Can a mass matrix be asymmetric? I am developing a mathematical model of a mechanical device consisting basically of coupled harmonic oscillators. It turns out that the system mass matrix is asymmetric. I seem to read somewhere that a mass matrix has to be symmetric, but I am not sure. So I would like to know whether it is possible for a mass matrix in this case to be asymmetric. If it can't, what are the physical implications of an asymmetric mass matrix in this case?
 A: I agree with the answer of Lubos (and ja72 is right that there is also a positivity criterion). However, since you repeated the question again to ja72, let me explain the answer of Lubos more explicitly and with more details.
Suppose that in your system of harmonic oscillators the kinetic term is defined with a mass matrix $M$, i.e. (as in the answer of ja72):
$$K=\frac{1}{2} \dot{q}^T M \dot{q}$$
Now define the following to matrices $M_s=\frac{1}{2}(M+M^T)$ and $M_a=\frac{1}{2}(M-M^T)$. It is very easy to verify that $M=M_s+M_a$, $M_s^T=M_s$ and $M_a^T=-M_a$. Because of these properties, $M_s$ is called the symmetric part of $M$, while $M_a$ is called the antisymmetric part (this is also the terms Lubos used).
Now, let us notice that since $M_a$ is antisymmetric the identity $v^T M_a v=0$ holds for any vector $v$. Writing up the kinetic term again, we obtain
$$K=\frac{1}{2} \dot{q}^T M \dot{q}= \frac{1}{2} \dot{q}^T M_s \dot{q} + \frac{1}{2} \dot{q}^T M_a \dot{q}= \frac{1}{2} \dot{q}^T M_s \dot{q}$$
Hence $M_s$ defines exactly the same kinetic term as $M$! This means that without loss of generality we can assume that the mass matrix is symmetric.
Or in other words: whenever you write down a kinetic energy with a a non-symmetric mass matrix $M$, you can just forget about the antisymmetric part, and write down the same kinetic energy with $M_s$ - it simply leads to the same system.
I hope that this more detailed explanation helps in understanding the answer (which was also previously given). Don't hesitate to ask, if something is still not clear.
Best,
Zoltan
A: Whenever the mass is expressed as a real matrix $M$, the mass term or something else that matters is a bilinear expression. For example, the effective mass in condensed matter physics is the mass matrix $M$ so that the kinetic term of the Hamiltonian is written as
$$ E_k = \frac{\hbar^2}{2m} \vec k \cdot M^{-1} \cdot \vec k $$
where $M^{-1}$ is the inverse matrix. One may always divide the latter matrix to the symmetric and antisymmetric part. The antisymmetric part doesn't affect the Hamiltonian (doesn't affect the physics) at all because it is being contracted with the symmetric tensor $k_i k_j$. So without a loss of generality, we may demand the matrix to be symmetric, and everyone does so.
In some contexts, like the mass terms for complex fields in quantum field theory, it is only the Hermitian part of the mass matrix that affects the Lagrangian or the Hamiltonian because the term is $\bar\psi \cdot M \cdot \psi$ with an extra bar (complex conjugation). In that case, we assume that the mass matrix is Hermitian for a reason analogous to the previous paragraph; the anti-Hermitian part doesn't contribute.
A: In the world of robotics and dynamical systems the mass matrix is always symmetric. It is also positive definite, a result of kinetic energy
$$ K=\frac{1}{2} \dot{q}^\top M \dot{q} $$
being always positive.
