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Thread: Complex Matrix

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

    The question reads:

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

    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 $\displaystyle z$ such that $\displaystyle z^{\top}y=0$, so that we have to solve $\displaystyle z^{\top}Bx=0$, in other words, show that there exists a vector $\displaystyle z$ that is orthogonal to $\displaystyle x$ with respect to the bilinear form $\displaystyle B$ (and then maybe use the fact that the eigenvectors of $\displaystyle B$ 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 $\displaystyle x,y$ be vectors in $\displaystyle \mathbb{C}^n$, and assume that $\displaystyle x\neq0$. Prove that there is a symmetric matrix $\displaystyle B$ such that $\displaystyle Bx=y$.

    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 $\displaystyle z$ such that $\displaystyle z^{\top}y=0$, so that we have to solve $\displaystyle z^{\top}Bx=0$, in other words, show that there exists a vector $\displaystyle z$ that is orthogonal to $\displaystyle x$ with respect to the bilinear form $\displaystyle B$ (and then maybe use the fact that the eigenvectors of $\displaystyle B$ 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 $\displaystyle y=0$ then take $\displaystyle B=0$ ;

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

    3) In general, $\displaystyle \left(\begin{array}{rrcr}a_{11}&a_{12}&...&a_{1n}\ \a_{12}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{1n} &a_{2n}&...&a_{nn}\end{array}\right)\left(\begin{a rray}{c}x_1\\x_2\\...\\x_n\end{array}\right)$ $\displaystyle =\left(\begin{array}{c}y_1\\y_2\\...\\y_n\end{arra y}\right)$ . You have to solve here $\displaystyle n$ equations in $\displaystyle \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 $\displaystyle y=0$ then take $\displaystyle B=0$ ;

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

    3) In general, $\displaystyle \left(\begin{array}{rrcr}a_{11}&a_{12}&...&a_{1n}\ \a_{12}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{1n} &a_{2n}&...&a_{nn}\end{array}\right)\left(\begin{a rray}{c}x_1\\x_2\\...\\x_n\end{array}\right)$ $\displaystyle =\left(\begin{array}{c}y_1\\y_2\\...\\y_n\end{arra y}\right)$ . You have to solve here $\displaystyle n$ equations in $\displaystyle \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 $\displaystyle \mathbb{C}^{n\times n}$ be the $\displaystyle n\times n$ complex matrix space, $\displaystyle \mathbb{C}^n$ be the complex vector space.
    Define $\displaystyle <B,x>=Bx$ 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; Nov 22nd 2009 at 07:40 PM. Reason: Latex error
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