[proofplan]
We first use [affine equivariance of the John ellipsoid](/theorems/4131) to reduce to the case in which the maximal-volume ellipsoid is the Euclidean unit ball $B(0,1)$. In that position, the contact point characterization of John's theorem gives finitely many boundary contact points $u_i \in \partial K \cap \partial B(0,1)$ and positive weights $c_i$ whose rank-one operators decompose the identity. The supporting inequalities at these contact points, together with central symmetry, bound every scalar projection $u_i \cdot x$ by $1$ for $x \in K$. Substituting the identity decomposition into $|x|^2$ then gives $|x|^2 \le n$, proving $K \subset B(0,\sqrt n)$ and hence the stated affine form.
[/proofplan]
[step:Reduce by affine equivariance to John position]
Let $E$ be the John ellipsoid of $K$. Since $K$ is centrally symmetric, the center of $E$ is $0$: if $E$ has center $a$, then $-E$ is also an ellipsoid contained in $K$ with the same volume, and uniqueness of the John ellipsoid gives $E=-E$, hence $a=0$.
Choose an invertible [linear map](/page/Linear%20Map) $T:\mathbb{R}^n \to \mathbb{R}^n$ such that
\begin{align*}
E = T(B(0,1)).
\end{align*}
Define the affine-normalized body
\begin{align*}
\widetilde K := T^{-1}(K) \subset \mathbb{R}^n.
\end{align*}
The affine equivariance of the John ellipsoid implies that the John ellipsoid of $\widetilde K$ is $B(0,1)$. Therefore it is enough to prove
\begin{align*}
\widetilde K \subset B(0,\sqrt n).
\end{align*}
Indeed, applying $T$ to this inclusion gives
\begin{align*}
K = T(\widetilde K) \subset T(B(0,\sqrt n)) = \sqrt n\,T(B(0,1)) = \sqrt n\,E.
\end{align*}
[guided]
The theorem is invariant under invertible linear changes of coordinates. Since $E$ is an ellipsoid centered at $0$, there is an invertible linear map $T:\mathbb{R}^n \to \mathbb{R}^n$ with
\begin{align*}
E = T(B(0,1)).
\end{align*}
We define
\begin{align*}
\widetilde K := T^{-1}(K).
\end{align*}
This set is again a centrally symmetric convex body, because invertible linear maps preserve compactness, convexity, nonempty interior, and the symmetry relation $K=-K$.
The affine equivariance of the John ellipsoid says that applying $T^{-1}$ sends the maximal-volume ellipsoid inside $K$ to the maximal-volume ellipsoid inside $T^{-1}(K)$. Hence the John ellipsoid of $\widetilde K$ is $B(0,1)$. Thus $\widetilde K$ is in John position.
Now suppose we prove the normalized statement
\begin{align*}
\widetilde K \subset B(0,\sqrt n).
\end{align*}
Applying $T$ to both sides gives
\begin{align*}
K = T(\widetilde K) \subset T(B(0,\sqrt n)).
\end{align*}
Since $T$ is linear,
\begin{align*}
T(B(0,\sqrt n)) = \sqrt n\,T(B(0,1)) = \sqrt n\,E.
\end{align*}
Thus the general inclusion follows from the John-position case.
[/guided]
[/step]
[step:Use the symmetric John decomposition at the contact points]
Assume from now on that $K$ is in John position, so that $B(0,1)$ is the John ellipsoid of $K$.
By the contact point characterization of John's theorem for centrally symmetric convex bodies (citing a result not yet in the wiki: John decomposition theorem for centrally symmetric convex bodies), there exist an integer $m \in \mathbb{N}$, contact points
\begin{align*}
u_i \in \partial K \cap \partial B(0,1) \subset \mathbb{R}^n,
\qquad 1 \le i \le m,
\end{align*}
and weights $c_i > 0$ such that
\begin{align*}
\sum_{i=1}^{m} c_i\, u_i \otimes u_i = I_n.
