Image of Balanced Set under Linear Transformation is Balanced

Theorem
Let $\Bbb F \in \set {\R, \C}$.

Let $X$ and $Y$ be vector spaces over $\Bbb F$.

Let $E \subseteq X$ be balanced.

Let $T : X \to Y$ be a linear transformation.

Then $\map T E \subseteq Y$ is balanced.

Proof
We aim to show that for all $s \in \R$ with $\cmod s \le 1$, we have:


 * $s \map T E \subseteq \map T E$

Let $y \in s \map T E$.

Then there exists $x \in E$ such that $y = s T x$.

From the linearity of $T$, we have:


 * $y = \map T {s x}$

Since $x \in E$, we have $s x \in s E$ from the definition of dilation of $E$ by $s$.

Since $E$ is balanced, we have $s E \subseteq E$ and so:


 * $s x \in E$

giving:


 * $y = \map T {s x} \in \map T E$

Since $y \in s \map T E$ was arbitrary, we have $s \map T E \subseteq \map T E$, giving that $\map T E$ is balanced.