Let $u,v,w \in V$ a vector space over a field F such that $u \neq v \neq w$. If $\{ u , v , w \}$ is a basis for $V$, then prove that $\{ u+v+w , v+w , w \}$ is also a basis for $V$.
Proof:
Let $u,v,w \in V$ a vector space over a field $F$ such that $u \neq v \neq w$. Let $\{ u , v , w \}$ be a basis for $V$. Because $\{ u , v , w \}$ is a basis, then $u,v,w$ are linearly independent and $ \langle \{ u , v , w \} \rangle = V$.
Let $x \in V$ be an arbitrary vector then $x$ can be uniquely expressed as a linear combination of $\{ u , v , w \}$. Let's suppose $x=au+bv+cw$ for some $a,b,c \in F$.
On the other hand, let us consider $\{ u+v+w , v+w , w \} \subseteq V$.
Then $ \begin{align*} \langle \{ u+v+w , v+w , w \} \rangle &= \{d(u+v+w) + e(v+w) + f(w) \mid d,e,f \in F\} \\ &= \{du + (d+e)v +(d+e+f)w \mid d,e,f \in F \} . \end{align*}$
If $x \in V$, then $x=du + (d+e)v +(d+e+f)w$ is another unique representation of $x \in V$ . Then for any arbitrary $x \in V$, we have $d=a$, $d+e=b $and $d+e+f=c \in F$.
Because $\{ u , v , w \}$ is a basis for $V$, then $\{ u+v+w , v+w , w \}$ must also be a basis for $V$.