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 $BX=Y$.

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. 
