I am trying to learn about decision surfaces in Bayesian decision theory but my linear algebra is a little rusty. There is a derivation for a discriminant function that assumes normal densities for the classes. Each with their own mean and the same covariance (the covariance matrix is symmetric). At one point there is this expression:
$g_i(\mathbf{x}) = -\dfrac{1}{2}(\mathbf{x}^T\mathbf{\Sigma}^{-1}\mathbf{x} + \mathbf{\mu}_i^T\mathbf{\Sigma}^{-1}\mathbf{\mu}_i - \mathbf{x}^T\mathbf{\Sigma}^{-1}\mathbf{\mu}_i - \mathbf{\mu}_i^T\mathbf{\Sigma}^{-1}\mathbf{x}) + \ln{P(w_i)}$
The $\mathbf{x}^T\mathbf{\Sigma}^{-1}\mathbf{x}$ is discarded because it has no impact. Then the expression is simplified as:
$g_i(\mathbf{x}) = -\dfrac{1}{2}\mathbf{\mu}_i^T\mathbf{\Sigma}^{-1}\mathbf{\mu}_i + (\mathbf{\Sigma}^{-1}\mathbf{\mu}_i)^T\mathbf{x} + \ln{P(w_i)}$
But how is $\mathbf{x}^T\mathbf{\Sigma}^{-1}\mathbf{\mu}_i = \mathbf{\mu}_i^T\mathbf{\Sigma}^{-1}\mathbf{x}$? If we have $\mathbf{A} = \mathbf{x}$ and $\mathbf{B} = \mathbf{\Sigma}^{-1}\mathbf{\mu}_i$ then isn't this saying that $\mathbf{A}^T\mathbf{B} = \mathbf{B}^T\mathbf{A}$?
${\bf x}^T {\bf \Sigma}^{-1} \mu_i$ is a scalar. Any scalar is equal to its transpose.