Antisymmetric components in Gaussian kernel In Bishop's Pattern Recognition and Machine Learning, he states on page 80 in reference to the squared Mahalanobis distance $(\mathbf{x} - \boldsymbol{\mu})^\top\boldsymbol{\Sigma}^{-1}(\mathbf{x} - \boldsymbol{\mu})$ that: 

the matrix $\boldsymbol{\Sigma}$ can be taken to be symmetric,
  without loss of generality, because any antisymmetric component would
  disappear from the exponent.

My understanding was that all covariance matrices must be symmetric, so could someone please explain what this statement means?
 A: write (2.44) as $\Delta^2 = (x-\mu)^T A (x-\mu)$, where $A = \Sigma^{-1}$. 
We know that $A = \frac{1}{2} (A + A^T) + \frac{1}{2} (A - A^T)$. Let $B = \frac{1}{2} (A + A^T), C = \frac{1}{2} (A - A^T)$, then $B$ is symmetric, and $C$ is anti-symmetric, $c_{ij} = - c_{ji}$. 
So $\Delta^2 = (x-\mu)^T B (x-\mu) + (x-\mu)^T C (x-\mu)$, in which $(x-\mu)^T C (x-\mu) = \sum_{i=1}^D \sum_{j=1}^D c_{ij} (x-\mu)_i (x-\mu)_j = \sum_{i=1}^D\sum_{j=i+1}^D (c_{ij}+c_{ji}) (x-\mu)_i (x-\mu)_j = 0$. 
So $\Delta^2 = (x-\mu)^T B (x-\mu)$, where $B = \frac{1}{2} (A + A^T)$ is a symmetric matrix. That is, if $\Sigma^{-1}$ isn't symmetric, then there's another symmetric matrix $B$ so that $\Delta^2 = (x-\mu)^T \Sigma^{-1} (x-\mu)$ is equal to $\Delta^2 = (x-\mu)^T B (x-\mu)$. 
A: By definition, a covariance matrix $\Sigma$ is symmetric (positive semidefinite).
Perhaps, the author refers to the concept of multivariate symmetry.
A spherical symmetric multivariate distribution is for example a multivariate normal with covariance $\Sigma = \sigma I$. Elliptical symmetry is what is assumed by the Mahalanobis distance, allowing for a covariance matrix $\Sigma$ with the properties you are assuming and with a characteristic function as defined here
Now, if your multinomial distribution is characterized by some anti-symmetric properties (f.ex skewness and kurtosis), those properties will not be captured by the Mahalanobis norm.
