Prove GDA decision boundary is linear
If you expand the terms, you can see that the quadratic terms cancel out.

\begin{align}
a &amp;= \ln \frac{P(C_0)}{P(C_1)} + \frac12(x - \mu_1)^T\Sigma^{-1}(x - \mu_1) - \frac12(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)\\
&amp;=\ln \frac{P(C_0)}{P(C_1)} + \frac12\left[x^T\Sigma^{-1}x-2x^T\Sigma^{-1}\mu_1+\mu_1^T\Sigma^{-1}\mu_1\right]\\&amp; - \frac12\left[x^T\Sigma^{-1}x-2x^T\Sigma^{-1}\mu_0+\mu_0^T\Sigma^{-1}\mu_0\right]\\
&amp;= (\mu_0-\mu_1)^T\Sigma^{-1}x+\ln \frac{P(C_0)}{P(C_1)} +\frac12\left[\mu_1^T\Sigma^{-1}\mu_1-\mu_0^T\Sigma^{-1}\mu_0\right] \\
\end{align}