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Math Help - Diagonal Matrix

  1. #1
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    Diagonal Matrix

    Show that the matrix is diagonalizable and find a diagonal matrix similar to the given matrix.
     <br />
\begin{pmatrix}0 & -2 & 1\\1 & 3 & -1\\0 & 0 & 1\end{pmatrix}<br />
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  2. #2
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    Quote Originally Posted by jennifer1004 View Post
    Show that the matrix is diagonalizable and find a diagonal matrix similar to the given matrix.
     <br />
\begin{pmatrix}0 & -2 & 1\\1 & 3 & -1\\0 & 0 & 1\end{pmatrix}<br />
    Use this same proceedure to find the matrix P

    http://www.mathhelpforum.com/math-he...ar-matrix.html

    actually you don't need the matrix P. You will find that the char poly has a repeated eigenvalue of \lambda =1

    You will need to show that \lambda =1 gives two eigen vectors.

    After you have shown this the similar matrix will be the matrix with the eigenvalues of the char. poly down the diagonal.
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    I got 2, 1, 1 as the eigenvalues from the characteristic polynomial x^{3}-4x^{2}+5x-2. Do the numbers of the diagonal follow any order? Would it be \begin{pmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 1\end{pmatrix} or \begin{pmatrix}2 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix} Thank you!
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  4. #4
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    The order would depend on how you make your matrix P. The eigenvalues will be in the same column as the corrisponding eigenvector. both are similar to the original.
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  5. #5
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    My eigenvector for 2 is <-1, 1, 0> if I'm right and the two eigenvectors for 1 are <-2, 1, 0> and <1, 0, 1>. So I think my diagonal matrices may be wrong even though I'm leaning towards the first one with the 2 in the center.
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  6. #6
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    Quote Originally Posted by jennifer1004 View Post
    My eigenvector for 2 is <-1, 1, 0> if I'm right and the two eigenvectors for 1 are <-2, 1, 0> and <1, 0, 1>. So I think my diagonal matrices may be wrong even though I'm leaning towards the first one with the 2 in the center.
    Your eigenvalues and vectors are correct .

    What I am saying is this

    If you make your matrix P such that PAP^{-1}=D

    like this this

    \begin{bmatrix}<br />
-1 & 1 & -2 \\<br />
1 & 0 & 1 \\<br />
0 & 1 & 0 \\<br />
\end{bmatrix} <br />

    Where the first column is the vector corrisponding to the eigenvalue 2 then the diagonal matrix will look like this


    \begin{bmatrix}<br />
2 & 0 & 0 \\<br />
0 & 1 & 0 \\<br />
0 & 0 & 1 \\<br />
\end{bmatrix} <br />

    If you move exchange the first two columns P would look like this


    \begin{bmatrix}<br />
1 & -1 & -2 \\<br />
0 & 1 & 1 \\<br />
1 & 0 & 0 \\<br />
\end{bmatrix} <br />

    Then D would look like this

    \begin{bmatrix}<br />
1 & 0 & 0 \\<br />
0 & 2 & 0 \\<br />
0 & 0 & 1 \\<br />
\end{bmatrix} <br />

    Both of the above are similar to the original matrix.

    I hope this clears it up.
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