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I want to know if the set $\{(1, 1, 1), (3, 2, 1), (1, 1, 0), (1, 0, 0)\}$ spans $\mathbb{R}^3$. I know that if it spans $\mathbb{R}^3$, then for any $x, y, z, \in \mathbb{R}$, there exist $c_1, c_2, c_3, c_4$ such that $(x, y, z) = c_1(1, 1, 1) + c_2(3, 2, 1) + c_3(1, 1, 0) + c_4(1, 0, 0)$.

I've looked around the internet, but all the answers I found involve setting up a matrix and finding the determinant, and I can't do that here because my matrix isn't square. What am I missing here?

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@Geoff: We have studied systems of linear equations, matrices and determinants, and vector spaces (including subspaces, lineal combinations, lineal (in)dependence, spans and bases). –  Javier Badia Aug 7 '11 at 22:19
    
OK Javier, thanks. The other answers now cover most of what I would have said. –  Geoff Robinson Aug 7 '11 at 22:53

3 Answers 3

up vote 30 down vote accepted

There are several of things you can do. Here's four:

  1. You can set up a matrix and use Gaussian elimination to figure out the dimension of the space they span. They span $\mathbb{R}^3$ if and only if the rank of the matrix is $3$. For example, you have $$\begin{align*} \left(\begin{array}{ccc} 1 & 1 & 1\\ 3 & 2 & 1\\ 1 & 1 & 0\\ 1 & 0 & 0 \end{array}\right) &\rightarrow \left(\begin{array}{ccc} 1 & 0 & 0\\ 3 & 2 & 1\\ 1 & 1 & 0\\ 1 & 1 & 1 \end{array}\right) &&\rightarrow \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 2 & 1\\ 0 & 1 & 0\\ 0 & 1 & 1 \end{array}\right)\\ &\rightarrow \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 2 & 1\\ 0 & 1 & 1 \end{array}\right) &&\rightarrow \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 1 \end{array}\right). \end{align*}$$ (Sequence of operations: exchanged rows 1 and 4; subtracted first row from other rows to make $0$s in first column; exchanged second and third rows; added multiples of the second row to third and fourth row to make $0$s in the second column).

    At this point, it is clear the rank of the matrix is $3$, so the vectors span a subspace of dimension $3$, hence they span $\mathbb{R}^3$.

  2. See if one of your vectors is a linear combination of the others. If so, you can drop it from the set and still get the same span; then you'll have three vectors and you can use the methods you found on the web. For example, you might notice that $(3,2,1) = (1,1,1)+(1,1,0)+(1,0,0)$; that means that $$\mathrm{span}\Bigl\{(1,1,1),\ (3,2,1),\ (1,1,0),\ (1,0,0)\Bigr\} = \mathrm{span}\Bigl\{(1,1,1),\ (1,1,0),\ (1,0,0)\Bigr\}.$$

  3. Determine if the vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ lie in the span (or any other set of three vectors that you already know span). In this case this is easy: $(1,0,0)$ is in your set; $(0,1,0) = (1,1,0)-(1,0,0)$, so $(0,1,0)$ is in the span; and $(0,0,1) = (1,1,1)-(1,1,0)$, so $(0,0,1)$ is also in the span. Since the span contains the standard basis for $\mathbb{R}^3$, it contains all of $\mathbb{R}^3$ (and hence is equal to $\mathbb{R}^3$).

  4. Solve the system of equations $$\alpha\left(\begin{array}{c}1\\1\\1\end{array}\right) + \beta\left(\begin{array}{c}3\\2\\1\end{array}\right) + \gamma\left(\begin{array}{c}1\\1\\0\end{array}\right) + \delta\left(\begin{array}{c}1\\0\\0\end{array}\right) = \left(\begin{array}{c}a\\b\\c\end{array}\right)$$ for arbitrary $a$, $b$, and $c$. If there is always a solution, then the vectors span $\mathbb{R}^3$; if there is a choice of $a,b,c$ for which the system is inconsistent, then the vectors do not span $\mathbb{R}^3$. You can use the same set of elementary row operations I used in 1, with the augmented matrix leaving the last column indicated as expressions of $a$, $b$, and $c$.

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Thank you. Option 1 is the one I was supposed to know from class, by the way. So the general rule is that the rank of the matrix of vectors is the dimension of the space I'm looking for? –  Javier Badia Aug 7 '11 at 23:34
2  
@Javier: By definition, the rank of a matrix is the dimension off the span of its rows (which is equal to the dimension of the span of its columns); elementary row operations do not change the row space, so doing Gaussian elimination does not change the rank, it only makes it easier to tell what the rank is (if you are doing it correctly, at any rate). In summary: yes, because you are computing the dimension of the span of the rows, and the rows are the vectors you started with. –  Arturo Magidin Aug 7 '11 at 23:36

Use Gaussian elimination and check whether there are 3 non-zero rows at the end.

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It appears your answer is missing at least one (key) word. –  cardinal Aug 7 '11 at 22:13
    
@cardinal, thanks! –  lhf Aug 7 '11 at 22:15

If you check throw away $(3,2,1)$, you are left with 3 easily checked vectors. In fact, if $a(1,1,1)+b(1,1,0)+c(1,0,0)=(0,0,0) $ then we must have $a=0$ because only the first vector has a last coordinate. The same argument again gives $b=0$.

Three linearly independent vectors in a 3-dimensional space spans the space.

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But how do you know which of the vectors can be discarded? I guess you could take each one and check if it's a linear combination of the others, but is there a shorter way? –  Javier Badia Aug 7 '11 at 22:17
    
@Javier: I threw away (3,2,1) because it was handy to do so. The three other vectors "looked" linearly independent, and the argument in my answer proved they indeed were. –  Fredrik Meyer Aug 7 '11 at 22:19
1  
@Javier: learning efficient ways to do computations, and developing the skill of picking a route through, sometimes requires doing a number of computations yourself and trying different things until you get a feel for that works. –  Mark Bennet Aug 8 '11 at 6:07

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