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

User Ente
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Answer:

See the proof below.

Explanation:

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
L \\eq \emptyset and for this case the
span(B) \\eq \emptyset 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
r,s \in span(B) we can write this like that:


r= a_1 r_1 + ...+a_n r_n


s = b_1 s_1 + ...+ b_m s_m

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

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


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

Let
\alpha a constant in R


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


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

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

User Hexfire
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