Null Space Contains Only Zero Vector iff Columns are Independent

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Theorem

Let:

\(\ds \mathbf A_{m \times n}\) \(=\) \(\ds \begin{bmatrix} \mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n \end{bmatrix}\)

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:

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

if and only if:

$\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$

where $\map {\mathrm N} {\mathbf A}$ 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:

\(\ds \mathbf x\) \(\in\) \(\ds \map {\mathrm N} {\mathbf A}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \mathbf A \mathbf x_{n \times 1}\) \(=\) \(\ds \mathbf 0_{m \times 1}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \begin{bmatrix} \mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\) \(=\) \(\ds \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \sum_{k \mathop = 1}^n x_k \mathbf a_k\) \(=\) \(\ds \mathbf 0\)


Sufficient Condition

Let $\set {\mathbf a_1, \mathbf a_2, \cdots, \mathbf a_n}$ 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:

$\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$

$\Box$


Necessary Condition

Let $\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$.

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 $\set {\mathbf a_1, \mathbf a_2, \cdots, \mathbf a_n}$ is linearly independent.

$\blacksquare$


Sources

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