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?

  • $\begingroup$ Generically you don't know without examing the presumed "basis" vectors. You do know that three vectors are sufficient (x,y,z) to span 3-space; any fourth vector must be a linear combination of (x,y,z). There is no more room. $\endgroup$
    – rrogers
    Oct 13, 2020 at 21:32

4 Answers 4


There are several things you can do. Here are 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 $\mathbb{R}^3$.

  1. 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 $$\mathrm{span}\Bigl\{(1,1,1),\ (3,2,1),\ (1,1,0),\ (1,0,0)\Bigr\} = \mathrm{span}\Bigl\{(1,1,1),\ (1,1,0),\ (1,0,0)\Bigr\}.$$

  2. 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 $\mathbb{R}^3$).

  3. 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 $c$.

  • 1
    $\begingroup$ 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? $\endgroup$
    – Javier
    Aug 7, 2011 at 23:34
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    $\begingroup$ @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. $\endgroup$ Aug 7, 2011 at 23:36
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    $\begingroup$ For option 1, why is the matrix 4x3 and not 3x4? $\endgroup$ Oct 12, 2017 at 16:43
  • 1
    $\begingroup$ @user3932000 Because I'm doing row reduction, not column reduction. It has the advantage that the linearly independent rows will give you a basis of the span. And because the vectors are row vectors, not column vectors. $\endgroup$ Oct 12, 2017 at 16:55
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    $\begingroup$ @CheeLoongSoon (1) the question is specifically about $\mathbb{R}^n$, hence so is the answer. (2) It works in any vector space, if you use coordinate vectors. It works in any field. (3) What you say about item three is already stated in 3, explicitly. $\endgroup$ Apr 4 at 12:32

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


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.

  • $\begingroup$ 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? $\endgroup$
    – Javier
    Aug 7, 2011 at 22:17
  • $\begingroup$ @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. $\endgroup$ Aug 7, 2011 at 22:19
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    $\begingroup$ @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. $\endgroup$ Aug 8, 2011 at 6:07

Another approach that has not yet already been mentioned. Assume we know {v1, v2, ... vn} spans the vector space. We know this vector space has dimension n since there are n linearly independent vectors that spans the vector space. This is from a proven theorem that all basis of a vector space has the same number of vectors that are both linearly independent and spans it. Hence, as long as you can find n linearly independent vectors in your new set, you know it is guaranteed to also span the vector space.


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