Suppose $T: R^3 \to R^3$ is an $R$-linear transformation and $u$, $v$, $w$ are unit vectors in $R^3$ such that $T(u) + u = T(v) + 2v = T(w) + 3w = 0$. Determine the vectors $(a,b,c) \in R^3$ such that $au + bv + cw = 0$.
I am confused about what the question is asking me. Do I need to find an ordered triple $(a,b,c) \in R^3$ and then make sure it satisfies $au + bv + cw = 0$? I think the trivial solution $(0,0,0)$ works because $0u + 0v + 0w = 0$. How can I find a nontrivial solution? I am also not sure how to use the information that $u$, $v$, and $w$ are unit vectors.
Hint. The relevant consequence of the fact that the vectors are unit is that they are non-zero.
Since they are eigenvectors of $T$ associated with three different eigenvalues, and since they are non-zero, they must form a basis of $\Bbb R^3$ (for you to prove).
Therefore $au+bv+cw=0$ implies $a=b=c=0$.