I came across a statement "Clearly, the rank of $A$ is the dimension of the range of the linear transformation $f :X \to Ax$", and I want the justification or proof?
My definition of rank: The rank of a matrix A is the common value of the row rank of A and the column rank of A.
Suppose that we are given that the $m \times n$ matrix $A$ has column-rank (and therefore coincident row-rank) $r$. That is, the maximal set of linearly independent columns of $A$ contains $r$ vectors. It follows that the span of the columns of $A$, henceforth the column space of $A$, is an $r$-dimensional subspace of $\Bbb R^m$.
It now suffices to show that the range of the linear transformation $f:x \mapsto Ax$ is precisely the same subspace. From there, we would see that saying "the column space is $r$-dimesional" is ("clearly") equivalent to saying that "the range of $f$ is $r$-dimensional". To see that this is the case, we proceed as follows.
Let $\mathbf a_1,\dots,\mathbf a_n$ denote the columns of $A$. Consider an arbitrary element $y$ inside the column space of $A$. By definition, this mean that there exist coefficients $x_1,\dots,x_n$ such that $$ y = x_1 \mathbf a_1 + x_2 \mathbf a_2 + \cdots + x_n \mathbf a_n. $$ Note that we can rewrite the above sum as a product. In particular, we have $$ y = \pmatrix{\mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n} \pmatrix{x_1\\ x_2 \\ \vdots \\ x_n} = Ax $$ where $x = (x_1,x_2,\dots,x_n)^T \in \Bbb R^{n}$. So, any element $y$ from the column space of $A$ is also an element of the range of $f$.
By following this idea in reverse, we see that any element $y$ in the range of $f$ has the form $y = f(x) = Ax$, which means that we have $y = x_1 \mathbf a_1 + \cdots + x_n \mathbf a_n$ like before. That is, $y$ is an element of the columns space of $A$.
