For the equation $\frac{\partial u}{\partial t}=-(-\Delta)u$ with zero boundary condition  using finite element method. Applying the corresponding weak formulation and taking $v=\phi_{j},j=1,2,...,N$, We obtain the following equation.$$\sum\limits_{i=1}^{N} \frac{du_{i}}{dt}\int _{\Omega} \phi_{i}\phi_{j}d\Omega=-\sum\limits_{i=1}^{N}u_{i} \int _{\Omega} \nabla\phi_{i}.\nabla\phi_{j}d\Omega\,. $$Introducing the mass $\mathcal {M}_{i,j}=\int _{\Omega} \phi_{i}\phi_{j}d\Omega$ and stiffness matrices $\mathcal{K}_{i,j}=\int _{\Omega} \nabla\phi_{i}.\nabla\phi_{j}d\Omega$.Then we have $$\frac{d\bf{u}}{dt}=-\mathcal{M}^{-1}\mathcal{K}\bf{u}\, ,$$so far, we have derived an approximation matrix representation of the Diffusion equation for the finite element method that.is $A=\mathcal{M}^{-1}\mathcal{K}$.So my question is why $A$ is not only nonsymmetric, also dense.Although both $\mathcal{M}$ and $\mathcal{K}$ are positive-semidefinite and sparse.
Why the product of symmetric-sparse matrices is not symmetric, or dense
linear algebra;finite element