# Show that a line in R2 is a subspace

1. Mar 11, 2014

### negation

1. The problem statement, all variables and given/known data

Show that a line in R2 is a subspace if and only if it passes through the origin (0,0)

3. The attempt at a solution

Let A set of vectors be the subset of the vector space R2.

What does it implies in context of this problem if it passes through the origin (0,0)? Does it means contain the zero vector?

S = {(x,y)} = (0,0)

Let u = u1,u2
Let w = w1,w2

u+w = (u1+w1, u2+w2)

for u1+w1,u2+w2 = 0
u1=-w1
if u1=1, w1 = -1

scalar:

k.u = (ku1,ku2)
ku1,ku2 = 0
ku1 = 0 if k = 0

Last edited: Mar 11, 2014
2. Mar 11, 2014

### Dick

Yes. That handles the 'only if' part of the proof. To show the 'if' part you have to say what 'line' means. Is it the graph of a expression ax+by=c?

3. Mar 11, 2014

### negation

I just input the solution seconds ago although I have not addressed the 'if' part. Check the OP.

4. Mar 11, 2014

### Dick

Not a solution. If L is ANY subset (not necessarily a line) of R2 and L is a subspace then L must contain the zero vector. It's in the definition of subspace. That's the 'only if' part. You don't have to even write anything vector like. It's just logic. The 'if' part means you need to define what a line is in R2.

Last edited: Mar 11, 2014
5. Mar 11, 2014

### negation

A line is, by definition, y = mx+c?

Or, (x,y) = (0,1) or (x,y) = (1,0)

y = mx+c
Since, (x,y) = (0,0)
then,
0=m(0) + c
0 = c

Last edited: Mar 11, 2014
6. Mar 11, 2014

### negation

That's all?
Check post #5

7. Mar 11, 2014

### Dick

That's all. Doesn't it make sense? 'Line' is a subspace ONLY IF 'Line' contains the zero vector. It doesn't even matter what 'Line' means. And ok, suppose 'Line' means y=mx+c (that's leaving out the vertical lines, but ok for now). Now prove IF 'Line' contains the zero vector, then 'Line' is a subspace. Start by telling me what c must be.

8. Mar 11, 2014

### negation

Yes it make sense. I just didn't though I could miss out a trivial clue.
L = {(x,y) | y = mx+c}

Since x and y are both 0, c = 0.

9. Mar 11, 2014

### Dick

Right. So if (x1,y1) satisfies y1=m*x1 and (x2,y2) satisfies y2=m*x2 (so they are on the line) then does (x1+x2,y1+y2)? This is closure under addition.

10. Mar 11, 2014

### negation

Let's see.

S = {(x,y)|y = mx+c}
then, y1 = mx1+c and y2 = mx2+c

For closure under addition to hold:
Let x = x1,x2 and y = y1,y2

x+y = (x1+y1,x2+y2)

x = x1+y1 y = x2+y2
then,
(x2+y2) = m(x1+y1) +c is true.

11. Mar 12, 2014

### Staff: Mentor

You're being very sloppy here (above). What does "x1,x2" mean? The elements in your set (i.e., on your line) are pairs of numbers (x0, y0) such that y0 = mx + c.

12. Mar 12, 2014

### negation

I meant it to be (x1,x2) but admittedly, it was sloppy.

Was my final answer correct?

(x2+y2) = m(x1+y1) +c

13. Mar 12, 2014

### Dick

It's not much of a proof if you haven't even convinced yourself. Slow down. I thought we agreed c=0? Put it to 0. And if y1=m*x1 and y2=m*x2 then the points on your line are (x1,y1) and (x2,y2), right? Try that proof again.

14. Mar 12, 2014

### negation

It's better that if I condense my steps.

S = {(x,y) | y = mx + c}

At the origin: 0 = m(0) + c
c = 0

x = (x1,x2) and y = (y1,y2)

S = {(x1,x2) , (y1,y2) | y = mx + 0 }

Closure by addition : x + y = (x1 + y1, x2 + y2)

(x2+y2) = m(x1+y1) I'm puzzled as to why this is not a valid step.

15. Mar 12, 2014

### Dick

Maybe it's just that you are using x and y in kind of a confusing way. (x2+y2) = m(x1+y1) is correct. Why is it correct? Can you spell it out?

16. Mar 12, 2014

### negation

well, it can be seen that y = mx

and under closure by addition, x = (x1 + y1) and y = x2 + y2

so,

(x2+y2) = m(x1+x2)

17. Mar 12, 2014

### Dick

I hope you mean (x2+y2) = m(x1+y1).

18. Mar 12, 2014

### negation

Woops. That was a slip. It can be pretty confusing with all the sub's.
Yes, you are right.