A question that often comes up is determining if two sets of vectors have the same span.
Let \(P = \{v_1,v_2\}\) and \(Q = \{u_1,u_2\}\) where \(v_1 = \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}\), \(v_2 = \begin{bmatrix} 2 \\1 \\-1\end{bmatrix}\), \(u_1 = \begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}\), \(u_2 = \begin{bmatrix} -2 \\-1 \\1\end{bmatrix}\). We give an outline that shows that \(\operatorname{span}(P)=\operatorname{span}(Q)\).
Since we are attempting to establish the equality of two sets of vectors, what we need to do is to show that either set is a subset of the other.
Recall that if a vector space \(V\) contains the vectors \(x_1,\ldots,x_k\), then \(\operatorname{span}(\{x_1,\ldots, x_k\})\) is a subspace of \(V\).
Hence, to show that \(\operatorname{span}(P)\subseteq\operatorname{span}(Q)\), it suffices to show that \(v_1,v_2 \in \operatorname{span}(Q)\). And to show that \(\operatorname{span}(Q)\subseteq\operatorname{span}(P)\), it suffices to show that \(u_1,u_2 \in \operatorname{span}(P)\).
We now give the details for showing \(v_1 \in \operatorname{span}(Q)\).
To show that \(v_1 \in \operatorname{span}(Q)\), we need to write \(v_1\) as a linear combination of \(u_1\) and \(u_2\). Hence, we need to find real numbers \(\alpha\) and \(\beta\) such that \[v_1 = \alpha u_1 + \beta u_2.\] In other words, we need to solve the tuple equation \[\alpha~\begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}+ \beta~\begin{bmatrix} -2 \\-1 \\1\end{bmatrix}= \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}.\] Rewriting this in matrix form, we get \[\begin{bmatrix} 1 & -2\\ 0 &-1 \\ -1& 1\end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix}= \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}.\]
Row reducing the augmented matrix gives \(\begin{bmatrix} 1 & 0 &-1\\ 0 &1 &-1 \\ 0& 0 &0\end{bmatrix}.\) Hence, \(\alpha = \beta = -1\) is a solution. Thus \(v_1\in \operatorname{span}(Q)\).
One can now go on to show that \(v_2\in \operatorname{span}(Q)\). That will give \(\operatorname{span}(P) \subseteq \operatorname{span}(Q)\).
To show that \(\operatorname{span}(P) \subseteq \operatorname{span}(Q)\), one shows that \(u_1\in \operatorname{span}(P)\) and \(u_2\in \operatorname{span}(P)\).The above looks like a tedious process. But an observation can lead to a simpler procedure.
To decide if \(u_1, u_2\) is in the span of \(P\), we are really determining if the system \(Ax = b\), where the columns of \(A\) are \(v_1, v_2\) and \(x = \begin{bmatrix} \alpha \\ \beta\end{bmatrix}\), has a solution for each choice of \(b \in \{u_1, u_2\}\). We can answer this by first row reducing the extended augmented matrix \[\left[ \begin{array}{rr|rr} 1 & 2 & 1 & -2 \\ 1 & 1 & 0 & -1 \\ 0 & -1 & -1 & 1 \end{array} \right]\] to obtain \[\left[ \begin{array}{rr|rr} 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right].\] This tells us that \(-v_1 + v_2 = u_1\) and \(-v_2 = u_2\).
Similary, to decide if \(v_1, v_2\) is in the span of \(Q\), we row reduce the extended augmented matrix \[\left[ \begin{array}{rr|rr} 1 & -2 & 1 & 2 \\ 0 & -1 & 1 & 1 \\ -1 & 1 & 0 & -1 \end{array} \right].\] The remaining details are left as an exercise.
Fill in the details of the example above.