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 $$ex F(u,v)=sef{x,u}sef{y,v}+sef{z,u}sef{w,v} eex$$ is elementary.
Solution.
(1). If $w=ky$ for some $kinbC$, then $$eex ea F(u,v)&=sef{x,u}sef{y,v}+sef{z,u}sef{ky,v}\ &=sef{x+kz,u}sef{y,v}, eea eeex$$ 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 $scrH$: $$ex u_1,cdots,u_n eex$$ where $u_1=x,u_2=y,u_3=z$. And for $uin scrH$, we denote by $u_i$ the coordinate of $u$ with respect to this basis. Since $F$ is elementary, there exist $a,bin scrH$ such that $$ex F(u,v)=sef{x,u}sef{y,v}+sef{z,u}sef{w,v} =sef{a,u}sef{b,v}. eex$$ Taking $u=u_1$ or $u_3$, $v=u_j$ for arbitrary $j$, we obtain $$ex F(u_1,u_j)=y_j=a_1b_j,quad F(u_3,u_j)=w_j=a_3b_j. eex$$ Consequently, if $a_3=0$, then $w=0=0y$; if $a_3 eq 0$s, then $$ex w_j=a_3b_j=frac{a_3}{a_1}b_j a w=frac{a_3}{a_1}y. eex$$ Here $a_1 eq 0$ (otherwise $y=0$).