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 $BX=Y$.

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.
