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?

  • $\begingroup$ Why in this problem is RangeT$=\text{Span}(w_1+\dots+w_m)$ I am really confused Since every vector in the range of the linear transformation is $w_1+\dots+w_m$ there are no scalar multiples of this vector? $\endgroup$
    – user736276
    Jan 1, 2020 at 19:31
  • 1
    $\begingroup$ Note that in the proof, we have only plugged in $v = v_i$ for $i = 1,\dots,m$. Taking linear combinations of these vectors will produce multiples of this vector. $\endgroup$ Jan 1, 2020 at 19:33

1 Answer 1


Any vector in $V$ can be written in the form $v = a_1 v_1 + \cdots a_n v_n$. With that, we see that $$ \begin{align} T(v) &= T(a_1 v_1 + \cdots + a_m v_m) = a_1 T(v_1) + \cdots + a_m T(v_m) \\ & = a_1(w_1 + \cdots + w_n) + \cdots + a_m(w_1 + \cdots + w_n) \\ & = (a_1 + \cdots + a_m)(w_1 + \cdots + w_n). \end{align} $$ So, we can indeed say that for any $v \in V$, $T(v)$ is a multiple of $(w_1+ \cdots + w_n)$ and thus in the span of $w_1 + \cdots + w_n$.


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