If $v_1,\ldots,v_m$ and $w_1,\ldots,w_m$ are linearly independent lists of vectors in V, then is $v_1+w_1,\ldots,v_m+w_m$ linearly independent?
For the proof of this (if it is true) I am considering $$c_1(v_1+w_1)+\cdots+c_m(v_m+w_m)=0$$ Collecting like terms I have: $$(c_1v_1+\cdots+c_mv_m)+(c_1w_1+\cdots+c_mw_m)=0$$
I think that since $v_1,\ldots,v_m$ and $w_1,\ldots,w_m$ are linearly independent, then each group above is only zero when the $c_i$ are each zero, by definition. However, is it possible for one group above to be the additive inverse of the other group, thus allowing us to get zero without all of the $c_i$ having to be zero?
You can always try small examples, and you will see there are plenty of ways, even avoiding the obvious ones $($like Weijun's suggested $w_n=-v_n)$, for it to fail.
For instance, let $n=2$. Regardless of the values of $v_1$ and $v_2$, if $w_1\neq 0$ is not a multiple of $v_1-v_2$, you make take $w_2=v_1+w_1-v_2$ and the result will fail.