What is the relationship between the column space of a matrix and its dimensionality? The following text is from page 36 of chapter 2 of the book Deep Learning by Ian Goodfellow et al.:

In order for the system $Ax = b$ to have a solution for all values of
$b \in \mathbb{R}^m$, we therefore require that the column space of
$A$ be all of $\mathbb{R}^m$. If any point in $\mathbb{R}^m$ is
excluded from the column space, that point is a potential value of $b$
that has no solution. The requirement that the column space of $A$ be
all of $\mathbb{R}^m$ implies immediately that $A$ must have at least
$m$ columns, that is, $n \ge m$. Otherwise, the dimensionality of the
column space would be less than $m$. For example, consider a $3 \times
 2$ matrix. The target $b$ is 3-D, but $x$ is only 2-D, so modifying
the value of $x$ at best enables us to trace out a 2-D plane within
$\mathbb{R}^3$. The equation has a solution if and only if $b$ lies on
that plane.

I'm not sure I understood this part:

The requirement that the column space of $A$ be all of $\mathbb{R}^m$ implies
immediately that $A$ must have at least $m$ columns, that is, $n \ge m$. Otherwise, the dimensionality of the column space would be less than $m$.

Considering a dataset here then does he mean that the number of observations must be at least the number of fields in that dataset. Otherwise the dimensionality or rank or number of pivot entries or number of linearly independent columns would turn out to be less than the number of components i.e. $m$ involved. What will happen if it is less than $m$?
Later he has given an example. Please comment if my understanding is correct.

The target $b$ is 3-D, but $x$ is only 2-D, so modifying the value of $x$
at best enables us to trace out a 2-D plane within $\mathbb{R}^3$.

For a $3 \times 2$ matrix, the target $b$ is 3-D then that means there is a data-set (say) consisting of 3 observations but by $3 \times 2$ matrix as per book and considering each observation as a column vector then there are total 2 observations with 3 properties each and so 2-D nature of $x$ should be justified.
Have I mixed things? Please help and let me know if there is anything else that you might require for more complete comprehension.
 A: What will happen if it is less than $m$?
Then the system will no longer be assured of having a solution since there will be more equations than the number of unknowns involved.
Answer to part 2:
Since for more equations than the number of unknowns involved one will have to try different values of $x_1$ & $x_2$ (taking these to be components of $x$) or linear combinations to trace out a plane spanned by $v_1$ & $v_2$ i.e. $3 \times
 2$ here and checking if $b$ lies on that plane.
A: I'll answer the What happens if $n$ is less than $m$? part differently. Say we have an $m \times n$ matrix $A$ consisting of columns $A_1, \ldots, A_n$. We can think of these columns as vectors in $\mathbb{R}^m$. That is, each column in an $m \times n$ matrix has precisely $m$ real entries. Let's ask the question: What is the column space of $A$? Well, hopefully you know that the column space is, by definition, the span of $\{A_1, \ldots, A_n\}$. That is, the column space of $A$ is the set $$B = \{a_1A_1 + \ldots + a_nA_n: a_i \in \mathbb{R} \ \text{for each} \ i=1, \ldots, n\}$$
Notice that since each of the columns $A_i$ live in $R^m$, the column space $B$ is a subset of $R^m$. However, we are told that not only is $B \subset \mathbb{R}^m$, we are told that $B = \mathbb{R}^m$ exactly. This means that a subset $X$ of $\{A_1, \ldots, A_n\}$ with cardinality $|X| = r \leq n$ is a basis for $\mathbb{R}^m$. If the columns $A_1, \ldots, A_n$ are all linearly independent, then we may simply let $X = \{A_1, \ldots, A_m\}$. Otherwise, remove columns from the set $\{A_1, \ldots, A_n\}$ until you DO get a set of linearly independent columns and call that set $X$. But any basis of $\mathbb{R}^m$ must have exactly $m$ elements. Since $|X| = r \leq n < m$, we have found a basis of $\mathbb{R}^m$ with less than $m$ elements. This is a contradiction.
A: Well, for the first part, consider $A=\left(\begin{matrix}|&|&...&|\\a_1&a_2&...&a_n\\|&|&...&|\end{matrix}\right)$ and $x=\left(\begin{matrix}x_1\\x_2\\...\\x_n\end{matrix}\right)$. Then $Ax=\left(\begin{matrix}|\\a_1\\|\end{matrix}\right)x_1+\left(\begin{matrix}|\\a_2\\|\end{matrix}\right)x_2+...+\left(\begin{matrix}|\\a_n\\|\end{matrix}\right)x_n$ where each $\left(\begin{matrix}|\\a\\|\end{matrix}\right)$ has $m$ entries. If $n<m$ it's not possible to span all of $\mathbb R^m$ with a linear combination of $n$ vectors, even if they're all linearly independent. If $n\ge m$, then it's possible to span all of $\mathbb R^m$ (but it's possible that it also doesn't, if the vectors are linearly dependent etc.) Now the $b$ vector in $Ax=b$ has to be in the column space of $A$ for it to be a solution and... I think you can make the connection.
Now look at the $3\times2$ example.
$Ax=\left(\begin{matrix}3&2\\1&2\\4&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}a\\b\\c\end{matrix}\right)$
$Ax=\left(\begin{matrix}3\\1\\4\end{matrix}\right)x+\left(\begin{matrix}2\\2\\5\end{matrix}\right)y=\left(\begin{matrix}a\\b\\c\end{matrix}\right)$
You can see that you will get a plane through the origin right? If the right hand point/vector is in this plane, then it's a solution. If it's anywhere not on this plane, then it's not a solution.

A general result is that $Ax=b$ has solutions if $dim(C(A))=m$, and may not have solutions if $dim(C(A))<m$
As for your question in the title, the rank of a matrix is the dimension of its column space.
