[proofplan] We show that $A$ carries ellipsoids contained in $K$ bijectively onto ellipsoids contained in $A(K)$. For ellipsoids, the affine map $A(x)=Lx+a$ multiplies $\mathcal{L}^n$-measure by the fixed factor $|\det L|$. Therefore any ellipsoid with larger volume inside $A(K)$ would pull back to an ellipsoid with larger volume inside $K$, contradicting the maximality of $E$. The uniqueness built into the definition of the John ellipsoid then identifies $A(E)$ as the John ellipsoid of $A(K)$. [/proofplan] [step:Write ellipsoids as affine images of the unit ball] Let \begin{align*} B(0,1) := \{x \in \mathbb{R}^n : |x| < 1\} \end{align*} denote the open Euclidean unit ball, and let \begin{align*} \alpha_n := \mathcal{L}^n(B(0,1)). \end{align*} An ellipsoid in $\mathbb{R}^n$ is a set of the form \begin{align*} F = c + T(B(0,1)), \end{align*} where $c \in \mathbb{R}^n$ and $T: \mathbb{R}^n \to \mathbb{R}^n$ is an invertible [linear map](/page/Linear%20Map). For such an ellipsoid, \begin{align*} \mathcal{L}^n(F) = |\det T| \, \alpha_n. \end{align*} This follows from the linear change of variables formula applied to the map $T$ and then translation by $c$, since translation preserves $\mathcal{L}^n$. [/step] [step:Show that the affine map gives a bijection between admissible ellipsoids] Define \begin{align*} \mathcal{E}(K) := \{F \subset K : F \text{ is an ellipsoid in } \mathbb{R}^n\} \end{align*} and \begin{align*} \mathcal{E}(A(K)) := \{G \subset A(K) : G \text{ is an ellipsoid in } \mathbb{R}^n\}. \end{align*} We claim that the map \begin{align*} \Phi: \mathcal{E}(K) &\to \mathcal{E}(A(K)) \\ F &\mapsto A(F) \end{align*} is a bijection. Indeed, if $F = c + T(B(0,1)) \in \mathcal{E}(K)$, then \begin{align*} A(F) &= \{Lx+a : x \in c + T(B(0,1))\} \\ &= Lc+a + (LT)(B(0,1)). \end{align*} Since $L$ and $T$ are invertible linear maps, $LT$ is invertible, so $A(F)$ is an ellipsoid. Also $F \subset K$ implies $A(F) \subset A(K)$, hence $A(F) \in \mathcal{E}(A(K))$. Conversely, let $G \in \mathcal{E}(A(K))$. Since $A$ is invertible, its inverse is the affine map \begin{align*} A^{-1}: \mathbb{R}^n &\to \mathbb{R}^n \\ y &\mapsto L^{-1}(y-a). \end{align*} The same computation applied to $A^{-1}$ shows that $A^{-1}(G)$ is an ellipsoid. Since $G \subset A(K)$, we have $A^{-1}(G) \subset K$. Therefore $A^{-1}(G) \in \mathcal{E}(K)$ and $G = A(A^{-1}(G))$. Thus $\Phi$ is surjective, and injectivity follows from the injectivity of $A$. [/step] [step:Compute the fixed volume scaling on admissible ellipsoids] Let $F \in \mathcal{E}(K)$, and write \begin{align*} F = c + T(B(0,1)), \end{align*} where $c \in \mathbb{R}^n$ and $T: \mathbb{R}^n \to \mathbb{R}^n$ is an invertible linear map. From the previous computation, \begin{align*} A(F) = Lc+a + (LT)(B(0,1)). \end{align*} Therefore \begin{align*} \mathcal{L}^n(A(F)) &= |\det(LT)|\,\alpha_n \\ &= |\det L|\,|\det T|\,\alpha_n \\ &= |\det L|\,\mathcal{L}^n(F). \end{align*} Thus $A$ multiplies the volume of every admissible ellipsoid by the same positive constant $|\det L|$. [guided] The point of