# Show that a line in R2 is a subspace

## Homework Statement

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

## 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

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Dick
Homework Helper

## Homework Statement

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

## 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?

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?

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?

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

Dick
Homework Helper
I just input the solution seconds ago. Check the OP

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.

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Not a solution. You have to say what a 'line' is.

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

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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.

That's all?
Check post #5

Dick
Homework Helper
That's all?
Check post #5

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.

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.

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.

Dick
Homework Helper
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.

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.

1 person
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.

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.

Mark44
Mentor
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
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.
x+y = (x1+y1,x2+y2)

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

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.

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

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

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

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

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.

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.

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.

Dick
Homework Helper
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.

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?

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?

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)

Dick
Homework Helper
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)

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

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

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