Prove that $\dim range T = 1$ if and only if there is a basis of $V$ and a basis of $W$ all entries of $M(T)$ equal $1$ Suppose $V$ and $W$ are finite dimensional and $T \in L(V,W)$ . Prove that $\dim range T = 1$ if and only if there is a basis of $V$ and a basis of $W$ such that with respect to these bases , all entries of $M(T)$ equal $1$. 
proof: Conversely suppose there is a basis of $V$ and a basis of $W$ such that with respect to these bases , all entries of $M(T)$ equal $1$. 
Then let $v_1,...,v_m$ and $w_1,...,w_n$ be basis of $V$ and $W$. And define a unique linear map $T: V →W$ by $Tv_i = w_i $ for $1 \leq i \leq m$.
Suppose $V$ and $W$ are finite dimensional and $T \in L(V,W)$ .
And let  $\dim range T = 1$. Then $\dim null T = \dim V -1$. Let $u_1,...,u_m $ be a basis of null$T$. Thus dim null $T$ = $m.$ 
Can someone please help? I am stuck.Thank you!
 A: Let $\{ v_1 , \ldots ,v_n\}$ be an arbitrary basis of $V$ and $w \in W$ a generator of $range(T)$. Without loss of generality $T v_i = \lambda_i w$ with $\lambda_i \neq 0$  (as for example you can replace all $v_i$ with $v_i + v$ where $T v = w$).
Write $w = a_1w_1 + \cdots + a_mw_m$ as a linear combination of some basis such that $a_i \neq 0$ for each $i = 1,\ldots , m$.
Then, take $\mathcal{B}_V = \{ \frac{v_1}{\lambda_1} , \cdots , \frac{v_n}{\lambda_n}\}$ and $\mathcal{B}_W = \{a_1w_1 , \cdots, a_mw_m\}$. 
A: I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
This exercise is Exercise 3.C.6 on p.79 in this book.
I solved this exercise as follows:
Let $m:=\dim V$.
Let $n:=\dim W$.
Then, by Fundamental Theorem of Linear Maps on p.63, $$\dim\operatorname{null}T=\dim V-\dim\operatorname{range}T=m-1.$$
Let $v_2,\dots,v_m$ be a basis of $\operatorname{null}T$.
Let $v_1,v_2,\dots,v_m$ be a basis of $V$.
Then, $v_1,v_2+v_1,\dots,v_m+v_1$ is linearly independent:
Proof:
Let $a_1v_1+a_2(v_2+v_1)+\dots+a_m(v_m+v_1)=0$.
Then, $(a_1+a_2+\dots+a_m)v_1+a_2v_2+\dots+a_mv_m=0$.
Then, $a_1+a_2+\dots+a_m=a_2=\dots=a_m=0$ since $v_1,v_2,\dots,v_m$ is linearly independent.
Then, $a_2=\dots=a_m=0$ and $a_1=-(a_2+\dots+a_m)=-(0+\dots+0)=-0=0$.
So, $v_1,v_2+v_1,\dots,v_m+v_1$ is linearly independent.
So, $v_1,v_2+v_1,\dots,v_m+v_1$ is a basis of $V$.
Since $v_2,\dots,v_m\in\operatorname{null}T$, $Tv_2=\dots=Tv_m=0$.
So, $Tv_1=T(v_2+v_1)=\dots=T(v_m+v_1)$.
Since $\dim\operatorname{range}T=1$, $T\neq 0$.
So, $Tv_1=T(v_2+v_1)=\dots=T(v_m+v_1)\neq 0$.
Let $w_1:=Tv_1$.
Let $w_1,w_2,\dots,w_n$ be a basis of $W$.
Then, $w_1-w_2-\dots-w_n,w_2,\dots,w_n$ is linearly independent:
Proof:
Let $b_1(w_1-w_2-\dots-w_n)+b_2w_2+\dots+b_nw_n=0$.
Then, $b_1w_1+(b_2-b_1)w_2+\dots+(b_n-b_1)w_n=0$.
Then, $b_1=b_2-b_1=\dots=b_n-b_1=0$ since $w_1,w_2,\dots,w_n$ is linearly independent.
Then, $b_1=0$ and $b_2=\dots=b_n=b_1=0$.
So, $w_1-w_2-\dots-w_n,w_2,\dots,w_n$ is linearly independent.
So, $w_1-w_2-\dots-w_n,w_2,\dots,w_n$ is a basis of $W$.
Then,
$Tv_1=w_1=1(w_1-w_2-\dots-w_n)+1w_2+\dots+1w_n$.
$T(v_2+v_1)=Tv_1=1(w_1-w_2-\dots-w_n)+1w_2+\dots+1w_n$.
$\dots$
$T(v_m+v_1)=Tv_1=1(w_1-w_2-\dots-w_n)+1w_2+\dots+1w_n$.
So, there exist a basis of $V$ and a basis of $W$ such that with respect to these bases, all entries of $\mathcal{M}(T)$ equal $1$.
