[proofplan]
Represent the Wishart matrix as a sum of independent Gaussian outer products. The principal block $W_{11}$ is obtained by projecting each Gaussian vector onto its first $q$ coordinates before taking the same outer-product sum. The projected vectors are independent $q$-variate Gaussian vectors with covariance matrix $\Sigma_{11}$, so the defining representation of the $q$-dimensional central Wishart distribution gives the result.
[/proofplan]
[step:Represent the Wishart matrix by independent Gaussian outer products]
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space on which the random vectors in the Wishart representation are realised. By the definition of the central Wishart distribution, there exist independent random vectors
\begin{align*}
X_i : (\Omega,\mathcal{F}) \to (\mathbb{R}^p,\mathcal{B}(\mathbb{R}^p)),
\qquad
1 \leq i \leq n,
\end{align*}
such that $X_i \sim \mathcal{N}_p(0,\Sigma)$ for each $i$, and
\begin{align*}
W \overset{d}{=} \sum_{i=1}^{n} X_i X_i^\top.
\end{align*}
Since the desired conclusion is distributional, it is enough to prove the statement for the random matrix
\begin{align*}
\widetilde{W} : (\Omega,\mathcal{F}) \to (\mathbb{R}^{p \times p},\mathcal{B}(\mathbb{R}^{p \times p})),
\qquad
\widetilde{W}(\omega) = \sum_{i=1}^{n} X_i(\omega)X_i(\omega)^\top.
\end{align*}
[/step]
[step:Identify the leading block as a sum of projected outer products]
Define the coordinate projection
\begin{align*}
P : \mathbb{R}^p &\to \mathbb{R}^q \\
(x_1,\dots,x_p) &\mapsto (x_1,\dots,x_q).
\end{align*}
For each $1 \leq i \leq n$, define
\begin{align*}
Y_i : \Omega &\to \mathbb{R}^q \\
\omega &\mapsto P X_i(\omega).
\end{align*}
Writing $X_i = (Y_i^\top,Z_i^\top)^\top$ with $Z_i : \Omega \to \mathbb{R}^{p-q}$ denoting the remaining coordinates, we have
\begin{align*}
X_i X_i^\top =
\begin{pmatrix}
Y_iY_i^\top & Y_iZ_i^\top \\
Z_iY_i^\top & Z_iZ_i^\top
\end{pmatrix}.
\end{align*}
Therefore the leading $q \times q$ block of $\widetilde{W}$ is
\begin{align*}
\widetilde{W}_{11}
= \sum_{i=1}^{n} Y_iY_i^\top.
\end{align*}
[/step]
[step:Compute the distribution of each projected Gaussian vector]
For each $1 \leq i \leq n$, the vector $Y_i = PX_i$ is multivariate normal because it is the image of a multivariate normal vector under the [linear map](/page/Linear%20Map) $P$. Its mean vector is
\begin{align*}
\mathbb{E}[Y_i]
= \mathbb{E}[PX_i]
= P\mathbb{E}[X_i]
= P0
= 0,
\end{align*}
and its covariance matrix is
\begin{align*}
\operatorname{Cov}(Y_i)
= \operatorname{Cov}(PX_i)
= P\Sigma P^\top
= \Sigma_{11}.
\end{align*}
Hence
\begin{align*}
Y_i \sim \mathcal{N}_q(0,\Sigma_{11})
\end{align*}
for every $1 \leq i \leq n$.
Because $Y_i = PX_i$ is a measurable function of $X_i$, and the random vectors $X_1,\dots,X_n$ are independent, the random vectors $Y_1,\dots,Y_n$ are independent.
[/step]
[step:Apply the defining representation of the lower-dimensional Wishart distribution]
The preceding steps show that $\widetilde{W}_{11}$ has the representation
\begin{align*}
\widetilde{W}_{11}
= \sum_{i=1}^{n} Y_iY_i^\top,
\end{align*}
where $Y_1,\dots,Y_n$ are independent random vectors and each satisfies
\begin{align*}
Y_i \sim \mathcal{N}_q(0,\Sigma_{11}).
\end{align*}
By the definition of the central Wishart distribution in dimension $q$, this means
\begin{align*}
\widetilde{W}_{11} \sim W_q(n,\Sigma_{11}).
\end{align*}
Since $W \overset{d}{=} \widetilde{W}$ and taking the leading principal block is a measurable map from $\mathbb{R}^{p \times p}$ to $\mathbb{R}^{q \times q}$, we have $W_{11} \overset{d}{=} \widetilde{W}_{11}$. Therefore
\begin{align*}
W_{11} \sim W_q(n,\Sigma_{11}),
\end{align*}
which proves the theorem.
[/step]
