1
$\begingroup$

I have been trying assignment questions of linear algebra and I am unable to solve this particular question

Let $x=\left(x_{1}, x_{2}, x_{3}\right), y=\left(y_{1}, y_{2}, y_{3}\right) \in \mathbb{R}^{3}$ be linearly independent. Let $\delta_{1}=x_{2} y_{3}-y_{2} x_{3}, \delta_{2}=x_{1} y_{3}-y_{1} x_{3}$ $\delta_{3}=x_{1} y_{2}-y_{1} x_{2} .$ If $V$ is the span of $x, y$ then

  1. $V=\left\{(u, v, w): \delta_{1} u-\delta_{2} v+\delta_{3} w=0\right\}$

  2. $V=\left\{(u, v, w):-\delta_{1} u+\delta_{2} v+\delta_{3} w=0\right\}$

  3. $V=\left\{(u, v, w): \delta_{1} u+\delta_{2} v-\delta_{3} w=0\right\}$

  4. $V=\left\{(u, v, w): \delta_{1} u+\delta_{2} v+\delta_{3} w=0\right\}$

I know the definitions of Linearly Independent and Span of vectors but I don't know how to solve this problem due to $\delta_{1}$, $\delta_{2}$,$\delta_{3}$ as I am unable to write V in terms of $\delta_{i}$'s and $(u, v, w)$.

Any help will be really appreciated .

$\endgroup$
  • 1
    $\begingroup$ -@User, you can use Mathpix Snip to convert mathematical terms in images to $\LaTeX$ code. $\endgroup$ – SarGe Aug 1 at 13:22
1
$\begingroup$

Given, $x=(x_1,x_2,x_3), y=(y_1,y_2,y_3) \in \mathbb{R}^3 $ are linearly independent.

Also, given, $V$ is the span of that two linearly independent vectors $x,y\in \mathbb{R}^3 $

It means clearly that each $(u,v,w)\in V $ can be uniquely spanned by the linearly independent vectors $x,y\in \mathbb{R}^3 $

That means for each $(u,v,w)\in V $, $$ \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ u & v & w \\ \end{vmatrix} =0 $$ $\implies u(x_2y_3-y_2x_3)-v(x_1y_3-y_1x_3)+w(x_1y_2-x_2y_1) = 0 \implies \delta_{1} u-\delta_{2} v+\delta_{3} w=0 $

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

Hint: Note that the cross-product of $x$ and $y$ is given by $$ x \times y = (\delta_1,-\delta_2,\delta_3). $$

| 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.