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
- $\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$
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
- For a video presentation of the contents of this page, visit the Khan Academy.