Null Space Closed under Scalar Multiplication
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Theorem
Let:
- $\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf {A x} = \mathbf 0}$
be the null space of $\mathbf A$, where:
- $ \mathbf A_{m \times n} = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end {bmatrix}$, $\mathbf x_{n \times 1} = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$, $\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$
are matrices where each column is an element of a real vector space.
Then $\map {\mathrm N} {\mathbf A}$ is closed under scalar multiplication:
- $\forall \mathbf v \in \map {\mathrm N} {\mathbf A} ,\forall \lambda \in \R: \lambda \mathbf v \in \map {\mathrm N} {\mathbf A}$
Proof
Let $\mathbf v \in \map {\mathrm N} {\mathbf A}$, $\lambda \in \R$.
By the definition of null space:
| \(\ds \mathbf {A v}\) | \(=\) | \(\ds \mathbf {0}\) |
Observe that:
| \(\ds \mathbf A \paren {\lambda \mathbf v}\) | \(=\) | \(\ds \lambda \paren {\mathbf {A v} }\) | Matrix Multiplication is Homogeneous of Degree $1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lambda \mathbf 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \mathbf 0\) |
Hence the result, by the definition of null space.
$\blacksquare$
Also see
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.