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Math Help - Complex Matrix

  1. #1
    Super Member redsoxfan325's Avatar
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    Complex Matrix

    The question reads:

    Let x,y be vectors in \mathbb{C}^n, and assume that x\neq0. Prove that there is a symmetric matrix B such that Bx=y.

    I don't really know where to begin on this. It's in the chapter on bilinear forms, but I don't see what this has to do with that. Maybe find some vector z such that z^{\top}y=0, so that we have to solve z^{\top}Bx=0, in other words, show that there exists a vector z that is orthogonal to x with respect to the bilinear form B (and then maybe use the fact that the eigenvectors of B form an orthogonal basis).

    But other than that speculation, I got nothing.

    Help is appreciated!
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  2. #2
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    Quote Originally Posted by redsoxfan325 View Post
    The question reads:

    Let x,y be vectors in \mathbb{C}^n, and assume that x\neq0. Prove that there is a symmetric matrix B such that Bx=y.

    I don't really know where to begin on this. It's in the chapter on bilinear forms, but I don't see what this has to do with that. Maybe find some vector z such that z^{\top}y=0, so that we have to solve z^{\top}Bx=0, in other words, show that there exists a vector z that is orthogonal to x with respect to the bilinear form B (and then maybe use the fact that the eigenvectors of B form an orthogonal basis).

    But other than that speculation, I got nothing.

    Help is appreciated!

    I'm almost sure there must be a shorter way to do this, but the following seems pretty simple:

    1) if y=0 then take B=0 ;

    2) if y=\lambda x , take B=\lambda I_n ;

    3) In general, \left(\begin{array}{rrcr}a_{11}&a_{12}&...&a_{1n}\  \a_{12}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{1n}  &a_{2n}&...&a_{nn}\end{array}\right)\left(\begin{a  rray}{c}x_1\\x_2\\...\\x_n\end{array}\right) =\left(\begin{array}{c}y_1\\y_2\\...\\y_n\end{arra  y}\right) . You have to solve here n equations in \frac{n(n+1)}{2}>n\,\mbox{ (for } n> 1) unknowns, and such a system has always solution.

    Tonio
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by tonio View Post
    I'm almost sure there must be a shorter way to do this, but the following seems pretty simple:

    1) if y=0 then take B=0 ;

    2) if y=\lambda x , take B=\lambda I_n ;

    3) In general, \left(\begin{array}{rrcr}a_{11}&a_{12}&...&a_{1n}\  \a_{12}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{1n}  &a_{2n}&...&a_{nn}\end{array}\right)\left(\begin{a  rray}{c}x_1\\x_2\\...\\x_n\end{array}\right) =\left(\begin{array}{c}y_1\\y_2\\...\\y_n\end{arra  y}\right) . You have to solve here n equations in \frac{n(n+1)}{2}>n\,\mbox{ (for } n> 1) unknowns, and such a system has always solution.

    Tonio
    I agree. That is simple. It's strange that they put this in a section on bilinear forms and marked it as a challenge problem.

    Nonetheless, I'm more than happy to see that this has a simple solution.

    Thanks for the help!

    If anyone knows how to do this using bilinear forms, I'd love to know.
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  4. #4
    Senior Member Shanks's Avatar
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    Bilinear form

    Let \mathbb{C}^{n\times n} be the n\times n complex matrix space, \mathbb{C}^n be the complex vector space.
    Define <B,x>=Bx for any B in the matrix space, x in the vector space. It is obviously a biliear function.
    For any fixed B, it defines a liear transform in the complex vector space;
    For any fixed x, it defines a action (linear transform) on the complex matrix space, Since x is not 0, there is a base which contains x, Obviously there is a symmetric matrix B such that Bx=y, if we consider B as the matrix of a symmetric linear transform with respect to the given base.
    Last edited by Shanks; November 22nd 2009 at 07:40 PM. Reason: Latex error
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