]
     
    Last edited: Dec 5, 2006
  6. Dec 5, 2006 #5
    Yes thats how I got it. And Im not sure why they must equal 0.
     
  7. Dec 5, 2006 #6
    No one cares that your an english major..have you pointed that out in every post?
     
  8. Dec 6, 2006 #7

    Hootenanny

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    Okay, so you've got your matrix A, solved the characteristic equation and obtained your eigenvalues. You've then applied your eigenvalues to your characteristic equation of A and obtained the following;

    for [itex]\lambda =1 [/itex]

    [tex]( A - (1)I ) = \left(\begin{array}{cc}1&1\\0&0\end{array}\right)[/tex]

    Now the definition of an eigenvector (v) of matrix A is a non-zero vector such that [itex]A\underline{v}=\lambda\underline{v}[/itex], where [itex]\lambda[/itex] is your eigenvalue. We can rewrite the above relationship as

    [tex]A\underline{v} - \lambda\underline{v} = 0[/tex]

    Using the fact the [itex]I\underline{v} = \underline{v}[/itex] (I being the Identity matrix) we obatin;

    [tex]A\underline{v} - \lambda I \underline{v} = 0[/tex]

    [tex]\therefore \; (A-\lambda I)\underline{v} = 0[/tex]

    Does that make sense? In R2 (as in your case);

    [tex]\underline{v} = \left(\begin{array}{c}x\\y\end{array}\right)[/tex]

    which is where the equation I posted comes from. In fact the characteristic equation follows directly from this relationship.
     
  9. Dec 6, 2006 #8
    ohhh, Ok, that makes sense,

    Thanks
     
  10. Dec 6, 2006 #9

    Hootenanny

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    No problem, so what do you obtain for your eigenvector corresponding to [itex]\lambda=1[/itex]?
     
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