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

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.

• How would a linear map span a subspace? What is S in all this? – Bernard Jan 22 '17 at 1:59
• What is $S$? It seems to have just shown up out of nowhere in this question. – The Count Jan 22 '17 at 1:59
• I've edited your question. Please check that this is what you meant. – user378947 Jan 22 '17 at 2:02
• Yes, thank you for editing, that better says what I was trying to say. – Snackbreak Jan 22 '17 at 5:01

$S$ 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.

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.

• By the way, saying that a subset $S$ of a vector space $V$ spans $V$ means that for every $v\in V$ there are a finite number of vectors $v_1,...,v_k\in S$ and scalars $\lambda_1,...,\lambda_k$ such that $v=\lambda_1v_1+\cdots+\lambda_kv_k$. In words, that every vector in $V$ can be expressed as a finite linear combination of elements in $S$. – user378947 Jan 22 '17 at 2:07
• Okay, I think the lambdas were throwing me off in the Wikipedia definition, I wasn't sure how to express anything useful with a sum of (λ sub i)(v sub i), but you breaking it down like that helps significantly. Thank you! @ mathbeing @David Holden – Snackbreak Jan 22 '17 at 5:05