Confused with working out the span of a set of vectors in R^2

by gilabert1985
Tags: confused, span, vectors, working
gilabert1985 is offline
Apr27-12, 04:30 AM
P: 7
Hi everyone!

I have the following problem which I don't understand... It is already solved, but there are three questions I have regarding it.

The problem says: "Let S be the set of all vectors [itex]x=(x_{1}, x_{2})[/itex] in [itex]\Re^{2}[/itex] such that [itex]x_{1}=1[/itex]. What is the span of S?"

And here is the answer that has me so confused...

"[itex]span S = \Re^{2}[/itex] because [itex](x_{1}, x_{2})=x_{1}(1, x^{-1}_{1}x_{2}[/itex] when [itex]x_{1}\neq0[/itex]
and [itex](x_{1}, x_{2})=(1, 0)-(1, -x_{2})[/itex] when [itex]x_{1}=0[/itex]."

But I don't understand the first line... why does it say when [itex]x_{1}\neq0[/itex] if [itex]x_{1}[/itex] is supposed to be equal to 1?

And in the second line, the same... why is [itex]x_{1}=0[/itex]?

So yeah, I understand they are linear combinations and all that, but for the condition given ([itex]x_{1}=1[/itex]), I don't understand how this answer satisfies it.
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Office_Shredder is offline
Apr27-12, 05:26 AM
P: 4,499
The points (x1,x2) in the lines that are confusing you are not points in S, but points in R2, which it then shows how to represent as linear combinations of points in S
Fredrik is online now
Apr27-12, 08:05 AM
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Since you can't rewrite ##x_2## as ##x_1 x_1{}^{-1}x_2## when ##x_1=0##, you have to consider the case ##x_1=0## separately.

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