Preimage of Dilation of Set under Linear Transformation is Dilation of Preimage

Theorem
Let $K$ be a field.

Let $X$ and $Y$ be vector spaces over $K$.

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

Let $E \subseteq X$ be a non-empty set.

Let $\lambda \in K$.

Then:
 * $T^{-1} \sqbrk {\lambda E} = \lambda T^{-1} \sqbrk E$

where $\lambda E$ denotes the dilation of $E$ by $\lambda$.

Proof
The result is immediate when $\lambda = 0_K$, since $T^{-1} \sqbrk {\set { {\mathbf 0}_X} } = \set { {\mathbf 0}_X}$.

Now take $\lambda \ne 0_K$.

Let $x \in X$.

We have:
 * $x \in T^{-1} \sqbrk {\lambda E}$


 * $T x \in \lambda E$
 * $T x \in \lambda E$

From linearity, this is the case :
 * $\map T {\lambda^{-1} x} \in E$

This is equivalent to:
 * $\lambda^{-1} x \in T^{-1} \sqbrk E$

Which in turn is equivalent to:
 * $x \in \lambda T^{-1} \sqbrk E$

So we obtain:
 * $T^{-1} \sqbrk {\lambda E} = \lambda T^{-1} \sqbrk E$