Solution: Let $B = \{ v_1, v_2, \dots, v_k \} $ be linearly independent set in $V$. We need to prove that $T\left( B \right) = \{ T(v_1), T(v_2),\ldots , T(v_k) \} $ is also linearly independent. Note that as $T$ is $1-1$, the kernel of $T$ is trivial, that is $T(v) = 0$ if and only if $v = 0$. Let \begin{align*} \sum_{i=1}^{k} \alpha _i T(v_i) = 0 & \implies \sum_{i=1}^{k} T\left( \alpha _i v_i \right) = 0 \\ & \implies T \left( \sum_{i=1}^{k} \alpha _i v_i \right) = 0 \\ & \implies \sum_{i=1}^{k} \alpha _i v_i = 0 \\ & \implies \alpha _i = 0 \quad i = 1,2,\ldots k, \end{align*} where in the last step we used that $\{ v_1,v_2,\ldots ,v_k \} $ is linearly independent. Thus, $T(B)$ is linearly independent.

If $T$ is onto, then image of a spanning set of $V$ will be a spanning set of $W$. This is easy to prove. If $B = \{ v_1,v_2,\ldots ,v_k \} $ is a spanning set of $V$, then we claim that $T(B) = \{ T(v_1), T(v_2),\ldots ,T(v_k) \} $ is a spanning set of $W$. Let $w \in W$. Since $T$ is onto, we can find $v\in V$ such that $T(v) = w$. As $B$ is a spanning set of $V$, there exists $\alpha _i \in \mathbb{R}, i = 1,2,\ldots ,k $, such that \begin{align*} v = \sum_{i=1}^{k} \alpha _i v_i \implies T(v) & = T\left( \sum_{i=1}^{k} \alpha _i v_i \right) \\ & =\sum_{i=1}^{k} T\left( \alpha _i v_i \right) \\ & = \sum_{i=1}^{k} \alpha _i T(v_i). \end{align*} Thus, $T(B)$ is a spanning set of $W$.