Let $u, v, w$ be real numbers. Prove that $span (u, v, w) = span (u + v + w, w – u, 2w)$
Hey guys. I've been struggling with this question for a while now.
I believe I'm supposed to use linear independence to prove this, however I don't believe any of my solutions have made sense.
Things I've tried have included proving linear independence of the span on the right-hand side, followed by setting the linear combination ($a(u) + b(v) + c(w) = x(u+v+w) + y(w – u) + z(2w)$ equal to one another. However any progress I made with this made no sense.
Any help would be greatly appreciated. Thank you!
Best Answer
Clearly the latter is contained in the former, since it is a linear combination of elements of the former. For containment the other way, we can essentially invert the transformation: $$ \begin{align} u &= -(w-u)-\tfrac{1}{2}(2w), \\ v &= (u+v+w)+(w-u)-(2w), \\ w &= \tfrac{1}{2}(2w). \end{align} $$ So any vector in the former can be written as a linear combination of vectors in the latter.
Since we have containment in both directions, we must have equality.