What is the difference between a linearly independent set and a set that spans $\Bbb{R}^m$? This is more of a conceptual question. Here's what I know about a linearly independent set of vectors:
A set of vectors $\{v_1, ..., v_p\}$ is linearly independent if the equation
$$x_1v_1 + x_2v_2 + ... + x_pv_p = 0$$
has only the solution $x = 0$, the trivial solution.
Here's what I know about $span$:
$S = \{v_1, ..., v_p\}$ is the set of all linear combinations of $v_1, ..., v_p$
Given a set of vectors $S = \{v_1, v_2, . . . , v_k\}$ in a vector space $V$ , S is said to span $V$
if $span(S) = V$
The columns of a matrix A span $\Bbb{R}^m$ if and only if A has a pivot position in every row.
Also, a matrix A spans $\Bbb{R}^m$ if the number of nonzero rows $\textbf{when in reduced row echelon form}$ equals $m$
My confusion is that these two things - being linearly independent and spanning $\Bbb{R}^m$ - seem like the same thing. For example, if a matrix A has a pivot position in every row, doesn't that mean A is also linearly independent? The only example I can think of that proves they are not the same is if a given matrix has no free variables (linearly independent) but has some zero rows and so does not span $\Bbb{R}^m$
I am just trying to really grasp the concepts here since it seems they correlate to many different topics in linear algebra, any further explanation would be much appreciated! Thanks!
 A: Let $ S = \{(1, 0, 0), (0, 1, 0)\} $. Then, $ S $ is linearly independent, as is easily seen, on the other hand $ (0, 0, 1) \notin \textrm{span}\, S $, therefore $ S $ does not span $ \mathbb{R}^3 $.
On the other hand, your intuition is partly correct due to the following result:
Theorem. Let $ V $ be a vector space, $ L $ a linearly independent subset and $ S $ a subset that spans $ V $. Then, $ |L| \leq |S| $.
Proof. Let $ S = S_0 = \{ s_i : 1 \leq i \leq n \} $ and let $ L = \{ b_i : 1 \leq i \leq m \} $. We construct a sequence of spanning sets. Given $ S_k $, construct the set $ S_{k+1} $ as follows: $ S_k $ is a spanning set, therefore we may write $ b_k = \sum c_i {s_k}_i $ where $ {s_k}_i \in S_k $ and the $ c_i $ are members of the field of scalars. As $ L $ is linearly independent, there must be a vector on the right hand side which is not an element of $ L $, let one such vector be ${s_k}_j$. Therefore, removing $ {s_k}_j $ from the set $S_k$ and replacing it with $ b_k $ gives us a spanning set with the same number of elements as $ S_k $ (as ${s_k}_j$ can be expressed as a linear combination of the elements of this new set). Define this set to be $ S_{k+1} $.
With this construction, a new element of $ L $ is added to the sets $ S_i $ at each step, however the cardinality of the sets remains unchanged. The construction halts at $ S_m $, which contains all elements of $ L $, therefore $ L \subseteq S_m $ and $|L| \leq |S_m| = |S|$, which establishes the result.
Corollary. Let $ V $ have dimension $ n $ over its field of scalars and let $ L $ be a linearly independent subset of $ V $ which has $ n $ elements. Then, $ L $ is a basis of $ V $.
Proof. Let $ B $ be a basis for $ V $, then $ |B| = n $. Consider the set $ L' = L \cup \{v\} $ for any $ v \in V $ and $ v \notin L $. This set has $ n+1 $ elements. However, any linearly independent subset of $ V $ can have at most $ n $ elements by the above theorem, as $ B $ is a spanning subset. Therefore, $ L' $ is linearly dependent, and in particular $ v $ can be expressed as a linear combination of the elements of $ L $ (otherwise $ L $ would be linearly dependent), which establishes that $\textrm{span}\, L = V $. By definition of a basis, $ L $ is a basis of $ V $.
Therefore, if your linearly independent subset has as many elements as the dimension of your vector space. then it has to span your space.
A: For spaning, every vector is a linear combination of vectors from the set, such a set must have at least $m$ vectors, but may have more. For linearly independent there are no linear combinations within the set, it can have size at most size $m$, but could also be smaller.
A: The easiest way I have found is to understand the two terms in reference to subspaces.  Being linearly independent means, in a sense, you have "just enough" for the span of that set of vectors; add any more to the set and you either a) are still within the same span and so now you have "too many" vectors and are linearly dependent--at least one is superfluous and can be gotten by a linear combination of the others, or b) you're still linearly independent, but you've broken through and you now span a higher-dimension subspace. Note, that for linear independence, "too few" is not a problem.  If I take away any vectors from a linearly independent set, the new set is still linearly independent.  "Too many" is the issue.
Span is not the same thing, exactly.  It's not "just enough" vectors, like linear independence; it's "enough" vectors, meaning linear combinations of my vectors will get me anywhere I want in this subspace.  These vectors "span" the subspace.  And this time "too many" vectors is not a problem for a spanning set, provided that I have enough in the first place.  Add more vectors to my set and I will still span.  However, now it is "too few" that is the problem:  start taking away vectors from my set and I may no longer span the same subspace (I will still span SOME subspace, but the dimension might have gone down).
There is a sweet spot where your number of vectors is neither too many nor too few; you span the whole subspace (not too few; enough to get everywhere) but are also linearly independent (not too many to be redundant nor have linear dependence within the set). Then you have a basis for the subspace. (Yay!)
One other difference is that in a spanning set, I'm not concerned with whether I have unique ways to get anywhere--I just want SOME linear combination that gets me where I want to go.  So EXISTENCE of a linear combination is the key.  If no linear combination exists, I need to add more vectors to my set to get solutions of my linear combination.  Span is an existence concept.
Whereas in linear independence, UNIQUENESS of the linear combinations is the key--if I have more than one linear combination for any place I want to go, then there's some redundancy that I can remove by reducing my set of vectors.  Linear independence is a uniqueness concept.
So linear independence and span are not the same thing, but they are very closely related.
