Show that a line in R2 is a subspace

In summary: It's better that if I condense my steps.S = {(x,y) | y = mx + c}At the origin: 0 = m(0) + c c = 0x = (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 solution?Well, what you've written is "proof by example". You gave an example of
  • #1
negation
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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)addition:
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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  • #2
negation said:

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?
 
  • #3
Dick said:
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.
 
  • #4
negation said:
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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  • #5
Dick said:
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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  • #6
Dick said:
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
 
  • #7
negation said:
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.
 
  • #8
Dick said:
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.
 
  • #9
negation said:
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.
 
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  • #10
Dick said:
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.
 
  • #11
negation said:
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.
negation said:
x+y = (x1+y1,x2+y2)

x = x1+y1 y = x2+y2
then,
(x2+y2) = m(x1+y1) +c is true.
 
  • #12
Mark44 said:
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.

Was my final answer correct?

(x2+y2) = m(x1+y1) +c
 
  • #13
negation said:
I meant it to be (x1,x2) but admittedly, it was sloppy.

Was my final answer correct?

(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.
 
  • #14
Dick said:
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.
 
  • #15
negation said:
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?
 
  • #16
Dick said:
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)
 
  • #17
negation said:
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).
 
  • #18
Dick said:
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.
 

What is a subspace in R2?

A subspace in R2 is a subset of R2 that satisfies three conditions: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication.

How do you show that a line in R2 is a subspace?

To show that a line in R2 is a subspace, you must first demonstrate that it contains the zero vector. Then, you can show that it is closed under vector addition by adding two arbitrary vectors on the line and showing that the result is also on the line. Finally, you can show that it is closed under scalar multiplication by multiplying a vector on the line by an arbitrary scalar and showing that the result is also on the line.

Can a line in R2 be a subspace if it does not pass through the origin?

No, a line in R2 must pass through the origin to be a subspace. This is because the zero vector is a necessary condition for a subspace, and if a line does not pass through the origin, it cannot contain the zero vector.

Can a line in R2 be a subspace if it is not a straight line?

No, a line in R2 must be a straight line to be a subspace. This is because in order for a subspace to be closed under vector addition and scalar multiplication, it must be a straight line with a constant slope.

Why is it important to determine if a line in R2 is a subspace?

Determining if a line in R2 is a subspace is important because it allows us to understand the properties and behaviors of the line. It also allows us to apply mathematical operations and transformations to the line, knowing that it will still remain a subspace.

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