[proofplan] We encode infinitesimal ellipsoids inside $K$ by a translation vector and a symmetric linear perturbation of the identity. Maximality of the unit ball in [John position](/page/John%20Position) says that no admissible first-order perturbation can have positive trace, because positive trace increases volume to first order. The [finite-dimensional separating hyperplane theorem](/page/Separating%20Hyperplane%20Theorem) therefore forces $(I_n,0)$ to lie in the cone generated by the active contact constraints $(u\otimes u,u)$. Finally, the [conic Carathéodory theorem](/page/Carath%C3%A9odory%27s%20Theorem) gives finitely many contact points, and the [centrally symmetric](/page/Centrally%20Symmetric%20Set) case follows by suppressing translations and pairing opposite directions. [/proofplan]

[step:Describe the active supporting constraints of $K$]Let $S^{n-1}:=\partial B(0,1)$, and define the [support function](/page/Support%20Function) \begin{align*} h_K:S^{n-1} &\to \mathbb{R},\\ v &\mapsto \sup_{x\in K} v\cdot x. \end{align*} Since $B(0,1)\subset K$, we have $h_K(v)\geq 1$ for every $v\in S^{n-1}$. Define the active set \begin{align*} A:=\{v\in S^{n-1}:h_K(v)=1\}. \end{align*} For every $v\in A$, the half-space $\{x\in\mathbb{R}^n:v\cdot x\leq 1\}$ contains $K$, and the point $v\in S^{n-1}$ belongs to $K$ because $B(0,1)\subset K$. Hence $v\in \partial K\cap \partial B(0,1)$. Thus every active normal is a contact point.[/step]

[guided]The [support function](/page/Support%20Function) records all supporting half-spaces of the [convex body](/page/Convex%20Body). We declare \begin{align*} h_K:S^{n-1} &\to \mathbb{R},\\ v &\mapsto \sup_{x\in K} v\cdot x. \end{align*} Because the unit ball is contained in $K$, the supremum of $v\cdot x$ over $K$ is at least the supremum over $B(0,1)$, which is $1$ for $v\in S^{n-1}$. Hence $h_K(v)\geq 1$. The relevant directions are those where this lower bound is sharp. Define \begin{align*} A:=\{v\in S^{n-1}:h_K(v)=1\}. \end{align*} If $v\in A$, then by definition every $x\in K$ satisfies $v\cdot x\leq 1$, so $K$ is contained in the supporting half-space $\{x:v\cdot x\leq 1\}$. Since $v\in B(0,1)\subset K$ and $v\cdot v=1$, the point $v$ lies on that supporting hyperplane. Therefore $v\in \partial K\cap \partial B(0,1)$. These are precisely the contact points that will appear in the decomposition.[/guided]

[step:Convert failure of the decomposition into an improving infinitesimal ellipsoid]Let $\operatorname{Sym}_n$ denote the real [vector space](/page/Vector%20Space) of symmetric $n\times n$ matrices, equipped with the trace pairing $(M,N)\mapsto \operatorname{tr}(MN)$. Set \begin{align*} V:=\operatorname{Sym}_n\times \mathbb{R}^n. \end{align*} Define the constraint map \begin{align*} q:A&\to V,\\ u&\mapsto (u\otimes u,u). \end{align*} Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$. We claim that \begin{align*} (I_n,0)\in \operatorname{cone}\{q(u):u\in A\}. \end{align*} Suppose not. Since $A$ is compact and $q$ is continuous, $q(A)$ is compact. Moreover $0\notin q(A)$, because $u\in S^{n-1}$ implies $u\otimes u\neq 0$. Therefore the cone generated by $q(A)$ is closed in the finite-dimensional space $V$: indeed, the generators lie in the compact set $q(A)$ bounded away from $0$, so every convergent sequence of conic combinations has bounded total coefficient after normalising the generators by any fixed Euclidean norm on $V$. Since this cone is convex and closed and $(I_n,0)$ is outside it, the [finite-dimensional separating hyperplane theorem](/page/Separating%20Hyperplane%20Theorem) gives a linear functional on $V$ which is non-positive on the cone and positive at $(I_n,0)$. Using the trace pairing to identify $V^*$ with $\operatorname{Sym}_n\times\mathbb{R}^n$, there exist $H\in\operatorname{Sym}_n$ and $b\in\mathbb{R}^n$ such that \begin{align*} \operatorname{tr}(H)&>0,\\ u\cdot Hu+b\cdot u&\leq 