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!

  • $\begingroup$ What do you mean by "Also, a matrix $A$ spans $\mathbb{R}^m$ if the number of nonzero rows equals $m$".? What happens when you take a $m\times m$-matrix that's completely filled with $1$'s? $\endgroup$ – jazzinsilhouette Apr 30 '16 at 22:43
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    $\begingroup$ if you have n linearly independent vectors with n (real) components, they span $\mathbb R^n$. However if you have fewer then n vectors they are still linearly independent, but they do not span R^n. $\endgroup$ – Doug M Apr 30 '16 at 22:43
  • $\begingroup$ @jazzinsilhouette To clarify - $A$ spans $R^m$ if the number of nonzero rows when in row reduced in row echelon form equals $m$ $\endgroup$ – brdeav39 Apr 30 '16 at 22:51

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.


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.


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