\end{align*}
Here $u_i \otimes u_i$ denotes the rank-one linear map $\mathbb{R}^n \to \mathbb{R}^n$ defined by
\begin{align*}
(u_i \otimes u_i)(x) = (u_i \cdot x)u_i.
\end{align*}
Taking traces in the identity decomposition gives
\begin{align*}
\sum_{i=1}^{m} c_i
= \sum_{i=1}^{m} c_i\,\operatorname{tr}(u_i \otimes u_i)
= \operatorname{tr}(I_n)
= n,
\end{align*}
because $|u_i|=1$ implies $\operatorname{tr}(u_i \otimes u_i)=u_i \cdot u_i=1$.
[guided]
The normalized form of John's theorem gives more than the inclusion $B(0,1)\subset K$: it gives a finite identity decomposition built from contact points. Precisely, the contact point characterization for centrally symmetric bodies gives contact points
\begin{align*}
u_i \in \partial K \cap \partial B(0,1),
\qquad 1 \le i \le m,
\end{align*}
and positive weights $c_i>0$ satisfying
\begin{align*}
\sum_{i=1}^{m} c_i\,u_i \otimes u_i = I_n.
\end{align*}
The operator $u_i \otimes u_i:\mathbb{R}^n \to \mathbb{R}^n$ is the rank-one map
\begin{align*}
(u_i \otimes u_i)(x) = (u_i \cdot x)u_i.
\end{align*}
This decomposition is the mechanism that converts one-dimensional support inequalities into a Euclidean norm bound.
We also need the total mass of the weights. Taking traces gives it exactly:
\begin{align*}
\operatorname{tr}\left(\sum_{i=1}^{m} c_i\,u_i \otimes u_i\right)
=
\operatorname{tr}(I_n).
\end{align*}
Linearity of trace gives
\begin{align*}
\sum_{i=1}^{m} c_i\,\operatorname{tr}(u_i \otimes u_i) = n.
\end{align*}
Since $u_i \in \partial B(0,1)$, we have $|u_i|=1$, and the rank-one operator $u_i \otimes u_i$ has trace $u_i \cdot u_i=1$. Therefore
\begin{align*}
\sum_{i=1}^{m} c_i = n.
\end{align*}
[/guided]
[/step]
[step:Convert contact support inequalities into projection bounds]
Fix $x \in K$. For each $1 \le i \le m$, the hyperplane
\begin{align*}
H_i := \{y \in \mathbb{R}^n : u_i \cdot y = 1\}
\end{align*}
supports $K$ at the contact point $u_i$, so
\begin{align*}
u_i \cdot x \le 1.
\end{align*}
Since $K=-K$, we also have $-x \in K$. Applying the same support inequality to $-x$ gives
\begin{align*}
u_i \cdot (-x) \le 1,
\end{align*}
and hence
\begin{align*}
u_i \cdot x \ge -1.
\end{align*}
Therefore
\begin{align*}
|u_i \cdot x| \le 1
\end{align*}
for every $1 \le i \le m$.
[guided]
Fix a point $x \in K$. We want to bound $|x|$, and the identity decomposition expresses $|x|^2$ using the scalar quantities $u_i \cdot x$. Thus the next task is to prove that each scalar projection is bounded by $1$ in absolute value.
For each contact point $u_i$, the hyperplane
\begin{align*}
H_i := \{y \in \mathbb{R}^n : u_i \cdot y = 1\}
\end{align*}
is the supporting hyperplane to $K$ at $u_i$. Therefore every point of $K$ lies in the closed half-space
\begin{align*}
\{y \in \mathbb{R}^n : u_i \cdot y \le 1\}.
\end{align*}
Since $x \in K$, this gives
\begin{align*}
u_i \cdot x \le 1.
