[proofplan] We prove equality by proving membership equivalence for an arbitrary vector $y \in \mathbb{R}^m$. The definition of the polar of $TK$ turns the condition $y \in (TK)^\circ$ into an inequality over all vectors of the form $Tx$ with $x \in K$. The transpose identity $(Tx)\cdot y = x \cdot (T^\top y)$ then converts that inequality exactly into the defining condition for $T^\top y \in K^\circ$. [/proofplan] [step:Rewrite membership in the polar of the image using the definition of $TK$] Let $y \in \mathbb{R}^m$ be arbitrary. By definition of the polar in $\mathbb{R}^m$, \begin{align*} y \in (TK)^\circ \end{align*} holds exactly when \begin{align*} w \cdot y \leq 1 \quad \text{for every } w \in TK. \end{align*} Since $TK = \{Tx : x \in K\}$, this is equivalent to \begin{align*} (Tx)\cdot y \leq 1 \quad \text{for every } x \in K. \end{align*} [/step] [step:Use the transpose identity to move $T$ onto $y$] The transpose map $T^\top:\mathbb{R}^m \to \mathbb{R}^n$ is characterized by the Euclidean adjoint identity \begin{align*} (Tx)\cdot y = x \cdot (T^\top y) \end{align*} for every $x \in \mathbb{R}^n$ and every $y \in \mathbb{R}^m$. Therefore \begin{align*} (Tx)\cdot y \leq 1 \quad \text{for every } x \in K \end{align*} is equivalent to \begin{align*} x \cdot (T^\top y) \leq 1 \quad \text{for every } x \in K. \end{align*} [/step] [step:Identify the resulting condition as membership in $K^\circ$] By the definition of $K^\circ \subset \mathbb{R}^n$, \begin{align*} x \cdot (T^\top y) \leq 1 \quad \text{for every } x \in K \end{align*} holds exactly when \begin{align*} T^\top y \in K^\circ. \end{align*} Combining the equivalences above, for every $y \in \mathbb{R}^m$, \begin{align*} y \in (TK)^\circ \quad \Longleftrightarrow \quad T^\top y \in K^\circ. \end{align*} Hence \begin{align*} (TK)^\circ = \{y \in \mathbb{R}^m : T^\top y \in K^\circ\}, \end{align*} as claimed. [/step]