Properties of Balanced Sets of Vectors

Properties of Balanced Sets of Vectors

Proposition 1: Let $E$ be a vector space. If $A \subseteq E$ is a balanced set then for all $\lambda \in \mathbf{F}$, $\lambda A = |\lambda| A$.
  • Proof: Clearly, if $\lambda = 0$ then the conclusion follows immediately.
  • Suppose that $\lambda \neq 0$. Then observe that $\displaystyle{\frac{\lambda}{|\lambda|}, \frac{|\lambda|}{\lambda} \in \mathbf{F}}$ are such that $\displaystyle{\left | \frac{\lambda}{|\lambda|} \right | = 1}$ and $\displaystyle{\left | \frac{|\lambda|}{\lambda} \right | = 1}$. Since $A$ is balanced, we have that:
(1)
\begin{align} \quad \frac{\lambda}{|\lambda|} A \subseteq A \quad \mathrm{and} \quad \frac{|\lambda|}{\lambda} A \subseteq A \end{align}
  • Therefore:
(2)
\begin{align} \lambda A \subseteq |\lambda| A \quad \mathrm{and} \quad |\lambda| A \subseteq \lambda A \end{align}
  • So $\lambda A = |\lambda| A$. $\blacksquare$
Proposition 2: Let $E$ be a vector space. If $A \subseteq X$ is balanced and if $\lambda, \mu \in \mathbf{F}$ are such that $|\lambda| \leq |\mu|$ then $\lambda A \subseteq \mu A$.
  • The conclusion is trivially true when either $\lambda = 0$ or $\mu = 0$.
  • So suppose that $\lambda, \mu \neq 0$. Since $\mu A$ is balanced and since $\left | \frac{\lambda}{\mu} \right | \leq 1$, we have that:
(3)
\begin{align} \lambda A = \frac{\lambda}{\mu} (\mu A) \subseteq \mu A \quad \blacksquare \end{align}
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