# Find the span of a set of vectors

In general, what is the most straightforward way to find the span of a set of vectors? I'm trying to find the span of these three vectors:

[1, 3, 3]

[0, 0, 1]

[1, 3, 1]

-
I would probably start by writing the set of vectors as a system of linear equations, then writing the system as an augmented matrix, and then converting to reduced row echelon form - is this the correct procedure? – Anderson Green Dec 7 '12 at 20:14
Do you know the definition of span? – Math_Illiterate Dec 7 '12 at 20:19
@Ockham Yes - the span of a set of vectors is the set of all linear combinations of a set of vectors. How can I find the set of all linear combinations of a set of vectors? – Anderson Green Dec 7 '12 at 20:20
I think I understand it now (at least partially). The span would be a[1, 3, 3] + b[0, 0, 1] + c[1, 3, 1], which would be [a + c, 3a + 3c, 3a + b + c], where a, b, and c are arbitrary constants. But how can I use this information to determine whether the set of vectors spans R^3? – Anderson Green Dec 7 '12 at 20:23
Once you have the echelon form, the rows that are not identically zero (interpreted as vectors) will form the basis. – David Mitra Dec 7 '12 at 21:00

You're quite right that the span would be all vectors of the form $$[a+c,3a+3c,3a+b+c],$$ where $a,b,c$ are real. The question becomes how we can describe this using as few parameters as possible. The idea is to reduce your spanning set to a basis--that is, a spanning set that is linearly independent--by discarding superfluous vectors.

Let's suppose that $[a+c,3a+3c,3a+b+c]=[0,0,0]$ and see what we can determine about $a,b,c$. Well, since $a+c=0$, then we certainly have $c=-a$, and making the substitution $c=-a$ gives us $$[0,0,2a+b]=[0,0,0].$$ This holds exactly when $b=-2a$. Thus, $$a[1,3,1]-2a[0,0,1]-a[1,3,1]=[0,0,0],$$ regardless of what value of $a$ we choose, so your given spanning set is linearly dependent. Indeed, setting $a=1$, this means that $$[1,3,3]=2[0,0,1]+[1,3,1],$$ so the first vector is unnecessary to span the whole space, since it's a linear combination of the other two vectors. In particular, $$[a+c,3a+3c,3a+b+c]=(2a+b)[0,0,1]+(a+c)[1,3,1].$$

Thus, every vector in the generated space can be written in the form $$u[0,0,1]+v[1,3,1]=[v,3v,u+v]$$ for some real $u,v$. All that's left to do is confirm that $[0,0,1]$ and $[1,3,1]$ are linearly independent, which we can do by setting $[v,3v,u+v]=0$, and showing that this implies $u=v=0$.

-

If by finding the span you mean finding a set of linearly independent equations who define the subspace spanned by the given vectors, you can go like this. Suppose you are given $v_1,\ldots, v_k\in \mathbb{R}^n$; construct the matrix $$A=\begin{pmatrix}\phantom{hhh}v_1^t\phantom{hhh}\\\hline\vdots\\\\\hline\phantom{hhh}v_k^t\phantom{hhh}\end{pmatrix}$$ then find a basis for $\ker A$, say $\{w_1,\ldots, w_h\}$. The subspace spanned by your vectors is described by the equations $$w_j^tx=0\qquad j=1,\ldots, h$$ where $x=\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}$.

-

${\text{Span}}\{{\vec v_1}, {\vec v_2}, {\vec v_3}\}$ is the set of all vectors $(x, y, z) \in {{\mathbb R}^3}$ s.t. $(x, y, z) = {c_1}{\vec v_1} + {c_2}{\vec v_2} + {c_3}{\vec v_3}$. We need to find the values of $\vec x = (x, y, z)$ so that our system of equations has solutions for ${c_1}$, ${c_2}$ and ${c_3}$. We need to solve $$\left[{\begin{array}{ccc|c} {\vec v_1}&{\vec v_2}&{\vec v_3}&{\vec x} \end{array}}\right].$$ In your case $$\left[{\begin{array}{ccc|c} 1&0&1&x\\ 3&0&3&y\\ 3&1&1&z \end{array}}\right].$$

-
Does [v1, v2, v3, x] represent a set of vectors, or does it represent a system of equations? – Anderson Green Dec 7 '12 at 20:33
It is not a set. It is just a compact way of writing matrices. – glebovg Dec 7 '12 at 20:39

First you should investigate what is a linear independent set in your span. Then your set will be the linear combination of this set. For example, I will do your example

$$\left| \begin{array}{ccc} 1 & 3 & 3 \\ 1 & 3 & 1 \\ 0 & 0 & 1\\ \end{array} \right| = 0$$ So they are not linearly independent. In fact, $(1,3,3) = (1,3,1) +2(0,0,1)$. But as $$\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1\\ \end{array} \right| = 2 \neq 0$$ $(1,3,1),(0,0,1)$ are linearly independent. Then, \begin{eqnarray} \langle (1,3,3),(1,3,1),(0,0,1) \rangle &=& \langle (1,3,1),(0,0,1) \rangle \\ &=& \{ \alpha (1,3,1) + \beta (0,0,1) : \alpha , \beta \in \mathbb{R}\} \\ &=& \{ (\alpha ,3 \alpha , \alpha +\beta) : \alpha , \beta \in \mathbb{R}\} \end{eqnarray}

-