I got the proof for $\{T(v_1), T(v_2), . . . , T(v_n)\}$ spanning $W$:
Since the $v_i$ spans $V$, given any $v ∈ V$, we can write $v = a_1v_1 + . . . + a_nv_n$ for some scalars $a_i ∈ F$. Applying $T$ to both sides, we have $T(v) = T(a_1v_1 + . . . + a_nv_n) = a_1T(v_1) + . . . + a_nT(v_n)$, by linearity of $T$.
Now $T(v) = w$ for some arbitrary vector $w ∈ W$. Thus we have $w = T(v) = a_1T(v_1) + . . . + a_nT(v_n)$, for any arbitrary vector $w ∈ W$, so $(T(v_1), . . . , T(v_n))$ spans $W$.
But I'm confused about how to do it for $image(T)$. I know that the image of a linear transformation is the span of the vectors of the linear transformation, but I'm not sure how to use that.
Best Answer
Only the second part is correct. Consider the zero transformation, or one into a space of higher dimension. Say, for instance, $T:\Bbb R^4\to \Bbb R^5$. Then $T$ can't be surjective.
To prove the second part is easy by linearity and the fact that $\{v_1,\dots,v_n\}$ spans $V$. For the image of $T$ is nothing but $T(V)$.