1
$\begingroup$

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

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ 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
-1
$\begingroup$

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

| cite | improve this answer | | | | |
$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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*}

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.