How to tell if a set of vectors spans a space? 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?
 A: Use Gaussian elimination and check whether there are 3 non-zero rows at the end.
A: 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.
A: There are several things you can do. Here are four:

*

*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 $\mathbb{R}^3$.


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


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


*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 $c$.
A: 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.
