Image of Absorbing Set under Surjective Linear Transformation is Absorbing
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Theorem
Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\GF$.
Let $A \subseteq X$ be an absorbing set.
Let $T : X \to Y$ be a surjective linear transformation.
Then $T \sqbrk A$ is an absorbing set.
Proof
Let $y \in Y$.
Since $T$ is surjective, there exists $x \in X$ such that $y = T x$.
Since $A$ is absorbing, there exists $t \in \R_{> 0}$ such that:
- $x \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
Then:
- $y = T x \in T \sqbrk {\alpha A}$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have:
- $y \in \alpha T \sqbrk A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
So $T \sqbrk A$ is absorbing.
$\blacksquare$