Proof of Kernel and Image of a Linear Map Are Subspaces

Routine facts about mutual inclusion of classes are used without comment (Toolkit for AL2). By the linearity of $T$ (Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i)) and Elementary Identities in a Vector Space (ii), used in $V$ and in $W$,

$$T0=T(0\cdot0)=0\cdot T0=0,$$

and for $v\in V$, $T(-v)=T((-1)v)=(-1)Tv=-Tv$ by Elementary Identities in a Vector Space (iv). So $T$ is a map from $V$ to $W$ that fixes the zero vector and is additive (Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i)), and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (vi), applied to the map $i\mapsto\lambda_{i}v_{i}$ on $[n]$, gives

$$T\Bigl(\sum_{i=1}^{n}\lambda_{i}v_{i}\Bigr)=\sum_{i=1}^{n}T(\lambda_{i}v_{i})=\sum_{i=1}^{n}\lambda_{i}\,Tv_{i},$$

the last step again by Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i). This proves (i).

The kernel $\ker T$ is a subset of $V$ by Kernel and Image of a Linear Map (i), and it contains $0$ because $T0=0$. If $u,w\in\ker T$ and $\lambda\in K$, then $T(u+w)=Tu+Tw=0+0=0$ and $T(\lambda w)=\lambda Tw=\lambda0=0$ by Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i) and Elementary Identities in a Vector Space (iii). So $\ker T$ satisfies Linear Subspaces (i).

By The Image and the Preimage of a Class under a Class (i), $\operatorname{im}T$ consists of the values $Tv$ with $v\in V$, and it is a subset of $W$ by Kernel and Image of a Linear Map (ii). It contains $0=T0$, and it is closed under the operations, since $Tu+Tw=T(u+w)$ and $\lambda Tw=T(\lambda w)$ by Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i). So $\operatorname{im}T$ satisfies Linear Subspaces (i) as well, and together with the previous paragraph this proves (ii).

If $T$ is injective and $Tv=0=T0$, then $v=0$; since $0\in\ker T$, this gives $\ker T=\{0\}$. Conversely, let $\ker T=\{0\}$ and $Tu=Tw$. Then

$$T(u-w)=Tu+(-1)Tw=Tu-Tw=0$$

by Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i), The Zero Vector, Negatives and Differences in a Vector Space (iii), Elementary Identities in a Vector Space (iv) and Elementary Identities in a Vector Space (vii), so $u-w=0$, and $u=(u-w)+w=w$ by Elementary Identities in a Vector Space (viii). This proves (iii).

Since $\operatorname{im}T\subseteq W$, the equality $\operatorname{im}T=W$ says that every $w\in W$ is $Tv$ for some $v\in V$ (The Image and the Preimage of a Class under a Class (i)), which is what Injective, Surjective and Bijective Functions between Classes (ii) requires. This proves (iv).