Columns of A - Linear independence, span, rank. [GStrang P180, 3.5.19]

Question

The columns of $A$ are $n$ vectors from $\mathbb{R^m}$. If they're linearly independent, what's $rank(A)$? If they span $\mathbb{R^m},$ what's $rank(A)$? If they're a basis for $\mathbb{R^m},$ what's $rank(A)$?
Solution: $n$-independent columns $\Large{\color{red}{[}}$ implies rank $= n$ $\Large{{\color{red}{]}}}$.
Columns span $\mathbb{R^m}$ $\Large{\color{red}{[}}$ implies rank $= m$ $\Large{{\color{red}{]}}}$ .
Columns are basis for $\mathbb{R^m}$ implies rank = $m = n. \qquad \square$

P124: The column space consists of all linear combinations of the columns.
The combinations are all possible vectors $\mathbf{Ax}$.
P144: The rank of a matrix is its number of pivots.
P171: A set of vectors spans a space if their linear combinations fill the space.

$\Large{1.}$ I'm trying to complete the steps in this (terse) solution. Since the $n$ columns are linearly independent, thus $\left[\vec{\text{col 1 of A}} \ldots \vec{\text{col n of A}} \right]_{m \times n}\mathbf{x = 0} \implies \sum_{1 \le i \le n}x_i(\vec{\text{col i of A}}) = \mathbf{0} \implies \mathbf{x = 0}.$
How and why does this imply $rank(A) = n$?

$\Large{2.}$ Are there more intuitive, lucid arguments/proofs than #$1$?

$\Large{3.}$ Since the columns that span $\mathbb{R^m}$ might be linearly dependent, shouldn't this be rank $\le m$ ?
I then considered $L = \left[c \quad kc \right]_{1 \times 2}$ where $m = 2, n = 2$ and
by inspection, the number of pivots $ = 1$. For all $k \neq 0$, this is just the straight line $x_2 = (-1/k)x_1,$ which isn't all of $\mathbb{R^2}$. So $L$ doesn't span $\mathbb{R^2}$ ?

This question precedes dimension, dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations, all of which should please be omitted.

Answer 1

Re:1,2 : If you apply eliminations (on a matrix with n independent columns) to get reduced row echelon form (rref) then the result will be matrix with n pivots in n columns. Since rank equates with no. of pivots, in this case it'll be n.

Re:3 : It's not mentioned that if either n$\leq$m or n$\geq$m. Since, given n vectors are in $R^m$ and are linearly independent, n=m follows. Otherwise, if n < m then, they can't span whole of $R^m$ but just part of it. Moreover, if n > m then they can't be linearly independent which contradicts (1).
In your example you took linearly dependent columns ($L = [c\hspace1ex kc]$) that's why it has single pivot and rank=1. If you choose ($L = [c\hspace1ex d]$) where c and d are independent then you have two pivots and L has rank=2. Thus, n=m follows.