Decrementing a span

Dependencies:

  1. Vector Space
  2. Linear independence
  3. Span

Let $S = \{v_1, v_2, \ldots, v_n\}$ be a subset of vector space $V$ over a field $F$. Let $S$ be linearly dependent. Without loss of generality, assume $v_n$ is a linear combination of the rest of the elements of $S$. Then $\operatorname{span}(S - \{v_n\}) = \operatorname{span}(S)$.

Proof

Any linear combination of $S - \{v_n\}$ is also a linear combination of $S$. Therefore, $\operatorname{span}(S - \{v_n\}) \subseteq \operatorname{span}(S)$.

Let $v_n = \sum_{i=1}^{n-1} b_iv_i$.

\begin{align} & \textrm{Let } u \in \operatorname{span}(S) \\ &\Rightarrow u = \sum_{i=1}^n a_iv_i \tag{for some $(a_1, a_2, \ldots, a_n) \in F^n$} \\ &\Rightarrow u = \sum_{i=1}^{n-1} a_iv_i + a_n\left(\sum_{i=1}^{n-1} b_iv_i \right) \\ &\Rightarrow u = \sum_{i=1}^{n-1} (a_i + a_nb_i)v_i \tag{using commutativity of addition and distributivity} \\ &\Rightarrow u \in \operatorname{span}(S - \{v_n\}) \end{align}

Therefore, $\operatorname{span}(S) \subseteq \operatorname{span}(S - \{v_n\})$.

Dependency for:

  1. span(A)=span(B) & |A|=|B| & A is linindep ⟹ B is linindep
  2. Minimally spanning iff basis
  3. Spanning set of size dim(V) is a basis

Info:

Transitive dependencies:

  1. Group
  2. Ring
  3. Field
  4. Vector Space
  5. Linear independence
  6. Span