Bilinear functional to be elementary Let $V$ be a $n$ dimensional vector space over $\Bbb C$. We say a functional $f:V\times V\to \Bbb C$ is bilinear if $f$ is linear in each variable if the other variable is fixed. 
And $$f$$ is called elementary if there exists some $x,y\in V$ such that
$$f(u,v)=<x,u>\cdot <y,v>.$$
Here $<x,u>=x^*u$, $x^*$ is the conjugate transpose of $x$.
Now my problem is as follows. Let $x,y,z$ be three linearly independent vectors in $V$. What conditions on $w$ should satisfy to ensure the following functional
$$f(u,v)=<x,u>\cdot <y,v>+<z,u>\cdot <w,v>$$
to be elementary.
My first idea is as follows. If the above $f$ is elementary, then there exists some $a,b\in V$ such that 
$$f(u,v)=<a,u>\cdot <b,v>.$$
And thus
$$<x,u>\cdot <y,v>+<z,u>\cdot <w,v>=f(u,v)=<a,u>\cdot <b,v>$$
Since $f$ is bilinear, we need only to find conditions on $w$ such that there exists some $a,b\in\Bbb C^n$ such that 
$$\bar x_iy_j+\bar z_iw_j=\bar a_ib_j,\forall\ 1\leq i,j\leq n.$$
These are $n^2$ equations, but only with $2n$ variables, but I could not derive further..
I suspect that $w$ is a multiplier of $y$...
This is exercise I.4.1 of Bhatia: Matrix Analysis.
 A: A necessary and sufficient condition is $w=\lambda y$.
If your functional is identically zero then it is elementary, thus let us assume that $f(u,v)$ is not identically zero.
Suppose $\langle a,u\rangle\langle b,v\rangle=\langle x,u\rangle\langle y,v\rangle+\langle z,u\rangle\langle w,v\rangle$.
Thus, $\langle a,u\rangle\langle b,v\rangle=\langle \langle v,y\rangle x,u\rangle+\langle \langle v,w\rangle z,u\rangle=\langle \langle v,y\rangle x+\langle v,w\rangle z,u\rangle$. 
If $v\in\text{span}\{y,w\}^{\perp}$ then $\langle a,a\rangle\langle b,v\rangle=0$. But $\langle a,a\rangle\neq 0$ (otherwise $a=0$ and $f\equiv 0$) then $\langle b,v\rangle=0$ for every $v\in\text{span}\{y,w\}^{\perp}$. Thus, $b\in (\text{span}\{y,w\}^{\perp})^{\perp}=\text{span}\{y,w\}$.
If $y$ and $w$ are linear independent then exists 
$0 \neq v'\in\text{span}\{y,w\}$ such that $b\perp v'$.
Now let $u=\langle v',y\rangle x+\langle v',w\rangle z$ and $v=v'$. Thus, $0=\langle a,u\rangle\langle b,v\rangle=\langle \langle v',y\rangle x+\langle v',w\rangle z,\langle v',y\rangle x+\langle v',w\rangle z\rangle$.
Therefore, $\langle v',y\rangle x+\langle v',w\rangle z=0$. Since $x,z$ are l.i. then $\langle v',y\rangle=\langle v',w\rangle=0$. This is a contradiction with $0 \neq v'\in\text{span}\{y,w\}$. Thus, $y,w$ are linear dependent.
Of course if $w=\lambda y$ then $\langle x,u\rangle\langle y,v\rangle+\langle z,u\rangle\langle w,v\rangle=\langle x+\lambda z,u\rangle\langle y,v\rangle$.
