How would I go about showing that two spans are equal?
I was thinking that I could somehow prove that each is a subset of the other but I'm not sure how.
If sp(A) is a subset of sp(B), I could prove that sp(B) is a subset of sp(A). Could I extend this proof somehow to prove that each of the spans are subsets of the other?
What about showing that $$\dim\bigl(Sp(A)\bigr)=\dim\bigl(Sp(B)\bigr)=\dim\bigl(Sp(A\cup B)\bigr)?$$ Can be done easily by row reduction.
Given sets of vectors $A$ and $B$ we want to prove that $Sp(A)=Sp(B)$ or, to put it another way, that all vectors which can be represented as a linear combination of elements of $A$ can also be represented as a linear combination of elements of $B$, and vice versa.
By Gram-Schmidt Orthonormalization, we can construct a set of linearly independent, normalized vectors from both $A$ and $B$ which span $Sp(A)$ and $Sp(B)$, respectively - call these new sets $A'$ and $B'$, respectively. $A'$ and $B'$ are called the bases of their respective vector spaces.
[Note: If $M$ is an orthonormal basis which spans $Sp(M)$, then there exists an orthogonal transformation mapping $M$ to $M_0$, where $M_0$ is the single-component basis of $Sp(M)$ (i.e. the basis in which each basis vector has only one component which is non-zero, and that component is 1 since the basis is normalized).]
Let $A_0$ and $B_0$ be the single-component bases for $Sp(A)$ and $Sp(B)$ respectively. Let $\text{dim}(Sp(A))$ and $\text{dim}(Sp(B))$ be the dimension of of $Sp(A)$ and $Sp(B)$ respectively. It is obvious that any vector that is of the same dimension as a vector space can be represented as a linear combination of elements of a single-component basis of that vector space.
If the dimension of $Sp(A)$ and $Sp(B)$ (i.e. the number of elements of $A_0$ and $B_0$) are different, then it is a simple matter to construct a vector in the basis with the larger dimension that cannot be constructed in the basis with the smaller dimension. However, if the dimensions are equal, then the bases are equivalent (i.e. $Sp(A)=Sp(B)$) by the following proof.
Proof by Contradiction:
Suppose $A_0$ and $B_0$ are orthonormal, single-component bases which span the vector spaces $Sp(A)$ and $Sp(B)$ respectively. Suppose that $\text{dim}(Sp(A))=\text{dim}(Sp(B))=N$. Suppose $Sp(A)\neq Sp(B)$.
Since $Sp(A)\neq Sp(B)$, (without loss of generality) there exists a vector $a\in Sp(A)$ such that $a\notin Sp(B)$.
However, since $B_0$ is a single -component basis in $N$ dimensions, any $N$-dimensional vector can be represented as a linear combination of elements of $B_0$; therefore, $a$ can can be written as a linear combination of elements of $B_0$ because $a\in Sp(A)$ which implies that $a$ is an $N$-dimensional vector (since $\text{dim}(Sp(A))=N$).
Thus, we have a contradiction (namely, $a\notin Sp(B) \land a\in Sp(B)$), so it must be the case that $Sp(A)=Sp(B)$.
I find that it is less confusing to think of $A:=$ Span($S_A$), where $S_A$ is a set of vectors.
I will demonstrate that we just need that $\forall a \in S_A; a \in B$ and $\forall b \in S_B; b \in A$.
Let's do an example: Let $A:= {\rm Span}(\left\{ \pmatrix{-3\\1\\3} , \pmatrix{-1\\2\\2} \right\} )$ and $B:= {\rm Span}(\left\{ \pmatrix{1\\3\\1} , \pmatrix{3\\4\\0} \right\} )$.
None of the basis vectors are shared between the two spaces, but they are actually the same space: you can solve some equations to show that $\pmatrix{1\\3\\1} = -1 \times \pmatrix{-3\\1\\3} + 2 \times \pmatrix{-1\\2\\2}$ so that $b_1 \in A$ and similarly $b_2 = -2 \times a_1 + 3 \times a_2$, so $b_2 \in A$, too.
Since any element $b$ in $B$ can be written as a linear combination of $b_1$ and $b_2$, say $b= j \times b_1 + k \times b_2 = j \times (-a_1 + 2a_2) + k \times (-2a_1 + 3a_2) = (-j-2k)a_1 + (2j + 3k)a_2$, hence $b \in A$. This shows that $B \subseteq A$. We can follow a similar process to show $A \subseteq B$.
However, you will see in future study that we can simplify this process further: what is really going on is that we are finding a linear transformation between two spaces of the same dimension and so if you can just identify that (and show that it has only the zero vector in the kernel) we are done.
