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
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 $$f= (x_1,x_2,x_3) \left(\begin{array}{ccc} b_1c_1 & b_1c_2 & b_1c_3\\ b_2c_1 & b_2c_2 & b_2c_3\\ b_3c_1 & b_3c_2 & b_3c_3 \end{array}\right) \left(\begin{array}{c} y_1\\ y_2\\ y_3 \end{array}\right) .$$