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How can I determine whether the span of a set of vectors (such as $\mathrm{span}\{(3, 1), (4,1), (0,1)\}$ contains the span of another set of vector?

EDIT: I realize that my original question was too vague. If A and B are sets of vectors in $\mathbb{R}^3$, how can you determine whether $\mathrm{span}\{A\}$ contains $\mathrm{span}\{B\}$?

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Are there any additional steps involved? Is this sufficient for showing that the span of one set of vectors contains the span of another set of vectors? – Anderson Green Jun 3 '12 at 18:05
@Ananda In general checking linear independence won't be good enough to answer this kind of question. – Brett Frankel Jun 3 '12 at 18:22
@Brett Frankel : I have to change the answer with the Edit. – Theorem Jun 3 '12 at 18:25
BTW this question is closely related: Linear Algebra: determine whether the sets span the same subspace – Martin Sleziak Jun 3 '12 at 19:07

If $M(A)$ is a matrix with the vectors of $A$ as columns and $M(AB)$ is the matrix with the vectors of both $A$ and $B$ as columns, then $span(B) \subset span(A)$ if $rank(M(A))=rank(M(AB))$. Rank is after all the dimension of the column space of a matrix.

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Let's show one possible approach on an example. Suppose we are given a set of vectors $(1,-2,-2,2),(2,0,-1,1),(3,0,-4,-4)$ in $\mathbb R^4$. We want to know whether their span $V_A=[(1,-2,-2,2),(2,0,-1,1),(3,0,-4,-4)]$ contains the span $V_B=[(1,0,1,1),(1,1,0,0)]$ of the vectors $(1,0,1,1)$ and $(1,1,0,0)$. This is equivalent to finding out whether $(1,0,1,1)\in V_A$ and $(1,1,0,0)\in V_A$.

First we put the vectors from the set $A$ as a rows into a matrix and we use elementary row operations to get a row echelon form.

$ \begin{pmatrix} 1 & -2 & -2 & 2 \\ 2 & 0 & -1 & -1 \\ 3 & 0 & -4 & -4 \end{pmatrix} \overset{(1)}{\sim} \begin{pmatrix} 1 & -2 & -2 & 2 \\ 0 & 4 & 3 & -5 \\ 0 & 0 & -\frac52 & -\frac52 \end{pmatrix} \overset{(2)}{\sim} \begin{pmatrix} 1 & -2 & -2 & 2 \\ 0 & 4 & 3 & -5 \\ 0 & 0 & 1 & 1 \end{pmatrix} \overset{(3)}{\sim} \begin{pmatrix} 1 & -2 & -2 & 2 \\ 0 & 4 & 0 & -8 \\ 0 & 0 & 1 & 1 \end{pmatrix} \overset{(4)}{\sim} \begin{pmatrix} 1 & -2 & -2 & 2 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{pmatrix} \overset{(5)}{\sim} \begin{pmatrix} \boxed1 & 0 & 0 & 0 \\ 0 & \boxed1 & 0 & -2 \\ 0 & 0 & \boxed1 & 1 \end{pmatrix} $

Now we know that $V_A=[(1,0,0,0),(0,1,0,-2),(0,0,1,1)]$ since row operations do not change the row space.

But since we know have simpler vectors generating $V_A$, it is much easier to find out whether some given vector belongs to $V_A$.

If some vector $(a_1,a_2,a_3,a_4)$ belongs to $V_A$, it must fulfill $(a_1,a_2,a_3,a_4)=a_1(1,0,0,0)+a_2(0,1,0,-2)+a_3(0,0,1,1)$. (It suffices to look at the position of the leading coefficients of the row echelon form; in this case the first three positions. The leading coefficients are marked by boxes in the row echelon matrix.)

Now we want test the vectors $(1,0,1,1)$ and $(1,1,0,0)$.

For the first one we get $1\times(1,0,0,0)+1\times(0,0,1,1)=(1,0,1,1)$, so this vectors belongs to $V_A$.

For the second one we get $1\times(1,0,0,0)+1\times(0,1,0,-2)=(1,1,0,-2)\ne(1,1,0,0)$. Therefore $(1,1,0,0)\notin V_A$ and, consequently, $V_B\not\subseteq V_A$.

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The simple case is where $\text{span}(A)$ contains three linearly independent vectors in $\mathbb{R}^{3}$, which of course will form a basis set for the vector space (and therefore all vectors in $\mathbb{R}^{3}$ are contained within that spanning set).

In this case it would simply be sufficient to demonstrate that 3 of the vectors within the $\text{span}(A)$ are linearly independent.

If this is not the case, then it suffices to simply show that each of the linearly independent vectors of $\text{span}(B)$ is linearly dependent upon some set of vectors $S \in \text{span}(A)$. Of course if there are more linearly independent vectors in $\text{span}(B)$ then it is impossible for $\text{span}(A)$ to contain $\text{span}(B)$.

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You need to show that each vector in $B$ (or, if you prefer, a set of vectors whose span contains $B$) can be spanned by vectors in $A$. Then, given a vector in the span of $B$, you can write it as a linear combination of vectors in $B$, each of which can in turn be written as a linear combination of vectors in $A$, so that any vector in the span of $B$ is thus in the span of $A$.

Note that this sufficient condition, that every vector in $B$ is in the span of $A$, is also clearly necessary, since the span of $B$ contains each vector in $B$.

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