I've looked at the following solution but am unsure if it translates to this problem. Prove that, if $\{u,v,w\}$ is a basis for a vector space $V$, then so is $\{u+v, v+w, u+v+w\}$.
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Actually this statement is wrong without extra assumptions on the vector space or rather the underlying field. Take for example $u,v,w$ to be the unit vectors of $V=\Bbb F_2^3$. Then $$(u+v)+(u+w)+(v+w)=0$$ hence $u+v,u+w,v+w$ are linearly dependent and therefore cannot form a basis of $V$. If however the underlying field has characteristic $\ne2$, then the statement is true. – leoli1 Feb 04 '21 at 13:28
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It is true. Subtract one of the three from another, and add it to the third one. You will get the original set.
user2277550
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That problem is very similar to this one- if you read the proof, you can see that it uses linear independence of the first basis to prove linear independence of the other set, and the fact that the first basis spans $V$ to show that the second set does. You can just mimic the style of proof in the linked question to solve this one. To start you off, let $x$ be a vector in $V$. Then for some scalars $a,b,c, au+bv+cw=x$ since $\{u,v,w\}$ spans $V$. Can you find some scalars $a',b',c'$ such that $a'(u+v)+b'(v+w)+c'(u+w)=x$?
Kevin Long
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