Show if $S_1 \subseteq \text{span} S_2$ then span$(S_1) \subseteq $ span$(S_2)$

Question

I am proving this statement as an exercise, and wondering if my proof is correct or not. I am also wondering if there are any improvements that can be made to my writing style.

if $S_1 \subseteq \text{span} (S_2)$ then span$(S_1) \subseteq $ span$(S_2)$.

Edit: $S_1$ and $S_2$ are subsets of a finite dimensional vector space $V$

I have also assumed $S_1$ and $S_2$ are finite, I'm not sure whether or not I was meant to do this though.

Proof:

let $S_1 = \{\textbf{v}_1,..,\textbf{v}_n\}$, $S_2 = \{\textbf{s}_1,...,\textbf{s}_j\}$. Suppose $S_1 \subseteq \text{span} (S_2)$.

let $\textbf{y} \in \text{span}(S_1)$. This implies that we can write $$\textbf{y} = \alpha_1 \textbf{v}_1 + ... + \alpha_n \textbf{v}_n$$ But we know that $\textbf{v}_1,...,\textbf{v}_n \in \text{span}(S_2)$ Hence, we can write $$\textbf{y} = \alpha_1(\lambda_{1,1} \textbf{s}_1 + ... + \lambda_{1,j} \textbf{s}_j) + ... + \alpha_n (\lambda_{n,1} \textbf{s}_1 + ... + \lambda_{n,j} s_j)$$ Where $\lambda_{i,k} \in \mathbb{F}, i \in \{1,..,n\}, k \in \{1,..,j\}$

This implies that $$\textbf{y} = (\alpha_1 \lambda_{1,1} \textbf{s}_1 + ... + \alpha_1 \lambda_{1,j} \textbf{s}_j) + ... + (\alpha_n \lambda_{n,1} \textbf{s}_1 + ... + \alpha_n \lambda_{n,j} \textbf{s}_j)$$ By the vector distributive law.

Simplifying, we get $$\textbf{y} = (\alpha_1 \lambda_{1,1} + ... + \alpha_n \lambda_{n,1}) \textbf{s}_1 + (\alpha_n \lambda_{1,j} + ... + \alpha_n \lambda_{n,j}) \textbf{s}_j$$ By using associativity and commutativity of vector addition, then the scalar distributive law.

We have written $\textbf{y}$ as a linear combination of vectors in $S_2$. Hence, $\textbf{y} \in \text{span}(S_2)$.

Hence, the result.

Answer 1

Overall, I think the proof looks good. If you don't assume that $S_1$ and $S_2$ then your proof can be adapted.

One small suggestion: when you say that you can write $\textbf{y} = \alpha_1 \textbf{v}_1 + ... + \alpha_n \textbf{v}_n$, Say that there exists $\alpha_1,\dots,\alpha_n\in\Bbb{F}$ such that you can do this. Similar for the $\lambda_{i,j}$ (Here, you put that they are in $\Bbb{F}$, but say that they exist).