1.3 Exercise 20
Prove that if \(W\) is a subspace of a vector space \(V\) and \(w_1,w_2,...,w_n\) in \(W\), then \(a_1w_1 + a_2w_2 + ... + a_nw_n \in W\) for any scalars \(a_1, a_2,...,a_n\).


Proof: By induction on \(n\):

Base Case (\(n=1\)): If \(w_1 \in W\), then \(a_1w_1 \in W\) since \(W\) is a subspace and so it is closed under scalar multiplication.

Inductive Case: Suppose this is true for \(k > 1\) so if \(w_1,...,w_k \in W\), then \(a_1w_1+...+a_kw_k \in W\) for any scalars \(a_1,...,a_k\). We will show that this is true for \(k+1\). So suppose that \(w_1,...,w_k, w_{k+1} \in W\), and let \(a_1,...,a_{k+1}\) be any scalars. We know that \(a_{k+1}w_{k+1} \in W\) since \(W\) is closed under scalar multiplication. We also know that \(a_1w_1+...+a_kw_k\) is in \(W\) by the inductive hypothesis. Furthermore, \(a_1w_1+...+a_kw_k + a_{k+1}w_{k+1}\) is also in \(W\) since \(W\) is closed under addition.

Therefore, the statmeent is true for all \(n \geq 1. \ \blacksquare\).


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