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Let $X,Y$ be vectors in $\mathbb{C}^n$, and assume that $X\ne0$. Prove that there is a symmetric matrix $B$ such that $BX=Y$.

This is an exercise from a chapter about bilinear forms. So the intended solution should be somehow related to it.

Pre-multiplying both sides by $Y^t$, we get $Y^tBX=Y^tY$. The left hand side is a bilinear form $\langle Y,X\rangle $ with $B$ as the matrix of the form with respect to the standard basis. Am I correct here?

If so, then it suffices to find a bilinear form $\langle\cdot,\cdot\rangle\colon\mathbb{C}^n\times\mathbb{C}^n\rightarrow\mathbb{C}$ such that $\langle Y,X\rangle=Y^tY$. If $Y=0$, any bilinear form will do, because $\langle0,X\rangle=0\langle 0,X\rangle =0$ by linearity in the first variable. If $Y\ne0$, it suffices to find a bilinear form such that $\langle Y,X\rangle$ is nonzero, then we can multiply by the appropriate factor. This should be very near to a complete solution, but I can't figure out the rest.

Edit: Okay, my approach seems to be completely wrong. Using Phira's hint, I think I managed to make a complete proof.

Choose an orthonormal basis $(v_1,\ldots,v_n)$ such that $v_1=\frac{X}{\|X\|}$, which can be done by Gram-Schmidt process. Let $P$ be the $n\times n$ matrix whose $i$-th column is the vector $v_i$. Then $P$ is orthogonal. Let $P^{-1}Y=(a_1,\ldots,a_n)^t$. Choose such that the first column and the first row is the vector $\frac1{\|X\|}(a_1,\ldots,a_n)$, and 0 everywhere else. Clearly $M$ is symmetric and it's easy to check that $(PMP^{-1})X=Y$. So the desired matrix is $B=PMP^{-1}$, which is symmetric because $P$ is orthogonal. $\Box$

However, this solution does not make use of bilinear forms. So there might be a simpler way.

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    One problem with your approach is that $Y^tBX=Y^tY$ (a scalar equation) does not guarantee that $BX=Y$ (a vector equation).2012-03-22
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    I overlooked that.2012-03-22
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    I'd think there'd be a way to build $B$ (or $A$, as in the title) one row/column at a time. That is, you can make the first row of $B$ so its product with $X$ gives you the first entry of $Y$. That determines the first column of $B$, but can still fill out the rest of the second row of $B$ to get the second entry in $Y$ correct, etc., etc. Well, some care in the order in which you fill in the rows/columns may be needed.2012-03-23
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    That is certainly feasible, but I believe there should be a smart way to use bilinear forms.2012-03-23
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    I like the idea you had in your second paragraph. What if you multiple by $X^t$ instead of $Y^t$?2012-03-23
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    Alternatively, bilinear forms are specified by their action on a basis. So pick a basis containing $X$ (since it's nonzero), and stipulate that $B(U,X)=U^tY$ for all $U$.2012-03-23

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