Am I sub-indexing this linear algebra proof correctly?

Question

Proposition

Let $ S \subset V \text{ and } w \in V$

$\textbf{span}(S \cup \{w\}) = \textbf{span}(S) \iff w \in \textbf{span}(S)$

Proof

$\impliedby$

Assume $w \in \textbf{span}(S)$.

We will check that if $v \in \textbf{span}(S \cup \{w\}) \text{ then } v \in \textbf{span}(S)$.

$v \in \textbf{span}(S \cup \{w\}) \implies \exists u_1,...,u_k \in S \text{ and } \exists a_1,...,a_k,b \in F: v = a_1u_1+...+a_ku_k+bw$

$w \in \textbf{span}(S) \implies w \text{ is a linear combination of vectors} \in S \implies$

$\exists u_1,...,u_k+u_{k+1},...,u_m \in S \text{ and } b_1,...,b_k,b_{k+1},...,b_m \in F:$

$w = b_1u_1+..+b_ku_k+b_{k+1}u_{k+1}+...+b_mu_m$

$v = (a_1+bb_1)u_1+...+(a_k+bb_k)u_k+(bb_{k+1})u_{k+1}+...+bb_mu_m$

Thus, $V \in \textbf{span}(S)$

$\implies$

Assume $\textbf{span}(S \cup \{w\}) = \textbf{span}(S) \implies w \in \textbf{span}(S \cup \{ w \}) \subset \textbf{span}(S)$

Q.E.D.

Answer 1

Here it is an alternative approach to the case in which $S$ finite.

To begin with, let us reinforce the definition of a vector space spanned by a set. Given a finite dimensional vector space $V$ over the field $\textbf{F}$ and a set $S\subseteq V$, the vector space spanned by $S$ is the intersection of all subspaces from $V$ which contains $S$. That is the same as the set of all linear combinations of its elements. Having said that, we may proceed.

Let $S = \{s_{1},s_{2},\ldots,s_{n}\}$. If $\text{span}(S\cup\{w\}) = \text{span}(S)$, it means that any linear combination of the elements from the set $S\cup\{w\} = \{s_{1},s_{2},\ldots,s_{n},w\}$ also belongs to the set $\text{span}(S)$. In other words, we have that \begin{align*} & a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s_{n} + w = b_{1}s_{2} + b_{2}s_{2} + \ldots + b_{n}s_{n} \Longrightarrow\\\\ & w = (b_{1}-a_{1})s_{1} + (b_{2}-a_{2})s_{2} + \ldots + (b_{n}-a_{n})s_{n} \end{align*} from whence it results that $w\in\text{span}(S)$.

Conversely, the inclusion $\text{span}(S)\subseteq\text{span}(S\cup\{w\})$ is trivial. Indeed, if $s\in\text{span}(S)$, one has \begin{align*} s = c_{1}s_{1} + c_{2}s_{2} + \ldots + c_{n}s_{n} = c_{1}s_{1} + c_{2}s_{2} + \ldots + c_{n}s_{n} + 0w \Longrightarrow s\in\text{span}(S\cup\{w\}) \end{align*}

Let us now prove that $\text{span}(S\cup\{w\}) \subseteq \text{span}(S)$ based on the assumption $w\in\text{span}(S)$. To begin with, we should notice that \begin{align*} w = a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s_{n} \end{align*}

Consequently, if $s\in\text{span}(S\cup\{w\})$, we have that \begin{align*} s & = b_{1}s_{1} + b_{2}s_{2} + \ldots + b_{n}s_{n} + b_{n+1}w\\\\ & = b_{1}s_{1} + b_{2}s_{2} + \ldots + b_{n}s_{n} + b_{n+1}(a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s_{n})\\\\ & = (b_{1} + b_{n+1}a_{1})s_{1} + (b_{2} + b_{n+1}a_{2})s_{2} + \ldots + (b_{n} + b_{n+1}a_{n})s_{n} \Longrightarrow s\in\text{span}(S) \end{align*} and the result holds.

Answer 2

All ok for me. You probably noticed that the proof essentially boils down to the fact that $A\subseteq span(B)\iff span(A)\subseteq span(B)$.

Here is an alternative: \begin{align*} span(S\cup\{w\})= span(S)\iff&span(S\cup\{w\})\subseteq span(S)\\ &\text{(because the reverse inclusion is always valid)}\\ \iff&S\cup\{w\}\subseteq span(S)\\ &\text{(by the fact above)}\\ \iff&\left\{w\right\}\subseteq span(S)\\ &\text{(because $S\subseteq span(S)$ always holds)}\\ \iff&w\in span(S)\\ &\text{(basic set theory)} \end{align*}