On my lecture notes, I came across the statement: 'Inverting the mass matrix is significantly easier than solving a linear system involving the stiffness matrix.'

Can someone explain why this is the case?

The context behind this question is finite element method.

  • 2
    $\begingroup$ You're going to have to provide more context. What kind of equation are you solving? Can you give an example where it's easier to invert the mass matrix than to solve a system with the stiffness matrix? $\endgroup$
    – K. Jiang
    Feb 25 at 6:36
  • 1
    $\begingroup$ @K.Jiang The context is sort of given by the tag 'finite-element-method'. Regardless to the equation that is being solved, 'mass matrix' usually means matrix that corresponds to the $L^2$ scalar product, 'stiffness matrix' corresponds to the form that involves 1st derivatives like in linear elasticity or Laplacian operator. This is common jargon of FEM. $\endgroup$
    – Korf
    Feb 25 at 12:43
  • $\begingroup$ This claim makes not much sense (It seems to imply that the inverse of the stiffness matrix might be computed?). The mass matrix is a structured (at least for mostly regular grids) sparse matrix, its inverse is dense. It is more economic to pre-compute a sparse decomposition, like LU or QR, and use that to solve the linear system of the step equation. $\endgroup$ Feb 25 at 16:32

1 Answer 1


The claim that "Inverting the mass matrix is significantly easier than solving a linear system involving the stiffness matrix" is more of a slogan than a precise theorem. This answer gives the usual surface level explanation without going into the technical details.

To explain the observation that lies behind this slogan lets take a regular and quasi-uniform mesh of the computational domain that is characterized by $h > 0$ being a diameter of a smallest mesh element. Let $\kappa(A)$ denote the condition number of a matrix $A$ computed as $\kappa(A) = \frac{\lambda_{max}}{\lambda_{min}}$ where $\lambda_{max}$ and $\lambda_{min}$ are the highest and lowest eigenvalues of $A$. The lower the condition number is the 'easier' is to solve the equation with the matrix.

Let $M$ denote the mass matrix and $K$ denote the stiffness matrix. It can be shown that $\kappa(M) \sim 1$ with respect to $h$ and $\kappa(K) \sim h^{-2}$. So $h \to 0$ leads to $\kappa(A) \to \infty$ which supports the slogans statement.

Another way in which it is 'easier' to invert the mass matrix than stiffness matrix is through preconditioned Krylov space method. For mass matrix even a diagonal of the matrix can serve as a suitable preconditioner whereas constructing an efficient preconditioner for the stiffness matrix is much more convoluted.

  • $\begingroup$ Thank you for your reply. It clearly answers my question. The lecture note has mentioned something about the condition number but I have hard time in understanding why. $\endgroup$
    – Ariel So
    Feb 26 at 0:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .