Span of S over R^2

  • Thread starter AkilMAI
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Homework Statement


Let S be the set of all vectors x = (x1; x2) in R^2 such that x1 = 1: What is the span of S ?


Homework Equations


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The Attempt at a Solution


x,y from S where x=(1,x2) ,y=(1,y2).Let w be the span of S => (w1,w2)=c1x+c2y......the system looks something like this w1=c1+c2 and w2=c1x1+c2y2...how can I find the 2 constants?
 

Answers and Replies

  • #2
Dick
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Can you show me a vector that's NOT in the span?
 
  • #3
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thanks again for the reply.....ok (w1,w2)=(w1-w2)x+w2y=>w1=w1-w2+w2 which is true but w2=w1x2-w2x2+w2y2
 
  • #4
Dick
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thanks again for the reply.....ok (w1,w2)=(w1-w2)x+w2y=>w1=w1-w2+w2 which is true but w2=w1x2-w2x2+w2y2

I don't think you are thinking about this concretely enough. Is (2,5) in the span? Is (-1,1)? Is (2.67,10.32)? Is (0,1)? Is (1,0)?
 
  • #5
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wait...is span S=R^2?if so how can i prove it....the thing that I find confusing in the problem is the second coordonate of the each vector(ex. x2,y2......).
well (2,5)=c1x+c2y=>2=c1+c2,5=c1x2+c2y2...so if c1=c2=1 then x2=3 and y2=-3 .....hmm I'm not doing it right
 
  • #6
Dick
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wait...is span S=R^2?if so how can i prove it....the thing that I find confusing in the problem is the second coordonate of the each vector(ex. x2,y2......).
well (2,5)=c1x+c2y=>2=c1+c2,5=c1x2+c2y2...so if c1=c2=1 then x2=3 and y2=-3 .....hmm I'm not doing it right

Yes, the span is R^2. (2,5)=(1,1)+(1,4). (1,1) and (1,4) are in S so (2,5) is in the span. Or (2,5)=2*(1,5/2) and (1,5/2) is in S. Do the other ones. It's good practice.
 
  • #7
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Ok,I understand ,thank you.....one last question...is there any way to prove it generally without the use of concrete examples?
 
  • #8
Dick
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Ok,I understand ,thank you.....one last question...is there any way to prove it generally without the use of concrete examples?

If you prove (1,0) and (0,1) are in the span, that would prove it, wouldn't it?
 
  • #9
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Apparently I don't think I'm paying much attention to the problem.....I did prove (1,0) and are in the span (0,1)...but how will this prove for all x,y in R^2?
 
  • #10
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I mean,why just between the coordinates 0 and 1?
 
  • #11
Dick
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I mean,why just between the coordinates 0 and 1?

The span of (1,0) and (0,1) is R^2, isn't it? I picked because that's a standard basis. (a,b)=a*(1,0)+b*(0,1). So if they are in the span then all of R^2 is in the span, right?
 
  • #12
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yes...
 

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