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!
Best Answer
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 $\mathcal{B}_W = \{a_1w_1 , \cdots, a_mw_m\}$.