Let A1, A2, …, Ar be non-zero vectors such that Ai.Aj = o if .
Let c1, c2, …., cr be numbers such that c1A1+ c2A2+…..+ crAr = 0.
Show that ci = 0 for all i = 1,2, 3, …, r.
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.
To prove that ci = 0 for all i = 1, 2, 3, ..., r, given that c1A1 + c2A2 + ... + crAr = 0 and Ai.Aj = 0 if i ≠ j, we can use the dot product properties and the assumption that the vectors are non-zero.
Let's assume, for the sake of contradiction, that there exists at least one non-zero ci. Without loss of generality, let's say c1 ≠ 0.
Since c1A1 + c2A2 + ... + crAr = 0, we can express c1A1 as - (c2A2 + c3A3 + ... + crAr). Multiplying both sides of this equation by A1, we get:
(c1A1)·A1 = - (c2A2 + c3A3 + ... + crAr)·A1
c1(A1·A1) = - c2(A2·A1) - c3(A3·A1) - ... - cr(Ar·A1)
Now, we know that A1·A1 = 0 (given in the problem statement) because Ai·Aj = 0 if i ≠ j. Therefore:
c1(0) = - c2(A2·A1) - c3(A3·A1) - ... - cr(Ar·A1)
0 = - c2(A2·A1) - c3(A3·A1) - ... - cr(Ar·A1)
Since all the dot products on the right side are equal to zero, we can rewrite the equation as:
0 = 0 - 0 - ... - 0
0 = 0
This contradicts our assumption that c1 ≠ 0. Therefore, our initial assumption that there exists a non-zero ci must be false.
Hence, we can conclude that ci = 0 for all i = 1, 2, 3, ..., r.