Question

Help me with this, please.

Show that in R^n, the lines represented by su1 + u0 and tv1 + v0 are the same if and only if both v1 and v0 - u0 are numberical multiples of u1.

(u1, u0, v0, v1 are all vectors, btw).

Answer 1

Let
\[ \large L_1=\{s\cdot u_1+v_0:s\in\mathbb{R}\}\quad\text{and}\quad
L_2=\{t\cdot v_1+v_0:t\in\mathbb{R}\}. \]
First, let's suppose that there exist real numbers \(\alpha,\beta\) such that
\[ \large v_1=\alpha u_1\quad\text{and}\quad v_0-u_0=\beta u_1. \]

Answer 2

then, if \(v\in L_2\) we have that for some real \(t\)
\[ \large v=t\cdot v_1+v_0=t(\alpha u_1)+\beta u_1+u_0=
(t\alpha+\beta)u_1+u_0\in L_1. \]
Also, if (\w\in L_1\) we have that for some real \(s\)
\[ \large w=s\cdot u_1+u_0=s(1/\alpha\cdot v_1)+v_0-\beta u_1
=(s/\alpha-\beta)+v_0\in L_2 \].
Therefore \(L_1=L_2\).

Answer 3

if have a typo in my first post. it should be
\[ \large L_1=\{s\cdot u_1+u_0:s\in\mathbb{R}\}. \]

Answer 4

got it so far?

Answer 5

Not really, I'm trying to copy that and put in equation but it's not working...

Answer 6

use pencil and paper then

Answer 7

I got that, thanks.

Answer 8

try and do the reciprolycal by yourself. (it is an if-and-only-if theorem)

Answer 9

Yes, I think I will.