How to show the equivalence between full row rank of $A$ and the column space of $A$ being equal to $\mathbb{R}^m$?

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Suppose that $A \in \mathbb{R}^{m \times n}$, $m \leq n$

Is there a simple example that shows:

The following are equivalent

  1. the row rank of $A$ is $m$ ($A$ has full row rank)
  2. the column space of $A$ is equal to $\mathbb{R}^m$ (any $y \in \mathbb{R}^m, \exists x \in \mathbb{R}^n$ such that $Ax = y$)

A simple example I tried to show is that, let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

Then $$Ax = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} ax_1 + bx_2 \\ cx_1 + dx_2 \end{bmatrix}$$

Now suppose that $A$ does not have full row rank, then there exists some $x_1, x_2$, not all zero, such that $\begin{bmatrix} a \\ b \end{bmatrix} x_1 + \begin{bmatrix} c \\ d \end{bmatrix} x_2 = 0.$ However, this does not mean that the column space of $A$ is not $\mathbb{R}^2$, as it does not violate given any $y \in \mathbb{R}^2$, $\exists x \in \mathbb{R}^2$ such that $Ax = y$.

Can anyone check where I went wrong?

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Yes, it does violate that condition. The only case when your homogeneous system has a non-zero solution is when the columns $[a,b]$ and $[c,d]$ are proportional, and in that case it is impossible to obtain $y\in\mathbb{R}^2$ as their linear combination unless $y$ is in the straight line generated by either one of them.