# Prove GDA decision boundary is linear My attempt:
(a) I solved that $$a=\ln{\frac{P(X|C_0)P(C_0)}{P(X|C_1)P(C_1)}}$$

(b) Here is where I'm running into trouble. I'm plugging the distributions into $$\ln{\frac{P(X|C_0)P(C_0)}{P(X|C_1)P(C_1)}}$$ and I get $$a=\ln{\frac{P(C_0)}{P(C_1)}}+\frac{1}{2}(x-\mu_1)^T\Sigma^{-1}(x-\mu_1)-\frac{1}{2}(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)$$.
I can see that $$b=\ln{\frac{P(C_0)}{P(C_1)}}$$ and $$w^Tx=\frac{1}{2}(x-\mu_1)^T\Sigma^{-1}(x-\mu_1)-\frac{1}{2}(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)$$.
I'm not sure how to simplify $$w^Tx$$ so that I can solve for $$w$$. Or is there something that I did wrong?

\begin{align} a &= \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)\\ &=\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]\\& - \frac12\left[x^T\Sigma^{-1}x-2x^T\Sigma^{-1}\mu_0+\mu_0^T\Sigma^{-1}\mu_0\right]\\ &= (\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}