I stumbled across this while playing around and managed to quite confuse myself, I am hoping someone can show me how I am wrong. Consider some column vector $ \Psi \in \mathbb{C}^N$ some collection of matrices $ \sigma_\mu \in M(\mathbb{C},N)$ and some collection of linear operator $ A: M(\mathbb{C},N) \to M(\mathbb{C},N)$, where $A(\sigma_\mu)^\dagger = A(\sigma_\mu) = \sigma_\nu A^\nu{}_\mu$ with $A^\nu{}_\mu \in \mathbb{R}$.
Consider then the integral $$ S=\int_\Omega d^4x\text{ } \Psi^\dagger iA(\sigma_\mu)\partial^\mu \Psi $$ where $\Psi$ and $A$ are zero at the boundary $\partial \Omega$ of the space $\Omega$, $\partial^\mu$ is the partial derivative w.r.t the $\mu^{th}$ component of the position vector, and we employ summation convention of raised and lowered indices. By taking the hermitian conjugate and integrating by parts (ignoring boundary terms) we have $$ \begin{align} S &= \int_\Omega d^4x\text{ }\Psi^\dagger i\partial^\mu\left[A(\sigma_\mu) \Psi\right]\\\newline &= \int_\Omega d^4x\text{ }\Psi^\dagger i\partial^\mu\left[A(\sigma_\mu)\right] \Psi + S \end{align} $$ From which we find that $$ \int_\Omega d^4x\text{ }\Psi^\dagger\partial^\mu\left[A(\sigma_\mu)\right]\Psi = 0 $$ For (assuming certain to boundary and integrability conditions) arbitrary $\Psi$ and $A$. This seems bizarre that should hold for such arbitrary conditions and I assume I have made a mistake somewhere but cannot find it. If anyone can point out my errors I would be much obliged!