Prove that the bilinear form can be presented as a product of two linear forms 
Let $ f:\Bbb R^3 \times\Bbb R^3 \to \Bbb R $ be a bilinear form such the the rank of $f$ is 1.
Prove that $f$ can be presented as a product of the linear forms, such that:

$$f(x,y)=(b_{1}x_1+b_2x_2+b_3x_3)(c_{1}y_1+c_{2}y_2+c_3y_3)$$

For every $$x=(x_1,x_2,x_3), y=(y_1,y_2,y_3)$$

Actually I don't know how to approach this question. I have to use the fact that the rank=1. But how?
Thanks,
Alan
 A: The general shape of a bilinear form is
$$(v,w)\longmapsto v^{\top}Qw,$$
where $v$ and $w$ are vectors and $Q$ is a square matrix.
In the other hand for two linear forms (on $\Bbb R^3$) we have
$$\alpha(x_1,x_2,x_3)=[b_1,b_2,b_3]
\left(\begin{array}{c}
x_1\\
x_2\\
x_3
\end{array}\right)=b_1x_1+b_2x_2+b_3x_3,$$ 
and
$$\beta(y_1,y_2,y_3)=[c_1,c_2,c_3]
\left(\begin{array}{c}
y_1\\
y_2\\
y_3
\end{array}\right)=c_1y_1+c_2y_2+c_3y_3,$$
The two matrices for $\alpha$ and $\beta$ respectively are used to
construct their  tensor product  (or Kronecker product) which is
$$\alpha\otimes \beta=\left(\begin{array}{c}
b_1\\
b_2\\
b_3
\end{array}\right)(c_1,c_2,c_3)
=
\left(\begin{array}{ccc}
b_1c_1 & b_1c_2 & b_1c_3\\
b_2c_1 & b_2c_2 & b_2c_3\\
b_3c_1 & b_3c_2 & b_3c_3
\end{array}\right).
$$
Then your bilinear form would be
$$f=
(x_1,x_2,x_3)
\left(\begin{array}{ccc}
b_1c_1 & b_1c_2 & b_1c_3\\
b_2c_1 & b_2c_2 & b_2c_3\\
b_3c_1 & b_3c_2 & b_3c_3
\end{array}\right)
\left(\begin{array}{c}
y_1\\
y_2\\
y_3
\end{array}\right)
.$$
