Prove that if $\{u_1, u_2, u_3\}$ is an orthogonal set of nonzero vectors in $\mathbb{R}^n$ and we have $c_1u_1+ c_2u_2+c_3u_3 = 0$, then $c_i=0$. [duplicate]

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....

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

• 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? – user171177 Oct 27 '14 at 2:15