Is the following a valid proof for the question in the title? Let A = Span{u, v, w}, B = Span{u+v, v+w, w+u}. Two sets are equal if and only if each element of A is in B and vice versa.
u = 1/2[(u+v) - (v+w) + (w+u)]
v = 1/2[(v+w) - (w+u) + (u+v)]
w = 1/2[(w+u) - (u+v) + (v+w)]
vectors in A is a linear combination of vectors in B, implies A subset B. Similarly, vectors in B is a linear combination of vectors in A, implies B subset A, therefore A = B.
Is this a valid proof? Thank you.
Yes, the proof is correct. If you want to flesh it out a bit more, your expressions for $u$, $v$ and $w$ show that $$\{u,v,w\} \subseteq {\rm span}\{u+v,v+w,w+u\}.$$Since the right hand side is a vector (sub)space itself, we get $${\rm span}\{u,v,w\} \subseteq {\rm span}\{u+v,v+w,w+u\},$$and this is the non-trivial inclusion. On the other hand, we clearly have $$\{u+v,v+w,w+u\} \subseteq {\rm span}\{u,v,w\},$$and the right side is a vector (sub)space, so it follows that $${\rm span}\{u+v,v+w,w+u\} \subseteq {\rm span}\{u,v,w\}.$$So the spans are equal.