Show that if $T$ is surjective and spans $V$, then $T(S)$ spans $W$.

Question

Given that $T: V \to W$ is a linear transformation from $V$ to $W$. Show that if $T$ is surjective and $S\subset V$ spans $V$, then $T(S)$ spans $W$.

I think the main thing stumping me right now is how to use spanning to show anything or what I would need to prove in order to show that $W$ has been successfully spanned. I understand how to use the definition of linear transformation and surjective, but I can't even get started (OR CAN I???) without a clear understanding of how spanning works.

I understand span in concrete numbers, so I need help getting a more abstract concept of $\operatorname{span} (S)$, if possible. And I tried YouTube and reading definitions already, I am truly stuck.

Answer 1

$S$ spans $V$ simply means that every element $v \in V$ can be written (not necessarily uniquely) as a finite linear combination of elements of $S$.

so it is required to prove that every element $w \in W$ can be written as a finite linear combination of elements of $T(S)$

now, the fact that $T$ is surjective means that any $w \in W$ is the image (under the mapping $T$) of some (not necessarily unique) element of $V$. i.e. we can find $u \in V$ such that:

$$ w = T(u) $$

write $u$ as a finite linear combination of elements of $S$:

$$ u = \sum_k \lambda_k s_k $$ where the $\lambda_k$ are scalars and each $s_k \in S$

since the mapping $T$ is linear we have:

$$ w = T(u) = T\bigg(\sum_k \lambda_k s_k \bigg) = \sum_k \lambda_k T(s_k) $$ showing that $w$ is a linear combination of elements of $T(S)$ as required.

Answer 2

Let $w\in W$. As $T$ is surjective we know that there is some $v\in V$ such that $T(v)=w$. We also know that $S$ spans $V$, so there are $v_1,...,v_k\in S$ and scalars $\lambda_1,...,\lambda_k$ such that $v=\lambda_1v_1+\cdots+\lambda_kv_k.$ Applying $T$ to both sides of this equation yields the desired conclusion.