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Math Help - Diagonalizable Matrix

  1. #1
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    Diagonalizable Matrix

     A= \left(\begin{array}{cccccc}1&2&2^2&...&...&2^{n-1}\\2&2^2&2^3&...&...&2^{n}\\2^2&2^3&2^4&...&...&2  ^{n+1}\\.&.&.&&&.\\.&.&.&&&.\\2^{n-1}&2^n&2^{n+1}&...&...&2^{2n-2}\end{array}\right)

    Question: Prove \lambda=0 is an eigenvalue of A and find a basis for it's space.

    Solution: (My friend gave a different answer so I'd like to know which one is correct)

    \lambda=0 if matrix A is singular. Matrix A is clearly singular...

    To find a basis, I do the elementary operations on matrix A and I'm left with

     A=  \left(\begin{array}{cccccc}1&2&2^2&...&...&2^{n-1}\\0&0&0&...&...&0\\0&0&0&...&...&0\\.&.&.&&&.\\.  &.&.&&&.\\0&0&0&...&...&0\end{array}\right) = 0

    Matrix A has a rank of 1 ---- dimp=n-p(A) ---- dimp = n-1

    x_1=-2t_1-2^2t_2-...-2^{n-1}t_{n-1}
    x_2=t_1
    x_3=t_2
    .
    .
    .
    x_n = t_{n-1}


    t_1 \left(\begin{array}{cccccc}-2\\1\\0\\.\\.\\0\end{array}\right)+ t_2 \left(\begin{array}{cccccc}-2^2\\0\\1\\.\\.\\0\end{array}\right) +...+ t_{n-1} \left(\begin{array}{cccccc}-2^{n-1}\\0\\0\\.\\.\\1\end{array}\right)

    and those eigenvectors form the basis for that space.

    My friend got:

    \left(\begin{array}{cccccc}2\\-1\\0\\.\\.\\0\end{array}\right), \left(\begin{array}{cccccc}0\\2\\-1\\.\\.\\0\end{array}\right),..., \left(\begin{array}{cccccc}0\\0\\.\\.\\2\\-1\end{array}\right)

    So which is the correct answer? (or is it possible they are both correct...or incorrect)
    Last edited by jayshizwiz; October 19th 2010 at 11:16 PM.
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  2. #2
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    ** Did I find the correct basis?
    ** Also, is there a way to change the Thread topic? I don't need to call it Diagonalizable Matrix...(that's the third part of the question, but it's not hard...)
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  3. #3
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    Both bases are correct.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Both bases are correct.
    If it's not too troublesome, would you mind showing me how the other one is a basis.
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