Prove that a system with infinite solutions has linearly dependent columns

Question

Suppose $A\textbf{x} = \textbf{b}$, where $\textbf{x}$ and $\textbf{b}$ are vectors. Prove that if $A$ has infinite solutions, it has linearly dependent columns

I'm not sure how to go about this. Should I show it by setting an arbitrary $\textbf{x}$ and arbitrary free variable $t$?

Answer 1

Here's an answer without rank nullity.

Let matrix $A=(c_1, c_2 \ldots c_n)$, where $c_i$ is a column vector.

Suppose infinite solutions exist. Suppose $x_1, x_2 \ (x_1 \neq x_2), $ are solutions to $Ax=b$. Take $Ax_1 -Ax_2 = b-b=\vec0 \implies A(x_1-x_2)=\vec0$.

Let $x_1 -x_2 = \bar x.$ Then, $A\bar x=\vec 0 \implies \sum_kc_kx_k=\vec 0$, where $x_k \in \mathbb{R}$.

Note not all $x_k$'s are $0$, as $x$ is not the zero vector. So the columns must be linearly dependent, by the definition of linear independence.