Null Space Contains Only Zero Vector iff Columns are Independent

Theorem
Let:

be a matrix where:


 * $\forall i: 1 \le i \le n: \mathbf{a_i} = \begin{bmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi} \end{bmatrix} \in \R^m$.

are vectors.

Then:


 * $\left\{ \mathbf{a_1}, \mathbf{a_2}, \cdots, \mathbf{a_n} \right\}$ is a linearly independent set

iff


 * $\operatorname{N}\left({\mathbf A}\right) = \left\{ \mathbf {0}_{n \times 1}\right\}$

where $\operatorname{N}\left({\mathbf A}\right)$ is the null space of $\mathbf A$.

Proof
Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \R^m$.

Suppose $\left\{ \mathbf{a_1}, \mathbf{a_2}, \cdots, \mathbf{a_n} \right\}$ is independent.

Then by the definition of independence, $\forall k: 1 \le k \le n: x_k = 0 \iff \mathbf x = \mathbf 0_{n \times 1}$.

By the definition of null space, this means $\operatorname{N}\left({\mathbf A}\right) = \left\{ \mathbf {0}_{n \times 1}\right\}$.

Now suppose $\operatorname{N}\left({\mathbf A}\right) = \left\{ \mathbf {0}_{n \times 1}\right\}$.

Then by the definition of null space, $\mathbf x = \mathbf {0}_{n \times 1}$.

This means that $\forall k: 1 \le k \le n: x_k = 0$, from which it follows that $\left\{ \mathbf{a_1}, \mathbf{a_2}, \cdots, \mathbf{a_n} \right\}$ is linearly independent.