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 $c_i=0$.

Question

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....

Answer 1

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.