0 \qquad \text{for every }u\in A. \end{align*} Choose $\varepsilon>0$ such that $\operatorname{tr}(H)-n\varepsilon>0$, and define \begin{align*} G:=H-\varepsilon I_n\in\operatorname{Sym}_n. \end{align*} Then \begin{align*} \operatorname{tr}(G)&>0,\\ u\cdot Gu+b\cdot u&\leq -\varepsilon \qquad \text{for every }u\in A. \end{align*} For $t>0$, define the affine image of the unit ball \begin{align*} E_t:=tb+(I_n+tG)B(0,1). \end{align*} For sufficiently small $t>0$, the matrix $I_n+tG$ is positive definite, so $E_t$ is an ellipsoid. Its volume is \begin{align*} \mathcal{L}^n(E_t)=\det(I_n+tG)\mathcal{L}^n(B(0,1)). \end{align*} The determinant expansion at the identity gives \begin{align*} \det(I_n+tG)=1+t\operatorname{tr}(G)+O(t^2), \end{align*} so $\mathcal{L}^n(E_t)>\mathcal{L}^n(B(0,1))$ for all sufficiently small $t>0$. It remains to prove $E_t\subset K$ for sufficiently small $t>0$. For $v\in S^{n-1}$, the support function of $E_t$ is \begin{align*} h_{E_t}(v)=t b\cdot v+|(I_n+tG)v|. \end{align*} Expanding the Euclidean norm gives, uniformly in $v\in S^{n-1}$, \begin{align*} |(I_n+tG)v|=1+t\,v\cdot Gv+O(t^2). \end{align*} Thus \begin{align*} h_{E_t}(v)=1+t(b\cdot v+v\cdot Gv)+O(t^2). \end{align*} Let $\psi:S^{n-1}\to\mathbb{R}$ be the continuous function $\psi(v):=b\cdot v+v\cdot Gv$. Since $\psi\leq -\varepsilon$ on the compact set $A$, there is an open neighbourhood $N\subset S^{n-1}$ of $A$ such that $\psi(v)\leq -\varepsilon/2$ for every $v\in N$. The uniform $O(t^2)$ remainder gives $h_{E_t}(v)\leq 1\leq h_K(v)$ on $N$ for all sufficiently small $t>0$. On the compact set $S^{n-1}\setminus N$, the continuous function $h_K-1$ has a positive minimum $\delta>0$, because $A=\{v\in S^{n-1}:h_K(v)=1\}$. Since $\psi$ is bounded on $S^{n-1}$ and the remainder is uniform, choosing $t>0$ smaller if necessary gives $h_{E_t}(v)\leq 1+\delta\leq h_K(v)$ on $S^{n-1}\setminus N$. Therefore $h_{E_t}(v)\leq h_K(v)$ for every $v\in S^{n-1}$ and all sufficiently small $t>0$. Since $E_t$ and $K$ are compact convex subsets of $\mathbb{R}^n$, the support-function containment criterion applies: a compact convex set $C$ is contained in a compact convex set $D$ exactly when $h_C(v)\leq h_D(v)$ for every $v\in S^{n-1}$. Hence $E_t\subset K$. This contradicts the maximality of $B(0,1)$ in [John position](/page/John%20Position). Therefore $(I_n,0)$ belongs to the cone generated by $\{(u\otimes u,u):u\in A\}$.[/step]

[guided]We now turn maximality into a first-order optimality condition. Let $\operatorname{Sym}_n$ be the vector space of real symmetric $n\times n$ matrices, with pairing $(M,N)\mapsto\operatorname{tr}(MN)$, and set \begin{align*} V:=\operatorname{Sym}_n\times\mathbb{R}^n. \end{align*} Define the constraint map \begin{align*} q:A&\to V,\\ u&\mapsto (u\otimes u,u). \end{align*} The matrix component $u\otimes u$ measures the first-order effect of a linear perturbation in the supporting direction $u$, while the vector component $u$ measures the first-order effect of translation. Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$. Assume, toward a contradiction, that $(I_n,0)$ is not in the cone generated by the vectors $q(u)$. We want to apply the [finite-dimensional separating hyperplane theorem](/page/Separating%20Hyperplane%20Theorem), so we verify its hypotheses. The cone is convex by construction. For closedness, $A$ is compact and $q$ is continuous, so $q(A)$ is compact; also $0\notin q(A)$ because $u\in S^{n-1}$ gives $u\otimes u\neq 0$. Thus $q(A)$ is bounded away from $0$ in any fixed Euclidean norm on $V$, and a convergent sequence of conic combinations has bounded total coefficient after normalising the generators. Compactness then gives a convergent subsequence of the normalised generators, so the limit remains in the cone. Hence the cone generated by $q(A)$ is closed. The point $(I_n,0)$ lies outside this closed convex cone by assumption. The