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:


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?



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  • $\begingroup$ an expression like $b_1x_1+b_2x_2+b_3x_3$ is a linear form instead of bilinear $\endgroup$ – 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 $$f= (x_1,x_2,x_3) \left(\begin{array}{ccc} b_1c_1 & b_1c_2 & b_1c_3\\ b_2c_1 & b_2c_2 & b_2c_3\\ b_3c_1 & b_3c_2 & b_3c_3 \end{array}\right) \left(\begin{array}{c} y_1\\ y_2\\ y_3 \end{array}\right) .$$

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    $\begingroup$ 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 $\endgroup$ – janmarqz Oct 31 '15 at 15:42
  • $\begingroup$ You didn't answer the question at all. $\endgroup$ – Arnaud D. Jun 3 '17 at 16:20

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*}

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