# 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

• an expression like $b_1x_1+b_2x_2+b_3x_3$ is a linear form instead of bilinear – janmarqz Oct 31 '15 at 14:44

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 $$f= (x_1,x_2,x_3) \left(\begin{array}{ccc} b_1c_1 & b_1c_2 & b_1c_3\\ b_2c_1 & b_2c_2 & b_2c_3\\ b_3c_1 & b_3c_2 & b_3c_3 \end{array}\right) \left(\begin{array}{c} y_1\\ y_2\\ y_3 \end{array}\right) .$$
• About the rank that you are asking, probably rank one means that the two vectors that represent the covectors $\alpha$ and $\beta$ are linearly dependent, and then the matrix for dyadic $\alpha\otimes\beta$ will have rank one – janmarqz Oct 31 '15 at 15:42
Your bilinear form can be expressed as $$f(x,y) = x^t A y$$ where $$x^t$$ is a $$1 \times 3$$ matrix, $$y$$ is is a $$3 \times 1$$ matrix, and $$A$$ is the $$3 \times 3$$ matrix of $$f$$ in the standard ordered basis. Because the rank of $$f$$ is $$1$$, the rank $$A$$ is $$1$$. Then the row space of $$A$$ has dimension $$1$$, i.e., the rows of $$A$$ are multiples of some non-zero row. Such a matrix can be expressed as a product of a column matrix, say $$b = [b_1,b_2,b_3]^t$$, and a row matrix, say $$c = [c_1,c_2,c_3]$$. Then\begin{align*} f(x,y) & = x^t b c y\\ & = \begin{bmatrix} x_1 & x_2 & x_3\end{bmatrix} \begin{bmatrix} b_1\\ b_2\\ b_3\end{bmatrix} \begin{bmatrix} c_1 & c_2 & c_3\end{bmatrix} \begin{bmatrix} y_1\\ y_2\\ y_3\end{bmatrix}\\ & = (b_1 x_1 + b_2 x_2 + b_3 x_3)(c_1 y_1 + c_2 y_2 + c_3 y_3).\end{align*}