Prove that if W is a subspace of a vector space V Prove that if $W$ is a subspace of a vector space $V$ and $w_1, w_2, ..., w_n$ are in $W$, then $a_1w_1 + a_2w_2 + ... + a_nw_n \in W$ for any scalars $a_1, a_2, ..., a_n$. 
My solution is we have $a_iw_i \in W$ for all $i$. And we can get the conclusion that $a_1w_1, a_1w_1 + a_2w_2, a_1w_1 + a_2w_2 + a_3w_3$ are in $W$ inductively. 
Any ideas on how to improve this because I feel it is not enough.
Thank you in advance. 
 A: Your solution is correct. But let's try to make the induction more explicit: The base case is obvious, as if $w_1$ is in $W$, then so is $a_1w_1$, as vector spaces are closed under scalar multiplication. Now assume that $a_1w_1+\cdots+a_nw_n\in W$. Since $w_{n+1}\in W$ it follows that $a_{n+1}w_{n+1}$ is in $W$. Hence from the assumption, one has that $a_1w_1+\cdots+a_nw_n+a_{n+1}w_{n+1}$ is in $W$, which finishes the proof.
A: Since $W$ is a subspace of vector space $V$, then $W$ is closed under addition and scalar multiplication by the definition of subspace.
Now we can inductively prove the statement.
Base:
Let $w_1 \in W$ and let scalar $a_1 \in \digamma$ where $\digamma$ is the field over which $V$ is defined.
We have $a_1w_1 \in W$ because subspace $W$ is closed under scalar multiplication.
Inductive Step:
Let $w_1, w_2, ... , w_n \in W$ and $a_1, a_2, ... , a_n \in \digamma$
Assume $a_1w_1 + a_2w_2 + ... + a_nw_n \in W$ is true.
Let $w_{n+1} \in W$ and let scalar $a_{n+1} \in \digamma$.
We can show that $a_1w_1 + a_2w_2 + ... + a_nw_n + a_{n+1}w_{n+1} \in W$ by closure of addition of subspace $W$.
Therefore, we have established the inductive steps to prove the statement.
