Find the span of a set of vectors

Question

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:

$$\{[1, 3, 3], [0, 0, 1], [1, 3, 1]\}$$

Answer 1

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, $$a[1,3,3]-2a[0,0,1]-a[1,3,1]=[0,0,0],$$ 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 $u=v=0$.

One final step you can take is to prove that your span is the following: $$\bigl\{[x,y,z]\in\Bbb R^3:y=3x\bigr\}.$$

Answer 2

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

Answer 3

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 $x=\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}$.

Answer 4

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}