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$V$ is $K$-Vector Space and $0

$$\mathscr{L}((v_1,...,v_n)):=\{x|\exists (\alpha_1,...,\alpha_n)\in K^n:x=\sum_{i\in J_m:=\{1,...,n\}}\alpha_i\, v_i\}$$$$:=\operatorname{Span}(T):=\{x|\exists (w_1,...,w_n)\in T^n: x \in \mathscr{L}((w_1,...,w_n))\}$$

Is it correct?

  • 2
    The second set probably needs a $\exists n$ in it if you want to write it in this way. – 2017-01-23
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    $$\mathscr{L}((v_i)_{i \in J_n}):=\{x|\exists \alpha \in K^{J_n}:x=\sum_{i\in J_n}\alpha_i\, v_i\}$$ and $$:=\{x|\exists 0 – 2017-01-23
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    if $T=\emptyset$, $=?$.. – 2017-02-05

3 Answers 3

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This looks correct, though a little overboard, to me. There are more succinct ways to write this, namely "the span of $T$ is the set of all vectors that are linear combinations of elements of $T$"

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Another way to put it, which I prefer because it can easily be generalized, is the following : since the set of subspaces of $V$ is closed under arbitrary intersections, then there is a least (w.r.t. inclusion) subspace containing $T$. We let $$ or $Span(T)$ denote this subspace.

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Let be $m \in \Bbb{N}=\big \{1,2,3,...,n,n+1,...\big \}$, we can define $$J_m:=\big \{x \in \Bbb{N}\big | x \leq m \big \}=\big \{1,2,3,...,m \big \}$$ Let be $T \subseteq V$ and $V$ a $K$-vector Space, we can define $$\operatorname{Span}\big (T \big):=\mathscr{L}\big (T\big ):=$$$$:= \begin{cases}\big \{0_V \big \} & \text{ if } \,\,\,T=\emptyset \\ \bigg \{z\big| \exists m\in \Bbb{N}:\bigg(\exists v \in V^{J_m}, \exists\alpha \in K^{J_m}:\bigg( z= \displaystyle\sum_{i=1}^m \alpha_i\, v_i\bigg)\bigg)\bigg\} & \text{ otherwise}\end{cases}$$ $$\big \langle T \big \rangle := \bigcap \big \{x\subseteq V| T \subseteq x \wedge x \text{ is subspace von } V\big \} $$ Now you can prove that $\big \langle T \big \rangle=\operatorname{Span}\big (T \big ) $