The span of a list of vectors in V is the smallest subspace of V containing all vectors in the list.
Suppose \({latex.inline[v_{1}, ..., v_{m}](v_{1}, ..., v_{m})} is a list of vectors in V. First we show that span(\){latex.inlinev{1}, ..., v{m}}) is a subspace of V by using 1753142531 - Axler 1.34 Conditions for a subspace|1.34.
0 is in span This is obvious because ${latex.inline0 = 0v{1} + ... + 0v{m}}.
0 is closed under addition Suppose we have two vectors in the span: \({latex.inline[w_{1}, w_{2} \in span](w_{1}, w_{2} \in span)}. Then by definition we have \){latex.inlinew{1} = a{1}v{1} + ... + a{m}v_{m}} and \({latex.inline[w_{2} = b_{1}v_{1} + ... + b_{m}v_{m}](w_{2} = b_{1}v_{1} + ... + b_{m}v_{m})}. When we add those two together we get \){latex.inline(w{1} + w{2}) = (a{1} + b{1})v{1} + ... + (a{m} + b{m})v{m}}. But each a, b are in V so a + b is in V and we get what we need.
0 is closed under scalar addition Same proof as the addition one essentially
Each \({latex.inline[v_{k}](v_{k})} is contained within the span since we can set \){latex.inlinea_{k} = 1} and all other a’s = 0. Conversely, because subspaces are closed under addition and scalar multiplication, every subspace of V that contains each ${latex.inlinev_{k}} contains the span. Thus we get the required result.