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




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

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

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

We have that:

Sufficient Condition
Let $\left\{{\mathbf a_1, \mathbf a_2, \cdots, \mathbf a_n}\right\}$ be linearly independent.

Then by definition:
 * $\forall k: 1 \le k \le n: x_k = 0 \iff \mathbf x = \mathbf 0_{n \times 1}$

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

Necessary Condition
Let $\operatorname{N} \left({\mathbf A}\right) = \left\{ {\mathbf 0_{n \times 1} }\right\}$.

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.