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Math Help - Similar Matrices and Diagonalization

  1. #1
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    Similar Matrices and Diagonalization

    I have a question regarding two matrices. Are the following matrices similar and if so find a third matrix P such that:
    B = P^{-1}AP

    A = \left(\begin{array}{ccc}1&0&0\\0&2&0\\0&0&3\end{ar  ray}\right)

    B = \left(\begin{array}{ccc}3&0&0\\0&2&0\\0&0&1\end{ar  ray}\right)

    As far as I can tell they are similar because they have the same eigenvalues: 1, 2, 3. If I plug in 1 for
    \lambda I - A

    I get:

    \left(\begin{array}{ccc}0&0&0\\0&-1&0\\0&0&-2\end{array}\right)
    This row reduces to:
    \left(\begin{array}{ccc}0&1&0\\0&0&1\\0&0&0\end{ar  ray}\right)
    Doesn't that have the zero solution? My understanding of this topic regarding diagonalization isn't the best. Thanks for any help...I realize I have two more eigenvalues to find eigenvectors for.
    Last edited by Alterah; December 4th 2009 at 10:04 PM.
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  2. #2
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    How, exactly, did you row reduce it to that form? That is incorrect. The matrix will row-reduce to \left(\begin{array}{ccc}0&1&0\\0&0&1\\0&0&0\end{ar  ray}\right) meaning eigenvectors of eigenvalue 1 are of the form (x ~ 0 ~ 0)^T.
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  3. #3
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    Quote Originally Posted by Alterah View Post
    I have a question regarding two matrices. Are the following matrices similar and if so find a third matrix P such that:
    B = P^{-1}AP

    A = \left(\begin{array}{ccc}1&0&0\\0&2&0\\0&0&3\end{ar  ray}\right)

    B = \left(\begin{array}{ccc}3&0&0\\0&2&0\\0&0&1\end{ar  ray}\right)

    As far as I can tell they are similar because they have the same eigenvalues: 1, 2, 3. If I plug in 1 for
    \lambda I - A

    I get:

    \left(\begin{array}{ccc}0&0&0\\0&-1&0\\0&0&-2\end{array}\right)
    This row reduces to:
    \left(\begin{array}{ccc}0&1&0\\0&1&0\\0&0&0\end{ar  ray}\right)
    Doesn't that have the zero solution? My understanding of this topic regarding diagonalization isn't the best. Thanks for any help...I realize I have two more eigenvalues to find eigenvectors for.

    There are similar not because they have the same eignevalues (this would be, in general, false), but because they have the same eigenvalues with the same algebraic and geomtric multiplicity. In this particular case, we can say they're similar because they both are diagonal and have the same elements on the main diagonal.

    Now, to find P you'll do the following: find eigenvectors of the first matrix ; since the matrix is diagonal these eigenvectors are a basis of \mathbb{R}^3 (if we really are on the real space vector space). Now, form the matrix P by taking as its first column the eigenvector of 3, the 2nd. column the eigenvector of 2 and the third column the eigenvector of 1...and that's all!
    BTW, choosing wisely you get a very standard and easy basis of eigenvectors of A...

    Tonio
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  4. #4
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    I messed up in the original post when I posted the row reduced form. And...wow, you're basically saying that:

    P = \left(\begin{array}{ccc}3&0&0\\0&2&0\\0&0&1\end{ar  ray}\right)

    I guess I was making it out to be harder than it needed to be.
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