Null Space Contains Zero Vector/Proof 2
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Theorem
Let:
- $\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
- $ \mathbf A_{m \times n} = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n} \\ \end{bmatrix}$
is a matrix in the matrix space $\map {\MM_\R} {m, n}$.
Then the null space of $\mathbf A$ contains the zero vector:
- $\mathbf 0 \in \map {\mathrm N} {\mathbf A}$
where:
- $\mathbf 0 = \mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$
Proof
From Matrix Product as Linear Transformation, $\mathbf {Ax} = \mathbf 0$ defines a linear transformation from $\R^m$ to $\R^n$.
The result then follows from Linear Transformation Maps Zero Vector to Zero Vector.