How do I prove that if $A$ is a balanced subset of a vector space $E$ then $\lambda A = |\lambda| A$?

Question

I have to somehow use $\left|\frac{\lambda}{|\lambda|}\right|=1=\left|\frac{|\lambda|}{\lambda}\right|$, I don't know how, however.

Answer 1

All vectors in $\lambda A$ are of the form $\lambda u$ for some $u \in A$, and all vectors of that form are elements of $\lambda A$ (that's the definition of $\lambda A$). Same thing for $|\lambda |A$. You need to show that it is also the case that $|\lambda| u \in \lambda A$ and that $\lambda u \in |\lambda |A$.

I will prove $\lambda u \in |\lambda|A$, the other way is very similar. Also, assume that $\lambda \neq 0$, since the special case $\lambda = 0$ is trivial. We note that $$ \lambda u = \lambda\left(\frac{|\lambda|}{|\lambda|}u\right) = |\lambda|\left(\frac{\lambda}{|\lambda|}u\right) $$ and this is the part where we use that $\left|\frac{\lambda}{|\lambda|}\right| = 1$: Since $A$ is balanced, and $u \in A$, that means that $\frac{\lambda}{|\lambda|}u \in A$. We have therefore written $\lambda u$ on the form $|\lambda| v$ for the vector $v = \frac{\lambda}{|\lambda|}u$, which means that $\lambda u \in |\lambda|A$.