# Show that span {u, v, w}=span{u+v, u+w, v+w}! Let u,v and w denote vectors in a vector space V. We must show that the set of all linear combinations of u, v and w is equal to the set of all linear combinations of u+v, u+w and v+w.

Take a random element of span{u,v,w}, namely `c_1u+c_2v+c_3w` .  We must show that there exist constants `k_1, k_2, k_3` such...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

We must show that the set of all linear combinations of u, v and w is equal to the set of all linear combinations of u+v, u+w and v+w.

Take a random element of span{u,v,w}, namely `c_1u+c_2v+c_3w` .  We must show that there exist constants `k_1, k_2, k_3` such that `k_1(u+v)+k_2(u+w)+k_3(v+w) = c_1u+c_2v+c_3w` .

Notice that we may rewrite the left-hand side as

`k_1u+k_1v+k_2u+k_2w+k_3v+k_3w`

`=(k_1+k_2)u+(k_1+k+3)v+(k_2+k_3)w`

So our task is now equivalent to solving the system

`c_1=k_1+k_2`

`c_2=k_1+k_3`

`c_3=k_2+k_3`

Which gives us

`k_1=1/2(c_1+c_2-c_3)`

`k_2=1/2(c_1-c_2+c_3)`

`k_3=1/2(-c_1+c_2+c_3)`

Which was to be shown.  The proves that span{u,v,w} is a subset of span{u+v, u+w, v+w}, and a similar argument will show that the latter is a subset of the former, proving the sets congruent.

Approved by eNotes Editorial Team