No they won’t be.Consider the linear combination (1)(u – v) + (1) (v – w) + (-1)(u – w).This will add to 0. But the coefficients aren’t all 0.Therefore, those vectors aren’t linearly independent. You can try an example of this with (1, 0, 0), (0, 1, 0), and (0, 0, 1), the usual basis vectors of R3. That method relied on spotting the solution immediately.If you couldn’t see that, then there’s another approach to the problem. We know that u, v, w are linearly independent vectors.So if au + bv + cw = 0, then a, b, and c are all 0 by definition. Suppose we wanted to ask whether u – v, v – w, and u – w are linearly independent.Then we’d like to see if there are non-zero coefficients in the linear combinationd(u – v) + e(v – w) + f(u – w) = 0, where d, e, and f are scalars. Distributing, we get du – dv + ev – ew + fu – fw = 0.Then regrouping by vector: (d + f)u + (-d +e)v + (-e – f)w = 0. But now we have a linear combo of u, v, and w vectors.Therefore, all the coefficients must be 0.So d + f = 0, -d + e = 0, and –e – f = 0. It turns out that there’s a free variable in this solution.Say you let d be the free variable.Then we see f = -d and e = d. Then any solution of the form (d, e, f) = (d, d, -d) will make (d + f)u + (-d +e)v + (-e – f)w = 0 a true statement. Let d = 1 and you get our original solution. You can let d = 2, 3, or anything if you want.