\end{align*}
This is only a one-sided inequality. The central symmetry of $K$ supplies the other side. Because $K=-K$ and $x \in K$, we also have $-x \in K$. Applying the same supporting half-space inequality to $-x$ gives
\begin{align*}
u_i \cdot (-x) \le 1.
\end{align*}
Multiplying by $-1$ yields
\begin{align*}
u_i \cdot x \ge -1.
\end{align*}
Combining the two inequalities gives
\begin{align*}
|u_i \cdot x| \le 1
\end{align*}
for every contact point $u_i$.
[/guided]
[/step]
[step:Insert the identity decomposition to bound the Euclidean norm]
Using the identity decomposition and the definition of $u_i \otimes u_i$, we compute
\begin{align*}
|x|^2
= x \cdot I_n x
= x \cdot \left(\sum_{i=1}^{m} c_i\,u_i \otimes u_i\right)x
= \sum_{i=1}^{m} c_i\,x \cdot ((u_i \otimes u_i)(x))
= \sum_{i=1}^{m} c_i\,(u_i \cdot x)^2.
\end{align*}
By the projection bounds from the previous step and the positivity of the weights,
\begin{align*}
|x|^2
= \sum_{i=1}^{m} c_i\,(u_i \cdot x)^2
\le \sum_{i=1}^{m} c_i
= n.
\end{align*}
Hence $|x| \le \sqrt n$. Since $x \in K$ was arbitrary,
\begin{align*}
K \subset B(0,\sqrt n).
\end{align*}
By the affine reduction, this is equivalent to
\begin{align*}
E \subset K \subset \sqrt n\,E.
\end{align*}
This proves the theorem.
[guided]
We now use the identity decomposition exactly where it is useful: inside the quadratic form defining the Euclidean norm. Since
\begin{align*}
I_n = \sum_{i=1}^{m} c_i\,u_i \otimes u_i,
\end{align*}
we have
\begin{align*}
|x|^2
= x \cdot I_n x.
\end{align*}
Substituting the decomposition of $I_n$ gives
\begin{align*}
|x|^2
= x \cdot \left(\sum_{i=1}^{m} c_i\,u_i \otimes u_i\right)x.
\end{align*}
By linearity of the dot product and of the finite sum,
\begin{align*}
|x|^2
= \sum_{i=1}^{m} c_i\,x \cdot ((u_i \otimes u_i)(x)).
\end{align*}
The rank-one operator was defined by
\begin{align*}
(u_i \otimes u_i)(x) = (u_i \cdot x)u_i.
\end{align*}
Therefore
\begin{align*}
x \cdot ((u_i \otimes u_i)(x))
= x \cdot ((u_i \cdot x)u_i)
= (u_i \cdot x)(x \cdot u_i)
= (u_i \cdot x)^2.
\end{align*}
Thus
\begin{align*}
|x|^2 = \sum_{i=1}^{m} c_i\,(u_i \cdot x)^2.
\end{align*}
From the support inequalities, $|u_i \cdot x|\le 1$ for every $i$, so
\begin{align*}
(u_i \cdot x)^2 \le 1.
\end{align*}
Since each weight satisfies $c_i>0$, multiplying by $c_i$ preserves the inequality, and summing gives
\begin{align*}
|x|^2
= \sum_{i=1}^{m} c_i\,(u_i \cdot x)^2
\le \sum_{i=1}^{m} c_i.
\end{align*}
The trace computation showed that
\begin{align*}
\sum_{i=1}^{m} c_i = n.
\end{align*}
Hence
\begin{align*}
|x|^2 \le n,
\end{align*}
so $|x|\le \sqrt n$.
Because the point $x \in K$ was arbitrary, every point of $K$ lies in $B(0,\sqrt n)$:
\begin{align*}
K \subset B(0,\sqrt n).
\end{align*}
Finally, the affine normalization step converts this normalized inclusion back to
\begin{align*}
E \subset K \subset \sqrt n\,E.
\end{align*}
[/guided]
[/step]
