I have reviewed a solution to the problem for the implication in one direction: Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal{L}(V,W)$. Prove that dimrange$T=1$ if and only if 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.
proving the backwards direction
Let $v_1,...,v_m$ be a basis for $V$
Let $w_1,...,w_n$ be a basis for $W$
Assume all entries of $\mathcal{M}(T)$ are 1 with respect to these bases
Define a linear map $T:V\rightarrow W$ by
$T(v_i)=w_1+w_2+\ldots +w_n$ for each $i \in \{1,2,...,m\}$
Hence $\text{RangeT}=\text{Span}(w_1+w_2+ ... +w_n)$
and dimRange$T=1$
I understand the solution in the part where all entries are equal to $1$ since the coefficients for the linear combinations of the basis vectors of $W$ will all be $1$.
However I cannot understand why this implies Range$T=\text{Span}(w_1+\ldots+w_n)$ since the span of $w_1+\dots +w_n$ should consist of all scalar multiples of the vector. Can someone explain the reason for this?