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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Pratyush Sarkar Sep 30 '13 at 0:56
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    $\begingroup$ Induction is the way to go. $\endgroup$ – ncmathsadist Sep 30 '13 at 1:06

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.


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