Let $\displaystyle u_{1} ,u_{2} ,...,u_{k}$ be a basis for $\displaystyle V$.

$\displaystyle Tu_{1} ,\ Tu_{2} ,...,Tu_{k}$ span range T.

Now, if $\displaystyle Tu_{1} \neq Tu_{2}$ then rank T is atleast 2, which is a contradiction.



Hence all $\displaystyle Tu_{i}$'s should be the same.

Let $\displaystyle Tu_{i} =c_{1} w_{1} +c_{2} w_{2} +...+c_{n} w_{n}$, where $\displaystyle w_{i} '$s are basis of W.

The above is true for all $\displaystyle u_{i}$.

So matrix of $\displaystyle T=\begin{bmatrix}

Let $\displaystyle u_{i} '$s be a basis of null T. Then extending it to basis of V gives

$\displaystyle u_{1} ,u_{2} ,...,u_{m} ,\ v_{1} ,v_{2} ,...,v_{n}$

So $\displaystyle Tv_{1} ,\ Tv_{2} ,...,Tv_{m}$ span range T

$\displaystyle \sum c_{i} Tv_{i} =0\Longrightarrow T\sum c_{i} v_{i} =0\Longrightarrow \sum c_{i} v_{i} =\sum d_{i} u_{i} \Longrightarrow $all $\displaystyle c_{i}$'s are zero.



So $\displaystyle Tv_{i}$'s are LI. Hence $\displaystyle n=1$.

So basis of V is $\displaystyle u_{1} ,u_{2} ,...,\ u_{m} ,v$, where $\displaystyle Tv$ spans range T.

Let $\displaystyle u_{i} '$s be a basis of null T. Then extending it to basis of V gives


\begin{equation}
u_{1} ,u_{2} ,...,u_{m} ,\ v_{1} ,v_{2} ,...,v_{n} \tag{1}
\end{equation}


So $\displaystyle Tv_{1} ,\ Tv_{2} ,...,Tv_{m}$ span range T. (Because for any for $\displaystyle x\in V$, there exist some scalars $\displaystyle x_{i} '$s and $\displaystyle x_{i} '$'s \ such that $\displaystyle x=x_{1} u_{1} +...+x_{m} u_{m} +x_{1} 'v_{1} +...+x_{n} 'v_{n}$ hence $\displaystyle Tx=\sum _{i=1}^{n} x_{i} 'Tv_{i} \in $range T. So $\displaystyle \{Tx:\ x\in V\} =span( Tv_{1} ,Tv_{2} ,...,Tv_{n}…