How do you prove null $A$={0} $\iff$ {$c_1,...,c_n$} is independent?

Question

So $A$ is a $m \times n$ matrix with columns $c_1, c_2...c_n$. How do you prove null $A$={0} $\iff$ {$c_1,...,c_n$} is independent?

Since null $A$ ={$x$ in $ℝ^n$ |$Ax=0$}, how is it possible to say that $r_1c_1+...+r_nc_n=0 \implies c_1=...=c_n=0$?

Answer 1

Careful: You're not showing that all $c_i$ are $0$ — if nothing else, this would make $\{c_i \}$ a dependent set, which you won't find helpful. I'll offer a few hints; let me know if you'd like anything filled in.

Assume that $\{c_i \}$ is an independent set, and pick some $v \in \text{null}(A)$. What does $Av$ look like? It's essentially the linear combination you wrote in your question (assuming the $r_1$ are the entries in your nullspace vector). What does the independence of the $\{c_i \}$ vectors tell us?

Now assume that $\text{null}(A) = \{0\}$. Let's say we have two linear combinations of $\{c_i \}$ that are equal. How can we put these two vectors together so that the result is related to the nullspace of $A$? Let $v = v_1 + \ldots + v_n$ such that $c_1v_1 + \ldots + c_nv_n = 0$. How can we relate this linear combination back to $A$? What has to be true about the $v_i$ in order to make $\{c_i\}$ independent?

Mouseover for the rest of it:

Note that, if $c_1v_1 + \ldots + c_nv_n = 0$, then $Av = 0$, and so by assumption, $v=0$. So all $v_i$ are $0$; hence, $\{c_i\}$ is independent.