Let A1, A2, …, Ar be non-zero vectors such that Ai.Aj = o if .

Question

Let c1, c2, …., cr be numbers such that c1A1+ c2A2+…..+ crAr = 0.
Show that ci = 0 for all i = 1,2, 3, …, r.

Answer 1

To prove that ci = 0 for all i = 1,2,3,...,r, we can use a proof by contradiction.

Assume that there exists some ci ≠ 0 for some i (let's say ci ≠ 0 for i = k).

Then we can rearrange the equation c1A1 + c2A2 + ... + ckAk + ... + crAr = 0 to isolate Ak by subtracting the terms involving other vectors:

Ak = -(c1A1 + c2A2 + ... + ck-1Ak-1 + ck+1Ak+1 + ... + crAr) / ck

Now, we can take the dot product of both sides with Ai for any i ≠ k:

Ai . Ak = Ai . (-(c1A1 + c2A2 + ... + ck-1Ak-1 + ck+1Ak+1 + ... + crAr) / ck)

Since we know that Ai . Aj = 0 for all i ≠ j (given in the problem statement), the right-hand side of the equation becomes 0:

Ai . Ak = 0

However, we also know that Ai . Ak ≠ 0, since Ai ≠ 0 and Ak ≠ 0. This leads to a contradiction, as we have arrived at the conclusion that Ai . Ak = 0 and Ai . Ak ≠ 0 simultaneously.

Therefore, our assumption that ci ≠ 0 for some i (k in this case) is false.

Hence, we can conclude that ci = 0 for all i = 1,2,3,...,r.