[proofplan] We first translate $K$ by its barycenter, which makes the barycenter equal to $0$ without changing convexity or volume. For the centered body, we form its normalized covariance matrix and prove that it is positive definite by using the non-empty interior of $K$. Applying the inverse square root of this positive definite matrix makes the covariance matrix equal to the identity. A final scalar dilation normalizes the volume to $1$, and this dilation multiplies the covariance matrix by a scalar square. [/proofplan] [step:Translate the convex body to place its barycenter at the origin] Let $V := \mathcal{L}^n(K)$. Since $K$ is compact with non-empty interior, $0 < V < \infty$. Define the barycenter $b \in \mathbb{R}^n$ by \begin{align*} b := \frac{1}{V}\int_K x \, d\mathcal{L}^n(x). \end{align*} Define the translation map \begin{align*} \tau: \mathbb{R}^n &\to \mathbb{R}^n \\ x &\mapsto x - b, \end{align*} and set $K_0 := \tau(K) = K - b$. Translation invariance of Lebesgue measure gives \begin{align*} \mathcal{L}^n(K_0) = V. \end{align*} Moreover, under the substitution $y = x - b$, Lebesgue measure is unchanged, so \begin{align*} \int_{K_0} y \, d\mathcal{L}^n(y) &= \int_K (x - b) \, d\mathcal{L}^n(x) \\ &= \int_K x \, d\mathcal{L}^n(x) - b\,\mathcal{L}^n(K) \\ &= Vb - Vb \\ &= 0. \end{align*} Thus $K_0$ has barycenter $0$. [guided] The first operation is only a translation. Define $V := \mathcal{L}^n(K)$. Because $K$ is compact, its Lebesgue measure is finite; because it has non-empty interior, it contains a non-empty open ball, so $V > 0$. Therefore the barycenter is well-defined: \begin{align*} b := \frac{1}{V}\int_K x \, d\mathcal{L}^n(x). \end{align*} Now define the affine translation \begin{align*} \tau: \mathbb{R}^n &\to \mathbb{R}^n \\ x &\mapsto x - b, \end{align*} and write $K_0 := \tau(K)$. Translation does not change Lebesgue measure, so $\mathcal{L}^n(K_0) = V$. To compute the barycenter of $K_0$, use the substitution $y = x - b$. The measure transform is $d\mathcal{L}^n(y) = d\mathcal{L}^n(x)$, and the integration domain changes from $K_0$ back to $K$. Hence \begin{align*} \int_{K_0} y \, d\mathcal{L}^n(y) &= \int_K (x - b) \, d\mathcal{L}^n(x) \\ &= \int_K x \, d\mathcal{L}^n(x) - b\,\mathcal{L}^n(K) \\ &= Vb - Vb \\ &= 0. \end{align*} So the translated body has barycenter at the origin. This is the reason for translating first: once the barycenter is $0$, the covariance matrix is simply the second moment matrix. [/guided] [/step] [step:Show that the centered covariance matrix is positive definite] Define the normalized covariance matrix $C \in \mathbb{R}^{n \times n}$ of $K_0$ by \begin{align*} C := \frac{1}{V}\int_{K_0} y \otimes y \, d\mathcal{L}^n(y). \end{align*} For every $\xi \in \mathbb{R}^n$, \begin{align*} \xi^\top C \xi = \frac{1}{V}\int_{K_0} (\xi \cdot y)^2 \, d\mathcal{L}^n(y). \end{align*} Hence $C$ is symmetric and positive semidefinite. Let $\xi \in \mathbb{R}^n$ with $\xi \neq 0$. Since $K_0$ has non-empty interior, there exist $a \in K_0$ and $r > 0$ such that $B(a,r) \subset K_0$. The continuous linear functional \begin{align*} \ell_\xi: \mathbb{R}^n &\to \mathbb{R} \\ y &\mapsto \xi \cdot y \end{align*} does not vanish identically on $B(a,r)$, since $\xi \neq 0$. Therefore the continuous non-negative function $y \mapsto (\xi \cdot y)^2$ is positive on some non-empty open subset of $B(a,r)$. This open subset has positive Lebesgue measure, so \begin{align*} \int_{K_0} (\xi \cdot y)^2 \, d\mathcal{L}^n(y) > 0. \end{align*} Thus $\xi^\top C\xi > 0$ for all $\xi \neq 0$, so $C$ is positive definite. [guided] We now build the matrix that records the second moments of the centered body. Define \begin{align*} C := \frac{1}{V}\int_{K_0} y \otimes y \, d\mathcal{L}^n(y). \end{align*} Here $y \otimes y$ is the $n \times n$ matrix with entries $(y \otimes y)_{ij} = y_i y_j$. Since $K_0$ is