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Here's the question I'm trying to solve:

Determine whether the set $S$ spans $\mathbb{R^2}$:
$S={(1,2),(-2,-4),(1/2,1)}$

I know that the answer is that it does not span $\mathbb{R^2}$ because when I attempt to solve the resultant matrix, a row of zeros is produced - indicating that the vectors are linearly dependent and therefore negating the possibility of S being a spanning set.

To explain, a spanning set, specifically in this case, would have to satisfy $(a,b)=c_1(1,2)+c_2(2,-4)+c_3(1/2,1)$.

From here, a matrix can be formed to solve for $c_1,c_2,c_3$.

That is fine. However, the other part of the question is along the lines of, "if S does not span $\mathbb{R^2}$, give a geometric description of the subspace it does define."

To me, it is intuitive that this would describe a line in $\mathbb{R^2}$. This is because the vectors are clearly multiples of each other, which would indicate they can be found on the same line. Even though that's the case, would would be a mathematical process that would reach the same conclusion?

Any help will be appreciated.

Edit: I should add that I feel that it can describe a line if one concludes that the free variable that will be generated describes a line. For example: $c_1+c_2(t)$ where $t=c_3$.

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3 Answers 3

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Denote $v=(1,2)$. You deduced that the span of the vectors in the question equals to $$span\{v\}=\{vt|t\in\mathbb{R}\}$$

This is a line (by definition) and it is the line that connects $0$ and $v$ in $\mathbb{R}^{2}$.

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I think that is the conclusion I reached in my edit above. –  Mlagma Oct 9 '12 at 2:35
    
@Mlagma - this is a line by your definition for take $c_1=0$ –  Belgi Oct 9 '12 at 2:36
    
You may be stating it, and I'm just missing the point. I'm basing it off the solution to the matrix, where a row of zeros is produced through Gauss-Jordan elimination. On can see as well that $c_3$, if taken as the third column, is a free variable. By that, I parametrize the solution, which generates a line. And yes, $c_1$ would equal zero in this case. –  Mlagma Oct 9 '12 at 2:40
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One zero row (and two non-zero ones) does not mean it cannot span $\mathbb{R}^2$. It just means the vectors are linearly dependent, while they may or may not be spanning. Two zero rows (and one non-zero one) on the other hand means they are definitely not spanning. –  Shahab Oct 9 '12 at 3:50
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They're not just on the same line --- they're on the same line through the origin. That's the same as saying each one is a scalar multiple of the others, and that gives you a mathematical process to reach the conclusion.

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Construct the matrix $M$ composed of coordinates of the vectors in $S$: \begin{equation} M = \begin{bmatrix} 1 & 2 \\ -2 & -4 \\ 0.5 &1 \end{bmatrix} \end{equation}

What is $\textrm{rank}(M)$? What is $\mathbb{R}^\textrm{rank(M)}$?

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Um, wouldn't the matrix be a 2 by 3 for just the coefficient matrix? –  Mlagma Oct 9 '12 at 2:34
    
Sorry, that's what I meant. –  Ganesh Oct 9 '12 at 2:35
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