Linear mapping/Linear subspaces/Kernel/Introduction/Section

A typical property of a linear mapping is that it maps lines to lines (or to a point). More general is following statement.

Lemma

Let ${}K$ be a field, let ${}V$ and ${}W$ denote ${}K$-vector spaces and let

$\varphi \colon V\longrightarrow W$

be a

${}K$-linear mapping. Then the following hold.

1. For a linear subspace ${}S\subseteq V$, the image ${}\varphi (S)$ is a linear subspace of ${}W$.
2. In particular, the image ${}\operatorname {Im} \varphi =\varphi (V)$ of the mapping is a linear subspace of ${}W$.
3. For a linear subspace ${}T\subseteq W$, the preimage ${}\varphi ^{-1}(T)$ is a linear subspace of ${}V$.
4. In particular, ${}\varphi ^{-1}(0)$ is a linear subspace of ${}V$.

Proof

See exercise.

$\Box$

Definition

Let ${}K$ denote a field, let ${}V$ and ${}W$ denote ${}K$-vector spaces, and let

$\varphi \colon V\longrightarrow W$

denote a ${}K$-linear mapping. Then

${}\operatorname {kern} \varphi :=\varphi ^{-1}(0)={\left\{v\in V\mid \varphi (v)=0\right\}}\,$

is called the kernel of ${}\varphi$.

Due to the statement above, the kernel is a linear subspace of ${}V$.

Remark

For an ${}m\times n$-matrix ${}M$, the kernel of the linear mapping

$K^{n}\longrightarrow K^{m},x\longmapsto Mx,$

given by ${}M$ is just the solution space of the homogeneous linear system

${}Mx=0\,.$

The following criterion for injectivity is important.

Lemma

Let ${}K$ denote a field, let ${}V$ and ${}W$ denote ${}K$-vector spaces, and let

$\varphi \colon V\longrightarrow W$

denote a ${}K$-linear mapping. Then ${}\varphi$ is injective if and only if ${}\operatorname {kern} \varphi =0$

holds.

Proof

If the mapping is injective, then there can exist, apart from ${}0\in V$, no other vector ${}v\in V$ with ${}\varphi (v)=0$. Hence, ${}\varphi ^{-1}(0)=\{0\}$.
So suppose that ${}\operatorname {kern} \varphi =0$, and let ${}v_{1},v_{2}\in V$ be given with ${}\varphi (v_{1})=\varphi (v_{2})$. Then, due to linearity,

${}\varphi (v_{1}-v_{2})=\varphi (v_{1})-\varphi (v_{2})=0\,.$

Therefore, ${}v_{1}-v_{2}\in \operatorname {kern} \varphi$, and so ${}v_{1}=v_{2}$.

$\Box$