separating hyperplane theorem gives a linear functional on $V$ which is non-positive on the cone and positive at $(I_n,0)$. Because $V=\operatorname{Sym}_n\times\mathbb{R}^n$ is finite-dimensional and the trace pairing identifies $\operatorname{Sym}_n^*$ with $\operatorname{Sym}_n$, this functional has the form $(M,z)\mapsto \operatorname{tr}(HM)+b\cdot z$ for some $H\in\operatorname{Sym}_n$ and $b\in\mathbb{R}^n$. Therefore \begin{align*} \operatorname{tr}(H)&>0,\\ u\cdot Hu+b\cdot u&\leq 0 \qquad \text{for every }u\in A. \end{align*} The non-strict inequality is not enough to control the second-order error in the support function expansion, so we create strict first-order slack. Choose $\varepsilon>0$ such that $\operatorname{tr}(H)-n\varepsilon>0$, and define \begin{align*} G:=H-\varepsilon I_n\in\operatorname{Sym}_n. \end{align*} Then \begin{align*} \operatorname{tr}(G)&>0,\\ u\cdot Gu+b\cdot u&=u\cdot Hu+b\cdot u-\varepsilon\leq -\varepsilon \qquad \text{for every }u\in A. \end{align*} The strict negative margin on $A$ is the point of replacing $H$ by $G$: it will absorb the $O(t^2)$ term at active directions. For $t>0$, define \begin{align*} E_t:=tb+(I_n+tG)B(0,1). \end{align*} For small $t>0$, the matrix $I_n+tG$ remains positive definite, so $E_t$ is an ellipsoid. Its volume is the determinant of the linear part times the volume of the unit ball: \begin{align*} \mathcal{L}^n(E_t)=\det(I_n+tG)\mathcal{L}^n(B(0,1)). \end{align*} The derivative of the determinant at the identity in the direction $G$ is $\operatorname{tr}(G)$, so \begin{align*} \det(I_n+tG)=1+t\operatorname{tr}(G)+O(t^2). \end{align*} Since $\operatorname{tr}(G)>0$, this volume is larger than $\mathcal{L}^n(B(0,1))$ for all sufficiently small positive $t$. We must check containment in $K$. For $v\in S^{n-1}$, the support function of $E_t$ is \begin{align*} h_{E_t}(v)=t b\cdot v+|(I_n+tG)v|. \end{align*} The norm expansion is uniform in $v\in S^{n-1}$ because the sphere is compact: \begin{align*} |(I_n+tG)v|=1+t\,v\cdot Gv+O(t^2). \end{align*} Therefore \begin{align*} h_{E_t}(v)=1+t(b\cdot v+v\cdot Gv)+O(t^2). \end{align*} Define the continuous first-order coefficient \begin{align*} \psi:S^{n-1}&\to\mathbb{R},\\ v&\mapsto b\cdot v+v\cdot Gv. \end{align*} On the active set $A$, we have $\psi(v)\leq -\varepsilon$. By continuity, there is an open neighbourhood $N\subset S^{n-1}$ of $A$ such that $\psi(v)\leq -\varepsilon/2$ for every $v\in N$. The remainder in the expansion is uniform, so for all sufficiently small $t>0$ the negative term $-t\varepsilon/2$ dominates the $O(t^2)$ error. Hence \begin{align*} h_{E_t}(v)\leq 1\leq h_K(v)\qquad\text{for every }v\in N. \end{align*} Away from $A$, compactness supplies a genuine support-function gap. Since $S^{n-1}\setminus N$ is compact and does not meet $A=\{v:h_K(v)=1\}$, the continuous function $h_K-1$ has a positive minimum $\delta>0$ on $S^{n-1}\setminus N$. The function $\psi$ is bounded on $S^{n-1}$, and the remainder is uniform, so after reducing $t>0$ if necessary we have \begin{align*} h_{E_t}(v)\leq 1+\delta\leq h_K(v)\qquad\text{for every }v\in S^{n-1}\setminus N. \end{align*} Combining the estimates on $N$ and on $S^{n-1}\setminus N$ gives \begin{align*} h_{E_t}(v)\leq h_K(v)\qquad\text{for every }v\in S^{n-1}. \end{align*} We now use the support-function containment criterion for compact convex sets: if $C,D\subset\mathbb{R}^n$ are compact and convex, then $C\subset D$ exactly when $h_C(v)\leq h_D(v)$ for every $v\in S^{n-1}$. The sets $E_t$ and $K$ satisfy these hypotheses, so the support-function inequality implies $E_t\subset K$. We have produced an ellipsoid inside $K$ with larger volume than $B(0,1)$, contradicting [John position](/page/John%20Position). Thus \begin{align*} (I_n,0)\in \operatorname{cone}\{(u\otimes u,u):u\in A\}. \end{align*}[/guided]

