prove that if B=(P,P,...,P) is any subset of a linear space L, then M=Span B is a subspace of L.

We can proof this with the following theorem "Let B a subset of a vector space of L then the span(B) is a subspace of L , and is the samlles subspace containing B"

Proof

We can show that the span(B) is a subspace of L.

For this special case we assume that [tex] L \neq \emptyset[/tex] and for this case the [tex] span(B) \neq \emptyset[/tex] and we can see that B \subseteq M=span (B) [/tex]

In order to see that the span(B) is a subspace of L we can assume two elements let's say [tex] r,s \in span(B)[/tex] we can write this like that:

[tex] r= a_1 r_1 + ...+a_n r_n [/tex]

[tex] s = b_1 s_1 + ...+ b_m s_m[/tex]

For some vectors [tex] r_1,....,r_n , s_1,....,s_m[/tex] \in B and scalars [tex] a_1,...,a_n , b_1,....,b_m[/tex] in R

So then the linear combination r+s is defined as:

[tex] r+s = a_1 r_1 +...+ a_n r_n + b_1 s_1 + ...+b_m s_m \in M=span(B)[/tex]

Let [tex]\alpha[/tex] a constant in R

[tex] \alpha r = \alpha a_1 r_1 + ....+ \alpha a_n r_s \in M=span(B)[/tex]

[tex] \alpha s = \alpha b_1 s_1 + ....+ \alpha b_m s_m \in M=span(B)[/tex]

So then we have all the conditions satisifed and we can conclude that M=span(B) is a subspace of L.