Prove that $\{c_1,…,c_n\}$ is independent if and only if $\text{null}(A)=\{0\}$

Question

Prove that $\{c_1,…,c_n\}$ is independent if and only if $\text{null}(A)=\{0\}$

I understand that for the columns of the matrix $A$ to be independent, the linear combination has to be trivial, and we know that $Ax = 0$.

I am not sure how to go about synthesizing the proof.

Edit: Matrix $A is not a square matrix

Answer 1

Suppose that $A$ has columns $c_1, c_2, \ldots, c_n$, then

$$Ax = 0 \Leftrightarrow x_1c_1 + x_2c_2 + \ldots x_nc_n = 0$$

See for example matrix multiplication using columns, which is a neat way of writing down a matrix multiplication.

Now $c_1, \ldots, c_n$ are linearly independent if and only if the $x_i = 0$ for all $i$. Hence the nullspace of $A$ equals $\{0\}$.

Answer 2

We have: $\text{null}(A) = 0\iff A \text{ is invertible} \iff \text{det}(A) \neq 0\iff \text{ the columns of A are linearly independent}$.