diff --git a/src/op/system/schur.tex b/src/op/system/schur.tex
index d9fc367..390eaa8 100644
--- a/src/op/system/schur.tex
+++ b/src/op/system/schur.tex
@@ -84,7 +84,7 @@
 
     extends into a completely positive map on $C(\mathbb{T}; \complex)$. 
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 2.6]{Paulsen}}}. ]
+\begin{proof}[Proof, using trigonometric moment sequences, {{\cite[Theorem 2.6]{Paulsen}}}. ]
     Assume without loss of generality that $\norm{T}_{B(H)} < 1$. Using \autoref{proposition:trig-moment-sequence-operator}, it is sufficient to show that the Toeplitz matrix
     \[
     A = \sum_{i, j = 1}^n E_{ij} \otimes A_{i - j} = \begin{bmatrix}
@@ -120,4 +120,23 @@
     \end{align*}
     
 \end{proof}
+\begin{proof}[Proof, using von Neumann's inequality. ]
+    Let $p = \sum_{j = 0}^n \lambda_j z^j, q = \sum_{j = 0}^n \mu_j z^j \in \complex[z]$ such that $p + \ol q \ge 0$, then
+    \begin{align*}
+    p + \ol q = \text{Re}(p + \ol q) &= \frac{p + \ol p}{2} + \frac{q + \ol q}{2} \\
+    \sum_{j = 0}^n (\lambda_jz^j + \ol{\mu_j} z^{-j}) &= \frac{1}{2}\sum_{j = 0}^n (\lambda_j + \mu_j)z^j + (\ol \lambda_j + \ol \mu_j)z^{-j} 
+    \end{align*}
+
+    so $\lambda_j = (\lambda_j + \mu_j)/2 = \mu_j$ for all $1 \le j \le n$. After rebalancing $\lambda_0$ and $\mu_0$, assume without loss of generality that $q = p$. 
+
+    Let $c = \normn{p + \ol p}_{C(\mathbb{T}; \complex)}/2$, then $\normn{p + \ol p - c}_{C(\mathbb{T}; \complex)} = \norm{p + \ol p}_{C(\mathbb{T}; \complex)}/2$. By \hyperref[von Neumann's inequality]{theorem:von-neumann-inequality-polynomial}, 
+    \[
+    \normn{p(T) + p(T)^* - c}_{B(H)} \le \normn{p + \ol p - c}_{C(\mathbb{T}; \complex)} = \norm{p + \ol p}_{C(\mathbb{T}; \complex)}/2 = c
+    \]
+
+    As $p(T) + p(T)^* - cI$ is self-adjoint,
+    \[
+    0 \le p(T) + p(T)^* - c + \norm{p(T) + p(T)^* - cI}_{B(H)} \le p(T) + p(T)^*
+    \]
+\end{proof}