Let $x,y,z$ be linearly independent vectors in $\scrH$. Find a necessary and sufficient condition that a vector $w$ mush satisfy in order that the bilinear functional $$\bex F(u,v)=\sef{x,u}\sef{y,v}+\sef{z,u}\sef{w,v} \eex$$ is elementary.
Solution.
(1). If $w=ky$ for some $k\in\bbC$, then $$\beex \bea F(u,v)&=\sef{x,u}\sef{y,v}+\sef{z,u}\sef{ky,v}\\ &=\sef{x+kz,u}\sef{y,v}, \eea \eeex$$ and thus $F$ is elementary.
(2). We now show that the condition that $w$ is a multiplier of $y$ is necessary to ensure that $F$ is elementary. It can be proved as follows easily; however, when I have not got it, it really hindered me to go forward this fun journey of the matrix analysis. We choose a basis of $\scrH$: $$\bex u_1,\cdots,u_n \eex$$ where $u_1=x,u_2=y,u_3=z$. And for $u\in \scrH$, we denote by $u_i$ the coordinate of $u$ with respect to this basis. Since $F$ is elementary, there exist $a,b\in \scrH$ such that $$\bex F(u,v)=\sef{x,u}\sef{y,v}+\sef{z,u}\sef{w,v} =\sef{a,u}\sef{b,v}. \eex$$ Taking $u=u_1$ or $u_3$, $v=u_j$ for arbitrary $j$, we obtain $$\bex F(u_1,u_j)=y_j=a_1b_j,\quad F(u_3,u_j)=w_j=a_3b_j. \eex$$ Consequently, if $a_3=0$, then $w=0=0y$; if $a_3\neq 0$s, then $$\bex w_j=a_3b_j=\frac{a_3}{a_1}b_j\ra w=\frac{a_3}{a_1}y. \eex$$ Here $a_1\neq 0$ (otherwise $y=0$).