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This question already has an answer here:

Prove that if $\{u_1, u_2, u_3\}$ is an orthogonal set of nonzero vectors in $\mathbb{R}^n$ and we have scalars $c_1, c_2, c_3$ such that $c_1u_1+ c_2u_2+c_3u_3 = 0$, then each of the scalars is equal to zero.

I'm having trouble finding a way to prove that the scalars must be zero.

I started by showing that for the set to be orthogonal you must have

$u_1 \cdot u_2 = 0$

$u_1 \cdot u_3 = 0$

$u_2 \cdot u_3 = 0$

and that the above stated condition also says while $u1, u2, u3 \not = 0$.

then I wrote the hypothesis and expanded it:

if $c_1u_1 + c_2u_2 + c_3u_3 = 0$, then

$c_1(u_{11}, u_{12}, \ldots ,u_{1n}) + c_2(u_{21}, u_{22}, \ldots, u_{2n}) + c_3(u_{31}, u_{32}, \ldots , u_{3n}) = (0, 0, \ldots , 0)$

I can show that each of scalars being $= 0$ would give me the result, but I'm stuck finding out how to show its the only way. I've looked everywhere for help. help?

edit:

also, if I expanded each $c_1u_{1n}$ vector, then set the sum of each first element to $0$, and solve for $c_1,c_2,c_3$ simultaniously by plugging one into another, would it come out to an end result yielding $0$?

and since we haven't been taught linear independence yet in this class I believe I am supposed to write the proof without referencing it. so it's not a duplicate....

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marked as duplicate by PhoemueX, user147263, Joonas Ilmavirta, Eric Stucky, Claude Leibovici Oct 27 '14 at 8:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Just to check (I didn't want to change it cause I wasn't sure) when you wrote " $u_1,u_2,u_3=/=0.$" did you mean that none of the vectors are zero? $\endgroup$ – user171177 Oct 27 '14 at 2:15
  • $\begingroup$ yes. __________ $\endgroup$ – J L Oct 27 '14 at 2:21
  • $\begingroup$ Try a proof by contradiction. $\endgroup$ – dustin Oct 27 '14 at 2:22
  • $\begingroup$ @dustin I'm not sure how to apply that to this question? also, if I expanded each c1u1n vector, then set the sum of each first element to 0, and solve for c1,c2,c3 simultaniously by plugging one into another, would it come out to an end result yielding 0? $\endgroup$ – J L Oct 27 '14 at 2:45
  • $\begingroup$ @JL have you looked at the question (and answer) Phoemuex linked to? If so what part of the answer don't you understand? $\endgroup$ – user171177 Oct 27 '14 at 2:47
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u1 . (c1u1 + c2u2 + c3u3) = u1 . 0 => c1 * (u1 . u1) + c2 * 0 + c3 * 0 = 0. Since u1 is non-zero, (u1 . u1) is positive. Therefore, c1 must be 0. By symmetry, c2 and c3 must be 0. Hence c1, c2, and c3 are 0.

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