this step is that the comparison of volumes is not distorted differently for different ellipsoids. Take an arbitrary admissible ellipsoid $F \in \mathcal{E}(K)$ and write it in the standard form \begin{align*} F = c + T(B(0,1)), \end{align*} with $c \in \mathbb{R}^n$ and $T: \mathbb{R}^n \to \mathbb{R}^n$ invertible. Applying $A(x)=Lx+a$ gives \begin{align*} A(F) &= \{L(c+Tu)+a : u \in B(0,1)\} \\ &= Lc+a + (LT)(B(0,1)). \end{align*} The volume of $F$ is $|\det T|\alpha_n$, and the volume of $A(F)$ is $|\det(LT)|\alpha_n$. Since determinants are multiplicative, \begin{align*} |\det(LT)| = |\det L|\,|\det T|. \end{align*} Hence \begin{align*} \mathcal{L}^n(A(F)) &= |\det(LT)|\,\alpha_n \\ &= |\det L|\,|\det T|\,\alpha_n \\ &= |\det L|\,\mathcal{L}^n(F). \end{align*} Because $L$ is invertible, $|\det L|>0$. Thus applying $A$ preserves the ordering of ellipsoid volumes: one ellipsoid has larger $\mathcal{L}^n$-measure than another exactly when its image under $A$ has larger $\mathcal{L}^n$-measure. [/guided] [/step] [step:Transfer maximality from $E$ to $A(E)$] Since $E$ is the John ellipsoid of $K$, we have $E \in \mathcal{E}(K)$ and, for every $F \in \mathcal{E}(K)$, \begin{align*} \mathcal{L}^n(F) \leq \mathcal{L}^n(E). \end{align*} Let $G \in \mathcal{E}(A(K))$. By the bijection above, there exists $F \in \mathcal{E}(K)$ such that $G=A(F)$. Using the fixed volume scaling, \begin{align*} \mathcal{L}^n(G) &= \mathcal{L}^n(A(F)) \\ &= |\det L|\,\mathcal{L}^n(F) \\ &\leq |\det L|\,\mathcal{L}^n(E) \\ &= \mathcal{L}^n(A(E)). \end{align*} Therefore $A(E)$ is a volume-maximizing ellipsoid contained in $A(K)$. [guided] We now prove maximality after applying $A$. Since $E$ is the John ellipsoid of $K$, it is an ellipsoid contained in $K$ and satisfies \begin{align*} \mathcal{L}^n(F) \leq \mathcal{L}^n(E) \end{align*} for every ellipsoid $F \subset K$. Now choose an arbitrary ellipsoid $G \subset A(K)$. The bijection from the previous step says that $G$ has the form $G=A(F)$ for a unique ellipsoid $F \subset K$, namely $F=A^{-1}(G)$. Since $F$ is admissible for $K$, maximality of $E$ gives \begin{align*} \mathcal{L}^n(F) \leq \mathcal{L}^n(E). \end{align*} Multiplying both sides by the positive number $|\det L|$ preserves the inequality: \begin{align*} |\det L|\,\mathcal{L}^n(F) \leq |\det L|\,\mathcal{L}^n(E). \end{align*} Using the volume scaling formula for ellipsoids under $A$, this becomes \begin{align*} \mathcal{L}^n(G) &= \mathcal{L}^n(A(F)) \\ &= |\det L|\,\mathcal{L}^n(F) \\ &\leq |\det L|\,\mathcal{L}^n(E) \\ &= \mathcal{L}^n(A(E)). \end{align*} Since $G$ was an arbitrary ellipsoid contained in $A(K)$, no admissible ellipsoid in $A(K)$ has larger volume than $A(E)$. [/guided] [/step] [step:Use uniqueness to identify the John ellipsoid of $A(K)$] The set $A(E)$ is an ellipsoid contained in $A(K)$ and has maximal $\mathcal{L}^n$-measure among all ellipsoids contained in $A(K)$. By the defining uniqueness of the John ellipsoid, $A(E)$ is the John ellipsoid of $A(K)$. This proves the affine equivariance of the John ellipsoid. [/step]