# 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.

• You can write subscripts by typing \$w_1\$. Also, your solution is correct. But may be it is better to make the induction explicit and not just state it since that is essentially the main part of the proof. Commented Sep 30, 2013 at 0:56
• Induction is the way to go. Commented Sep 30, 2013 at 1:06
• While you feel this is not enough in my opinion it's almost too much. The definition of a subspace implies closure under linear combinations. It's nearly trivial. Commented May 28, 2022 at 23:01

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.

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.