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Math Help - complete matrices question

  1. #1
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    complete matrices question

    I need to prove that if A is complete then S-1AS is complete where S is a nonsingular matrix. A complete matrix is one whose eigenvectors span Rn. Can anyone look over what i've done below. i believe it's correct but i have no confidence. i need a little reassurance. thanks!


    S-1AS = B. A is complete. Every eigenvalue of A is also an eigenvalue of B. To show this we suppose Av = λv. Then (S-1A)v = λ(S-1)v. This is equivalent to (S-1ASS-1)v = (S-1AS)( S-1)v = λ(S-1)v. This shows that λ is also an eigenvalue of S-1AS and the eigenvector is S-1v. And so if v is an eigenvector of A the S-1v is an eigenvector of S-1AS. If we consider S-1V where V is the matrix whose columns are eigenvectors of A. Because by assumption A is complete V is invertible (since it's columns are linearly independent). But S-1 is invertible and so therefore is S-1V. The columns of S-1V are S-1vi for each eigenvector of A. But the S-1vi are the eigenvectors of S-1AS. Since S-1V is nonsingular it's columns are linearly independent and therefore span Rn. This is what it means for S-1AS to be complete and we are done.
    Last edited by kkoutsothodoros; July 4th 2007 at 07:45 AM.
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  2. #2
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    S-1 is the inverse of S. thus we are asked to show that any matrix similar to A is complete.
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  3. #3
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    Quote Originally Posted by kkoutsothodoros View Post
    I need to prove that if A is complete then S-1AS is complete where S is a nonsingular matrix. A complete matrix is one whose eigenvectors span Rn. Can anyone look over what i've done below. i believe it's correct but i have no confidence. i need a little reassurance. thanks!


    S-1AS = B. A is complete. Every eigenvalue of A is also an eigenvalue of B. To show this we suppose Av = λv. Then (S-1A)v = λ(S-1)v. This is equivalent to (S-1ASS-1)v = (S-1AS)( S-1)v = λ(S-1)v. This shows that λ is also an eigenvalue of S-1AS and the eigenvector is S-1v. And so if v is an eigenvector of A the S-1v is an eigenvector of S-1AS. If we consider S-1V where V is the matrix whose columns are eigenvectors of A. Because by assumption A is complete V is invertible (since it's columns are linearly independent). But S-1 is invertible and so therefore is S-1V. The columns of S-1V are S-1vi for each eigenvector of A. But the S-1vi are the eigenvectors of S-1AS. Since S-1V is nonsingular it's columns are linearly independent and therefore span Rn. This is what it means for S-1AS to be complete and we are done.
    I think your argument is correct. Good job! I tried summarizing the argument, but this is probably too short as a proof.

    Every eigenvalue \lambda of A with eigenvector v is also an eigenvalue of S^{-1}AS with eigenvector S^{-1}v. This follows from

    \begin{aligned}<br />
\lambda (S^{-1} v) &= S^{-1}(\lambda v) \\<br />
&= S^{-1}(Av) \\<br />
&= S^{-1}A(SS^{-1})v \\ <br />
&= (S^{-1}AS)( S^{-1}v). \end{aligned}

    Let V be the matrix whose columns are the eigenvectors of A. Then the above shows that W = S^{-1}V has columns that are the eigenvectors of S^{-1}AS.

    A matrix is complete if and only if its matrix of eigenvectors is invertible. So V is invertible and since S^{-1} is invertible, so is W= S^{-1}V. Hence S^{-1}AS is complete.
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  4. #4
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    thanks!

    how do you get the nice math editing?
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  5. #5
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    Quote Originally Posted by kkoutsothodoros View Post
    thanks!

    how do you get the nice math editing?
    You're welcome.

    See the LaTeX tutorial.
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