[step:Reduce the conic representation to finitely many contact points] Since $(I_n,0)$ belongs to the cone generated by $\{q(u):u\in A\}$, there are finitely many points $u_1,\dots,u_N\in A$ and coefficients $a_1,\dots,a_N>0$ such that \begin{align*} (I_n,0)=\sum_{i=1}^N a_i(u_i\otimes u_i,u_i). \end{align*} The [conic Carathéodory theorem](/page/Carath%C3%A9odory%27s%20Theorem) in the finite-dimensional vector space $V$ allows this representation to be chosen with at most \begin{align*} \dim V=\frac{n(n+1)}{2}+n=\frac{n(n+3)}{2} \end{align*} terms. Its hypotheses are satisfied because $V$ is a real vector space of finite dimension and the representation is a conic combination of elements of $V$: if more than $\dim V$ generators occur, they are linearly dependent, and one subtracts a suitable multiple of the dependence relation to remove at least one positive coefficient while keeping all coefficients non-negative. Removing zero coefficients and repeating terminates. Writing the two components of the resulting identity gives contact points $u_1,\dots,u_m\in\partial K\cap\partial B(0,1)$ and weights $c_i:=a_i>0$ with \begin{align*} m\leq \frac{n(n+3)}{2} \end{align*} such that \begin{align*} \sum_{i=1}^m c_i u_i&=0,\\ \sum_{i=1}^m c_i u_i\otimes u_i&=I_n. \end{align*} This proves the general decomposition. [/step]

[step:Use symmetry to remove the translation component] Assume now that $K$ is [centrally symmetric](/page/Centrally%20Symmetric%20Set). Then $h_K(v)=h_K(-v)$ for every $v\in S^{n-1}$, so the active set satisfies $A=-A$. Repeat the preceding separation argument in the smaller vector space $\operatorname{Sym}_n$, using only the generators $u\otimes u$ with $u\in A$. If $I_n$ were not in their cone, the same compactness argument shows that this cone is closed, and the [finite-dimensional separating hyperplane theorem](/page/Separating%20Hyperplane%20Theorem), applied with the trace pairing on $\operatorname{Sym}_n$, would give $H\in\operatorname{Sym}_n$ such that \begin{align*} \operatorname{tr}(H)>0,\qquad u\cdot Hu\leq 0\quad\text{for every }u\in A. \end{align*} Choose $\varepsilon>0$ such that $\operatorname{tr}(H)-n\varepsilon>0$, and set $G:=H-\varepsilon I_n$. Then \begin{align*} \operatorname{tr}(G)>0,\qquad u\cdot Gu\leq -\varepsilon\quad\text{for every }u\in A. \end{align*} For $t>0$, define the centered ellipsoid \begin{align*} E_t:=(I_n+tG)B(0,1). \end{align*} The determinant expansion gives \begin{align*} \mathcal{L}^n(E_t)=\det(I_n+tG)\mathcal{L}^n(B(0,1))>\mathcal{L}^n(B(0,1)) \end{align*} for all sufficiently small $t>0$. To prove containment, use the support-function expansion \begin{align*} h_{E_t}(v)=1+t\,v\cdot Gv+O(t^2), \end{align*} uniformly for $v\in S^{n-1}$. Since $v\cdot Gv\leq -\varepsilon$ on $A$, continuity gives a neighbourhood $N\subset S^{n-1}$ of $A$ on which $v\cdot Gv\leq -\varepsilon/2$, and the uniform remainder gives $h_{E_t}(v)\leq 1\leq h_K(v)$ on $N$ for all sufficiently small $t>0$. On $S^{n-1}\setminus N$, the continuous function $h_K-1$ has a positive minimum, so the bounded first-order term and the uniform remainder are absorbed by this gap after reducing $t>0$. Hence $h_{E_t}\leq h_K$ on $S^{n-1}$, so the support-function containment criterion for compact convex sets gives $E_t\subset K$. This contradicts [John position](/page/John%20Position). Therefore \begin{align*} I_n\in \operatorname{cone}\{u\otimes u:u\in A\}. \end{align*} Applying the [conic Carathéodory theorem](/page/Carath%C3%A9odory%27s%20Theorem) in $\operatorname{Sym}_n$ gives \begin{align*} m\leq \frac{n(n+1)}{2}, \end{align*} contact directions $u_1,\dots,u_m\in A\subset\partial K\cap\partial B(0,1)$, and weights $c_1,\dots,c_m>0$ such that \begin{align*} \sum_{i=1}^m c_i u_i\otimes u_i=I_n. \end{align*} Since $A=-A$, replacing each $u_i$ by the pair $u_i,-u_i$ with weights $c_i/2$ gives a symmetric contact-point decomposition, and the vector sum cancels pairwise. This completes the proof. [/step]