Linear Transformations, Rank-Nullity and Quotient Spaces

Linearity gives $T(0_V)=T(0_F0_V+0_F0_V)=T(0_V)+T(0_V)$; cancellation in $W$ gives $T(0_V)=0_W$.

Since $0_W=T(0_V)=T(v+(-v))=T(v)+T(-v)$, the inverse law gives $T(-v)=-T(v)$.

Using $u-v=u+(-v)$ and step 2.1, $T(u-v)=T(u)+T(-v)=T(u)-T(v)$.

The kernel contains $0_V$ by [L3]. If $u,v\in\ker T$ and $a\in F$, then $T(au+v)=aT(u)+T(v)=0_W$, so $au+v\in\ker T$.

The image contains $0_W=T(0_V)$. If $T(u),T(v)\in\operatorname{im}T$ and $a\in F$, then $aT(u)+T(v)=T(au+v)$ lies in the image.

If $T$ is injective and $v\in\ker T$, then $T(v)=0_W=T(0_V)$, so $v=0_V$ and the kernel is trivial.

Conversely, suppose $\ker T=\{0_V\}$ and $T(u)=T(v)$. Then $T(u-v)=0_W$ by [L3], so $u-v=0_V$ and $u=v$.

The subspace criterion proves that both $\ker T$ and $\operatorname{im}T$ are linear subspaces.

Thus $T$ is injective exactly when its kernel is trivial.

By [L1] and [L2], $\ker T$ is finite-dimensional; choose a basis $K$ of $\ker T$. It is linearly independent in $V$, so [L2] extends it to a basis $B$ of $V$. Put $C:=B\setminus K$. Since $B$ is finite, so is $C$, and $B=K\mathbin{\dot\cup}C$.

For every finite list $v:n\to V$ and every scalar list $a:n\to F$, linearity gives $T(\sum_{i<n}a_iv_i)=\sum_{i<n}a_iT(v_i)$. This follows by induction on $n$: the empty case follows by applying [L6] to $0_V=0_V+0_V$ and cancelling, and the successor step uses the append recursion and $T(x+a_nv_n)=T(x)+a_nT(v_n)$. The same recursion, now by induction on the length of the second list, splits a spliced sum into its first and second blocks.

Write $\nu:=|K|$ and $r:=|C|$, and choose bijections $k:\nu\to K$ and $c:r\to C$. Splicing the two lists gives $d:\nu+r\to B$, with $d_i=k_i$ for $i<\nu$ and $d_{\nu+j}=c_j$ for $j<r$. The disjointness in step 1.1 makes $d$ injective and its image is $B$, so it is an ordered basis of $V$.

The set $T[C]$ spans $\operatorname{im}T$. Indeed, if $y=T(v)$ lies in the image, [L4] gives coordinates $a:\nu+r\to F$ with $v=\sum_{i<\nu+r}a_id_i$. Steps 2.1 and 1.2 split this into the $K$-block and the $C$-block. Applying $T$ kills the first block because every $k_i$ lies in $\ker T$, leaving $y$ as a finite linear combination of $T(c_0),\ldots,T(c_{r-1})$.

The list $T\circ c:r\to W$ is linearly independent. Suppose $\sum_{j<r}b_jT(c_j)=0_W$. By step 1.2, $u:=\sum_{j<r}b_jc_j$ lies in $\ker T$, so [L4] supplies $a:\nu\to F$ with $u=\sum_{i<\nu}a_ik_i$. Thus the spliced ordered basis $d$ has a vanishing linear combination whose coefficients are $-a_i$ on the first block and $b_j$ on the second. Its linear independence forces every $b_j=0_F$.

Linear independence makes $T\circ c$ injective. It is surjective onto $T[C]$ by that set's definition, so it is a bijection $r\to T[C]$. Together with steps 3.1 and 3.2, this says that it is an ordered basis of $\operatorname{im}T$ and that $T|_C:C\to T[C]$ is bijective.

Step 1.1 supplies $K$ and $B$, and step 4.1 proves all the asserted properties of their complement $C$.

Choose $K,B,C$ as in [L2]. Since $B=K\mathbin{\dot\cup}C$, [L4] gives $|B|=|K|+|C|$. The bijection $T|_C:C\to T[C]$ gives $|C|=|T[C]|$.

Since $B$, $K$, and $T[C]$ are bases of $V$, $\ker T$, and $\operatorname{im}T$, respectively, [L1] and [L3] turn step 1.1 into $\dim_F V=\operatorname{nullity}T+\operatorname{rank}T$.

The displayed equivalent form follows by unfolding the definitions of rank and nullity.