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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:

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

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    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?2012-12-07
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    Do you know the definition of span?2012-12-07
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    @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?2012-12-07
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    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?2012-12-07
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    To simplify things a tad, write the matrix whose rows are the given vectors. Find a basis for the row space of this matrix by taking the non-zero rows of an echelon form. The vectors comprising this basis will have the same span as the set of original vectors. (Note you'll still be in the situation where you have to say "the span is the set of linear combinations of these vectors..."; but the aforementioned procedure will weed out any redundancies.)2012-12-07
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    That is correct2012-12-07
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    @DavidMitra What do you mean by "taking the non-zero rows?" Specifically, how can I find a basis by "taking" the rows?2012-12-07
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    Once you have the echelon form, the rows that are not identically zero (interpreted as vectors) will form the basis.2012-12-07

4 Answers 4

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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 $u=v=0$.

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

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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 $$\left[ \begin{array}{ccc|c} 1 & 0 & 1 & x \\ 3 & 0 & 3 & y \\ 3 & 1 & 1 & z \end{array} \right].$$

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    Does [v1, v2, v3, x] represent a set of vectors, or does it represent a system of equations?2012-12-07
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    It is not a set. It is just a compact way of writing matrices.2012-12-07
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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 $x=\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}$.

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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}

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    Why is (1, 1, 1) not linearly independent?2017-10-19