# Linear independence in a vector space question

I'm working out the following question:

If $\{u_1, u_2, u_3\}$ is a linearly independent set in some vector space. Explain why if $a_1u_1 + a_2u_2 + a_3u_3 = b_1u_1 + b_2u_2 + b_3u_3$, where $(a_1,a_2,a_3)$ and $(b_1, b_2, b_3)$ are scalars, then $a_1 = b_1, a_2 = b_2, a_3 = b_3$.

I think that if the set $\{u_1,u_2,u_3\}$ is linearly independent, then it has no nontrivial linear combination that maps it to the zero vector. It only has the trivial zero vector solution with $0$ as coefficients.

So, $a_1u_1 + a_2u_2 + a_3u_3$ must be $0\cdot u_1 + 0\cdot u_2 + 0\cdot u_3$.Because of linear independence, there are no other solutions but the zero vector. Thus, $(b_1,b_2,b_3)$ must equal $(a_1,a_2,a_3)$.

Do you think I am missing something in my explanation?

$$a_1u_1 + a_2u_2 + a_3u_3 = b_1u_1 + b_2u_2 + b_3u_3$$
$$(a_1-b_1)u_1 + (a_2-b_2)u_2 + (a_3-b_3)u_3 = 0.$$
If $a_i \ne b_i$ for some $i$, then this would contradict the linear independence.