$T:V\rightarrow W$ such that $N(T)=V'\subset V$ and $R(T)=W'\subset W$ How should one prove that there exists a linear map $T:V\rightarrow W$ such that $N(T)=V'\subset V$ and $R(T)=W'\subset W$ if $\dim(V')+\dim(W')=\dim(V)$, where $V$ and $W$ are finite-dimensional vector spaces?
My "answer" is just a guess really... It seems pocketed with holes. What do you think?
 A: Choose a basis $(b_1,..,b_k)$ for $V'$ and extend this arbitrarily to a basis of $V$, say $(b_1,..,b_k,c_1,..,c_s)$ (assumed that $\dim V'=k$ and $\dim V=k+s$). 


*

*Check that $Tc_1,..,Tc_s$ are linearly independent,

*and they span the whole $W'=\{w\,\mid\, w=Tv$ for some $v\in V\}$.


Now, if you want a concrete example, just consider any matrix $A$ and the mapping $v\mapsto A\,v$.
A: Here is my reasoning:

Let $V$ and $W$ be finite-dimensional vector spaces over $F$. Let $A=\{a_1,\dots,a_l\}$ be a basis for $V'\subset V$, let $B=\{a_1,\dots,a_l,b_1,\dots,b_m\}$ be a basis for $V$, let $C=\{c_1,\dots,c_n\}$ be a basis for $W'\subset W$, and let $D=\{c_1,\dots,c_n,d_1,\dots,d_p\}$ be a basis for $W$. Thus $N(T)=V'$ and $R(T)=W'$ means that
\begin{eqnarray}
T(a_1)&=&0+\cdots +0\\
T(a_2)&=&0+\cdots +0\\
\vdots\\
T(a_l)&=&0+\cdots +0\\
T(b_1)&=&k_{11}c_1+k_{21}c_2+\cdots+k_{n1}c_n\\
T(b_2)&=&k_{12}c_1+k_{22}c_2+\cdots+k_{n2}c_n\\
\vdots\\
T(b_n)&=&k_{n1}c_1+k_{n1}c_2+\cdots+k_{nm}c_n,
\end{eqnarray}
or simply
\begin{eqnarray}
\begin{pmatrix}
c_1&c_2&\cdots&c_n
\end{pmatrix} \begin{pmatrix}
0&\cdots&0&k_{11}&\cdots&k_{n1}\\
\vdots&&\vdots&\vdots&&\vdots\\
0&\cdots&0&k_{n1}&\cdots&k_{nm}
\end{pmatrix},
\end{eqnarray}
which is an $(l+n)\times (l+m)$ matrix. Thus, such a $T$ exists and has the above form.

As far as I understand, we're dealing with something that looks like this:

A: $\dim V'+\dim W'=\dim V\implies\dim V'(=p,\text{say}),\dim W'(=q,\text{say})\le\dim V(=n,\text{say})$
Let $\{v_1,v_2,...,v_p\}$ be a basis of $V'.$ Extend this to a basis $B=\{v_1,v_2,...,v_p,b_1,...b_q\}$ of $V.$
Consider a basis $\{w_1,w_2,...,w_q\}$ of $W'.$
Define $T':B\to W$ by $T(v_i)=0~\forall~i=1(1)p$ and $T(b_i)=w_i~\forall~i=1(1)q.$
$T'$ has a unique extension to a linear transformation $T:V\to W.$
Verify that $\ker T=V'$ and $\text{range }T=W'.$
