Kernel of Linear Transformation between Finite-Dimensional Normed Vector Spaces is Closed

Theorem

Let $m, n \in \N_{> 0}$ be natural numbers.

Let $A \in \R^{m \times n}$ be a matrix.

Let $\ker A = \set {\mathbf x \in \R^n : A \mathbf x = 0}$ be the kernel of $A$.

Then $\ker A$ is a closed subspace of $\R^n$.

Proof

Let $T_A : \R^n \to \R^m$ be the linear transformation such that:

$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$

By Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous, $T_A : \R^n \to \R^m$ is continuous.

We have that Singleton in Normed Vector Space is Closed.

Hence, $\set {\mathbf 0}$ is closed in $\R^n$.

By the corollary of Mapping is Continuous iff Inverse Images of Open Sets are Open, the inverse image of $\set {\mathbf 0}$ under $T_A$ is closed in $\R^n$.

However:

$\map {T^{-1}_A} {\set {\mathbf 0}} = \set {\mathbf x \in \R^n : A \mathbf x = 0} = \ker A$

Hence, $\ker A$ is closed in $\R^n$.

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$