Theorem

The span of a list of vectors in V is the smallest subspace of V containing all vectors in the list.

Proof

Suppose ${latex.inlinev_{1}, ..., v_{m}} is a list of vectors in V. First we show that span(${latex.inlinev_{1}, ..., v_{m}}) is a subspace of V by using 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.inlinew_{1}, w_{2} \in span}. Then by definition we have ${latex.inlinew_{1} = a_{1}v_{1} + ... + a_{m}v_{m}} and ${latex.inlinew_{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.inlinev_{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.