# 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?

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 $$R3\mathbb{R}^3$$ if and only if the rank of the matrix is $$33$$. For example, you have
(111321110100)→(100321110111)→(100021010011)→(100010021011)→(100010001001).\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 $$00$$s in first column; exchanged second and third rows; added multiples of the second row to third and fourth row to make $$00$$s in the second column).

At this point, it is clear the rank of the matrix is $$33$$, so the vectors span a subspace of dimension $$33$$, hence they span $$R3\mathbb{R}^3$$.

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

2. Determine if the vectors $$(1,0,0)(1,0,0)$$, $$(0,1,0)(0,1,0)$$, and $$(0,0,1)(0,0,1)$$ lie in the span (or any other set of three vectors that you already know span). In this case this is easy: $$(1,0,0)(1,0,0)$$ is in your set; $$(0,1,0)=(1,1,0)−(1,0,0)(0,1,0) = (1,1,0)-(1,0,0)$$, so $$(0,1,0)(0,1,0)$$ is in the span; and $$(0,0,1)=(1,1,1)−(1,1,0)(0,0,1) = (1,1,1)-(1,1,0)$$, so $$(0,0,1)(0,0,1)$$ is also in the span. Since the span contains the standard basis for $$R3\mathbb{R}^3$$, it contains all of $$R3\mathbb{R}^3$$ (and hence is equal to $$R3\mathbb{R}^3$$).

3. Solve the system of equations
$$α(111)+β(321)+γ(110)+δ(100)=(abc)\alpha\left(\begin{array}{c}1\\1\\1\end{array}\right) + \beta\left(\begin{array}{c}3\\2\\1\end{array}\right) + \gamma\left(\begin{array}{c}1\\1\\0\end{array}\right) + \delta\left(\begin{array}{c}1\\0\\0\end{array}\right) = \left(\begin{array}{c}a\\b\\c\end{array}\right)$$
for arbitrary $$aa$$, $$bb$$, and $$cc$$. If there is always a solution, then the vectors span $$R3\mathbb{R}^3$$; if there is a choice of $$a,b,ca,b,c$$ for which the system is inconsistent, then the vectors do not span $$R3\mathbb{R}^3$$. 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 $$aa$$, $$bb$$, and $$cc$$.