Question 455735
Let {{{w = v[1] + v[2]}}} +...+{{{v[k]}}}.

Now let

{{{a[1]*(v[1] + w) + a[2]*(v[2]  + w)}}}+...+{{{a[k]*(v[k] + w)}}} = {{{theta}}}, 

which for the purpose of contradiction, we assume that not all the a's are 0, (i.e., the set is linearly DEPENDENT).


==> ({{{2a[1] + a[2]}}}+...+{{{a[k]}}})*{{{v[1]}}}
+({{{a[1] + 2a[2]}}}+...+{{{a[k]}}})*{{{v[2]}}} + ...+({{{a[1] + a[2]}}}+...+{{{2a[k]}}})*{{{v[k]}}} = {{{theta}}}

By the hypothesis, {v1, v2, v3,..., vk } is linearly independent, and thus,
{{{2a[1] + a[2]}}}+...+{{{a[k] = 0}}}
{{{a[1] + 2a[2]}}}+...+{{{a[k] = 0}}}
...................................

{{{a[1] + a[2]}}}+...+{{{2a[k] = 0}}}.

We get a homogeneous system of equations.

Adding all corresponding sides of the system, we get

{{{(k+1)a[1] + (k+1)a[2]}}}+...+{{{(k+1)a[k] = 0}}}
OR,
{{{a[1] + a[2]}}}+...+{{{a[k] = 0}}}.

Subtracting this equation from each one of the equations in the system above, we obtain
{{{a[1] = a[2] = a[3]}}}=...={{{a[k] = 0}}},

CONTRARY to the initial assumption that not all a's are 0.

Hence {v1+w,v2+w,...,vk+w} must be linearly independent.