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In my lecture notes we have the following:

$K$ field

Extension of the affine space.

Relation between points and lines:

Two discrete points define an unique line and two discrete lines always intersect at one point.

Why does the part "two discrete lines always intersect at one point" stand? Can the lines not be parallel?

I found the proof, which is the following:

Let $$E_1 | a_1 x+b_1y+c_1 z=0, (a_1, b_1, c_1) \neq (0, 0, 0) \\ E_2 | a_2 x+b_2 y+c_2 z=0$$

$E_1,E_2$ are different.

That means that $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ don't belong at the same equivalence class ($ [a_1, b_1, c_1] \neq [a_2, b_2, c_2]$).

So $\{a_1, b_1, c_1\}$ and $\{a_2, b_2, c_2\}$ are $K-$ linear independent.

So the order of the array $$A= \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{bmatrix}$$ is $2$. The variables are $3$.

$\Rightarrow $ The system has an infinity of monoparametric solutions, that means taht the set of the solutions is $$\{\lambda (x_0, y_0, z_0) | \lambda \in K\setminus \{0\} , (x_0, y_0, z_0) \in K^3 \setminus \{(0, 0, 0)\}=\left [x_0, y_0,z_0\right ]$$

$$$$

Can you explain the part:

$\Rightarrow $ The system has an infinity of monoparametric solutions, that means taht the set of the solutions is $$\{\lambda (x_0, y_0, z_0) | \lambda \in K\setminus \{0\} , (x_0, y_0, z_0) \in K^3 \setminus \{(0, 0, 0)\}=\left [x_0, y_0,z_0\right ]$$

?

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    $\begingroup$ What does "extension of the affine space" mean? Projective space? What does the adjective "discrete" mean? Different? $\endgroup$ – Hagen Knaf Jan 22 '15 at 11:04
  • $\begingroup$ Yes, "extension of the affine space" means projective space and "discrete" means different @Hagen . $\endgroup$ – user175343 Jan 22 '15 at 18:08
  • $\begingroup$ @Hagen I found the proof and I added it above but I need some explanations. $\endgroup$ – user175343 Jan 22 '15 at 21:13
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More directly you want to solve the linear system

$a_1x+b_1y+c_1z=0$

$a_2x+b_2y+c_2z=0$.

You already know that the rank of the matrix formed by the coefficients of the system is $2$, since the row vectors of this matrix are linearly independent. The dimension formula of linear algebra states that the rank of the matrix (here $2$) plus the dimension of the space of solutions of the linear system (here unknown) must be equal to the number of colums (here $3$). Hence the space of solutions is $1$-dimensional, that is it is a line passing through the origin. Such a line corresponds to a point in projective space.

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  • $\begingroup$ What is the dimension of the space of solutions ??? Also how do we conclude that "it is a line passing through the origin. Such a line corresponds to a point in projective space" ??? $\endgroup$ – user175343 Feb 1 '15 at 17:07

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