# 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!

• 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? – jazzinsilhouette Apr 30 '16 at 22:43
• 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. – Doug M Apr 30 '16 at 22:43
• @jazzinsilhouette To clarify - $A$ spans $R^m$ if the number of nonzero rows when in row reduced in row echelon form equals $m$ – 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.