TITLE: If $\{u,v,w\}$ is a basis for $V$, then $\{u+v+w,v+w,w\}$ is also a basis: is this proof correct? 
QUESTION [8 upvotes]: 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 $V$.

REPLY [4 votes]: It is easily seen that each of your original basis elements are in the span of $\{u+v+w,v+w,w\}$. Since $\{u, v, w\}$ spans $V$, so does $\{u+v+w,v+w,w\}$ (if $A\subseteq {\rm span} \,(V)$, then ${\rm span}(A)\subseteq{\rm span }(V) \thinspace $).
So, we have a set of three vectors that span a space of dimension three. It follows that the set is independent and thus  a basis.