Column Space vs Span and minimum size of a set of vectors to guarantee a vector is in span? Given a set A of vectors, is there a minimum number of vectors in A to guarantee that a particular vector b is in span(A)?
If we have a m x n matrix A - what is the difference between the column space and the span? Everywhere I read makes it seem like they both equal the same thing.
What I understand:
Col(A) is the span of the columns of A - thus Col(A) must have n vectors in it, since A is of size m x n. What differentiates this from the span(A)?
Also: 
Given a set A of vectors, is there a minimum number of vectors in A to guarantee that a particular vector b is in span(A)?
Shouldn't A be of R$^n$ once again since there are n columns?
 A: The notation $\operatorname{span}(X)$ should be used for a set $X\subseteq V$ that is contained in some vector space $V$. It is then defined as the intersection of all subspaces of $V$ containing $X$ or equivalently as the set of all linear combinations of elements from $X$.
Writing "$\operatorname{span}(A)$" when $A$ is a matrix, doesn't make any sense in terms of this definition. However, it might be an abuse of notation to denote the span of the set of columns of $A$, that is, the column space of $A$. This space is usually denoted $\operatorname{col}(A)$ though.
The question about a "minimum number of vectors in a set $X$ to guarantee $b\in\operatorname{span}(X)$" is unclear to me. For example in the plane $V=\mathbb R^2$ you can take the infinite set
$$
X = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 0 \end{bmatrix}, \dots \right\}
$$
with span
$$
\operatorname{span}(X) = \left\{ \begin{bmatrix} x \\ 0 \end{bmatrix} \,\middle|\, x\in\mathbb R\,\right\}
$$
the does not contain $b=\begin{bmatrix}0 \\ 1\end{bmatrix}$ despite $X$ having infinitely many elements.
A: For example
Let A= 
\begin{bmatrix}
1 &0 &0&0\\
   0& 1 &0&1\\
   0 &0& 1&1\\
\end{bmatrix}.
Here $Col(A)= \mathbb{R}^3$ and observe that last column in A is sum of 2nd and 3rd columns. 
Note that $Span(A)=Col(A)=Row(A)$ 
