# Complex Matrix

• Nov 22nd 2009, 06:02 PM
redsoxfan325
Complex Matrix

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!
• Nov 22nd 2009, 06:57 PM
tonio
Quote:

Originally Posted by redsoxfan325

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
• Nov 22nd 2009, 07:15 PM
redsoxfan325
Quote:

Originally Posted by tonio
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.
• Nov 22nd 2009, 07:37 PM
Shanks
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 $=Bx$ 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.