Prove that there is a symmetric matrix B, such that BX=Y Let $X,Y$ be two vectors in ${\mathbb C}^n$ and assume that $X≠0$. Prove that there is a symmetric matrix $B$ such that $BX=Y$.
 A: Ok, the last answer was not detailed enough, so here is another approach, which is constructive.
Pick some orthogonal matrix $Q$, hence $QQ^T = I$ and build $B$ as $B = Q\Lambda Q^T$, where $\Lambda$ is diagonal with it's entries $\Lambda_{jj}$ still left to determine. Then, 
$BX =Y \Leftrightarrow Q\Lambda Q^TX =Y \Leftrightarrow \Lambda Q^TX = Q^TY  $, which component-wise gives you the solution for $\Lambda_{jj}$ as $\Lambda_{jj} = \frac{q^T_jY}{q^T_jX}$. ($q_j$ is the jth column of $Q$). 
This is a solution in case you do not mind, that $B\in \mathbb{C}^{n \times n}$. If you want $B\in \mathbb{R}^{n \times n}$, then even if $B$ is not symmetric, it might be impossible. Notice, that the case where $X = \bar{Y}$ provides a counter-example for that.
A: here is one way solve this. if $y = 0,$ then the zero matrix for $B$ would do. so we can assume that $y \neq 0$ let $$z ={|x| \over |y|} y, \text{ so that } |z| = |x|.$$
the orthogonal projection matrix $$H = I - 2a{\bar a}^T/{\bar a}^Ta \text{ where } a = x - z \text{ and a bar means complex conjugate}$$   you can verify that $H$ is unitary and $Hx = z$  now take a scalar matrix  $D = {|y| \over |x|}.$ 
the matrix $DH x = y$ and $DH$ is symmetric as you wanted. 
A: Hint 1: If $X$ has no zero entry, prove that you can find a diagonal $B$ so that $BX=Y$.
Hint 2: If $X$ has some zero entries, since $X \neq 0$ prove that you can find some orthogonal matrix $P$ so that $PX$ has no zero entry.
Define $X'=PX$ and $Y'=PY$. Find a diagonal matrix $B'$ so that $B'X'=Y'$ and use 
$$
B=P^{-1}B'P 
$$