compact, the coordinate functions are bounded on $K_0$, so all entries of this matrix integral are finite. To understand positivity, test $C$ against a vector $\xi \in \mathbb{R}^n$. By the definition of the outer product, \begin{align*} \xi^\top (y \otimes y)\xi = (\xi \cdot y)^2. \end{align*} Therefore \begin{align*} \xi^\top C \xi = \frac{1}{V}\int_{K_0} (\xi \cdot y)^2 \, d\mathcal{L}^n(y). \end{align*} This proves that $C$ is symmetric and positive semidefinite. The key point is to rule out degeneracy. Let $\xi \neq 0$. Because $K_0$ has non-empty interior, there are $a \in K_0$ and $r > 0$ such that $B(a,r) \subset K_0$. Define the linear functional \begin{align*} \ell_\xi: \mathbb{R}^n &\to \mathbb{R} \\ y &\mapsto \xi \cdot y. \end{align*} Since $\xi \neq 0$, this functional is not identically zero on any open ball. Hence there is a point $y_0 \in B(a,r)$ with $\ell_\xi(y_0) \neq 0$. By continuity of $\ell_\xi$, the function $y \mapsto (\xi \cdot y)^2$ is positive on some open ball around $y_0$ contained in $B(a,r)$. That open ball has positive Lebesgue measure. Since the integrand is non-negative on all of $K_0$, we obtain \begin{align*} \int_{K_0} (\xi \cdot y)^2 \, d\mathcal{L}^n(y) > 0. \end{align*} Thus $\xi^\top C\xi > 0$ for every non-zero $\xi$, which is exactly positive definiteness. [/guided] [/step] [step:Apply the inverse square root to make the covariance matrix the identity] Since $C$ is a real symmetric positive definite matrix, it has a real symmetric positive definite inverse square root $C^{-1/2} \in \mathbb{R}^{n \times n}$ satisfying \begin{align*} C^{-1/2} C C^{-1/2} = I_n. \end{align*} Define the invertible [linear map](/page/Linear%20Map) \begin{align*} A: \mathbb{R}^n &\to \mathbb{R}^n \\ y &\mapsto C^{-1/2}y, \end{align*} and set $K_1 := A(K_0)$. Since $A$ is linear, the barycenter remains $0$: \begin{align*} \frac{1}{\mathcal{L}^n(K_1)}\int_{K_1} z \, d\mathcal{L}^n(z) = 0. \end{align*} By the linear change of variables $z = Ay$, with \begin{align*} d\mathcal{L}^n(z) = |\det A|\,d\mathcal{L}^n(y), \end{align*} we have $\mathcal{L}^n(K_1) = |\det A|V$ and \begin{align*} \frac{1}{\mathcal{L}^n(K_1)}\int_{K_1} z \otimes z \, d\mathcal{L}^n(z) &= \frac{1}{|\det A|V}\int_{K_0} (Ay)\otimes(Ay)\,|\det A|\,d\mathcal{L}^n(y) \\ &= A\left(\frac{1}{V}\int_{K_0} y\otimes y\,d\mathcal{L}^n(y)\right)A^\top \\ &= A C A^\top \\ &= C^{-1/2} C C^{-1/2} \\ &= I_n. \end{align*} [/step] [step:Normalize the volume by a scalar dilation] Let $W := \mathcal{L}^n(K_1) = |\det A|V$, so $W > 0$. Define \begin{align*} \lambda := W^{-1/n}. \end{align*} Define the dilation \begin{align*} D_\lambda: \mathbb{R}^n &\to \mathbb{R}^n \\ z &\mapsto \lambda z, \end{align*} and set $K_2 := D_\lambda(K_1) = \lambda K_1$. Then \begin{align*} \mathcal{L}^n(K_2) = \lambda^n \mathcal{L}^n(K_1) = W^{-1}W = 1. \end{align*} The barycenter remains $0$ because $D_\lambda$ is linear and the barycenter of $K_1$ is $0$. The normalized covariance matrix of $K_2$ is \begin{align*} \frac{1}{\mathcal{L}^n(K_2)}\int_{K_2} x\otimes x \, d\mathcal{L}^n(x) &= \lambda^2 I_n. \end{align*} Since $\mathcal{L}^n(K_2)=1$, this is equivalently \begin{align*} \int_{K_2} x\otimes x \, d\mathcal{L}^n(x) = \lambda^2 I_n. \end{align*} Set $L_{K_2} := \lambda > 0$. [/step] [step:Assemble the affine map and verify the required properties] Define the affine map \begin{align*} T: \mathbb{R}^n &\to \mathbb{R}^n \\ x &\mapsto \lambda C^{-1/2}(x-b). \end{align*} This map is invertible because $\lambda > 0$ and $C^{-1/2}$ is invertible. By construction, \begin{align*} T(K) = K_2. \end{align*} The preceding steps show that \begin{align*} \mathcal{L}^n(T(K)) &= 1, \\ \int_{T(K)} x \, d\mathcal{L}^n(x) &= 0, \\ \int_{T(K)} x\otimes x \, d\mathcal{L}^n(x) &= L_{T(K)}^2 I_n, \end{align*} with $L_{T(K)} := \lambda > 0$. Thus $T(K)$ is in isotropic position, completing the proof. [/step]