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Math Help - diagonalized matrices

  1. #1
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    diagonalized matrices

    Hi guys, I'm hoping you can help me. I'm sure I've done something wrong, though I can't figure out what. The question I have to answer is this:

    Find invertible U and diagonal D such that A = UDU^{-1}. A is the matrix:

    2, -4, 2
    -4, 2, -2
    2, -2, -1

    I have found the three eigenvalues -2, -2, 7. I understand that these values, along the diagonal of a matrix otherwise filled with zeros, make up the D matrix.

    I have also found the eigenvectors that go with the eigenvalues. Respectively they are [0,1,2], [1,0,-2], [1,-1,1/2]. I understand that these three vectors together make up U.

    My problem is how to fit it all together. I don't understand what order I have to put the vectors in within U, or the values in within D, to prove that A = UDU^{-1}. I've tried to brute force it but I'm not getting anywhere and it's making me think I've done everything wrong.

    Can anyone explain my mistake, please?
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  2. #2
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    Define
     D = \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ <br />
0 & \lambda_2 & 0 & \cdots & \vdots \\ <br />
0 & 0 & \lambda_3 & \ddots & \vdots \\ <br />
\vdots & \cdots & \ddots & \ddots & 0 \\ <br />
0 & \cdots &\cdots &0& \lambda_n\end{bmatrix}
    and then
     U = \begin{bmatrix} v_1 & v_2 & v_3 & \cdots & v_n\end{bmatrix}
    where \lambda_i is the ith eigenvalue and v_i is the corresponding eigenvector. This is called the Jordan Decomposition of A.
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  3. #3
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    Yes, but I don't understand which eigenvalue is lambda1, which is lambda2 etc. I'm sure I don't just pick randomly?
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  4. #4
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    Well as long as you are consistent it it doesn't matter (number of eigenvalues is meaningless anyway). You just have to make sure that column 1 of matrix U corresponds to the eigenvector for the first eigenvalue in D, etc.
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  5. #5
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    So look at it this way
     AU = UD
    A \begin{bmatrix}v_1 & v_2 & v_3 & \cdots & v_n\end{bmatrix} =  \begin{bmatrix}v_1 & v_2 & v_3 & \cdots & v_n\end{bmatrix} \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ <br />
0 & \lambda_2 & 0 & \cdots & \vdots \\ <br />
0 & 0 & \lambda_3 & \ddots & \vdots \\ <br />
\vdots & \cdots & \ddots & \ddots & 0 \\ <br />
0 & \cdots &\cdots &0& \lambda_n\end{bmatrix}
    Which leads to
    \begin{bmatrix}\lambda_1 v_1 & \lambda_2 v_2 & \lambda_3 v_3 & \cdots & \lambda_n v_n\end{bmatrix} = \begin{bmatrix}\lambda_1 v_1 & \lambda_2 v_2 & \lambda_3 v_3 & \cdots & \lambda_n v_n\end{bmatrix}.
    So if you rearrange things you get the same result, but in a different order.
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  6. #6
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    Ok, thanks. It's sounding like I did everything right then got cold feet at the end!
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