# Physical meaning of eigenvectors of mass matrix

What is the physical meaning of the eigenvectors of the mass matrix?

If I consider a 2-dof system with one mass linked to two orthogonal springs and I write the equations in any orthogonal system of coordinates, and get a mass matrix $M$. The eigenvectors correspond to the direction of the springs and the eigenvalues of $M$ correspond to the mass. If I am not too stupid and I write the equations "naturally", I directly get a diagonal matrix $$\begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}$$

So in this simple case, the physical interpretation of the eigenvalues and eigenvectors of the mass matrix is obvious.

Now, instead of considering ponctual masses, I choose a mass density (for e.g. using FEM: the component $(i,j)$ will be given by $m_{ij}=\int \bar{m}\varphi_i(x)\varphi_j(x)\text{d}x$ with $\varphi$ basis functions). The mass matrix will be symmetric positive definite but no longer diagonal. But what is the physical meaning of its eigenvalues and eigenvectors in this case?

Note that I am asking this in the context of Newtonian mechanics, but it seems to be mostly used in other fields: cf google. Hopefully we will get several answers from several fields.

Example of construction of a non-diagonal mass matrix, as in finite-element method.

Consider two bar elements with mass density $\bar m$ and unit length. I assume that for each element, the displacement between the two nodes is linear, which gives to basis function: $$\varphi_1(x)=1-x \quad \varphi_2(x)=x$$

Then, it can be proven (using the principle of virtual work) that the mass matrix of each element is given by M_e, of entries $m_{ij}=\int_0^1 \varphi_i(x)\varphi_j(x)\text{d}x$. The calculation yields:

$$M_e=\left( \begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3} \\ \end{array} \right)$$

Altogother, for my two elements, the mass matrix is: $$M=\left( \begin{array}{ccc} \frac{2}{3} & \frac{1}{6} & 0 \\ \frac{1}{6} & \frac{2}{3} & \frac{1}{6} \\ 0 & \frac{1}{6} & \frac{1}{3} \\ \end{array} \right)$$

The eigenvalues of $M$ are $0.861681, 0.551851, 0.253134$ and the eigenvectors are $$(0.631781,0.739239,0.233192),(0.755789,-0.520657,-0.397113),(0.172148,-0.427132,0.88765)$$ What is their physical meaning?

• Surely it's the effective inertia of the natural modes, no? I hesitate to make this an answer because (a) I probably couldn't reproduce the math without a lot of time or recourse to some books and (b) it seem too easy to be the answer. – dmckee Oct 5 '15 at 22:11
• @dmckee That's what I thought, but the normal modes should depend on the couplings between the masses, not just the mass matrix. But I am not sure how his mass matrix is even defined. – Brian Moths Oct 5 '15 at 22:14
• @dmckee If your talking about normal modes of a linear oscillator, then they would be determined by $M^{-1}K$. But here I did not mention any $K$ :) – anderstood Oct 5 '15 at 22:15
• @NowIGetToLearnWhatAHeadIs I'll add an (even) more explicit calculation of the mass matrix. – anderstood Oct 5 '15 at 22:16
• See, already I'm in over my head. The structure of the normal modes (for two masses on a spring, one is translation and the other is anti-symmetric oscillation) doesn't depend on the strength of the couplings, only the manner in which the masses are coupled. So the modes can have an inertia independent on the coupling strength. Or is that totally borked? – dmckee Oct 5 '15 at 22:23

## 1 Answer

From your description, I think the quantity you are talking about is similar to the concept of inertia where the mass is abstracted into a matrix through its distributions in space. Mathematically, you can understand the matrix properties as following:

1. For a positive semi-definite matrix, the Eigen values are always real and non-negative which makes sure they are physically meaningful.

2. If you sort the Eigen values from the largest to the smallest ones, they are actually the possible range you can obtain through observing them in arbitrary directions in space. The corresponding Eigen vectors correspond to those directions which give you those unique values.

3. In 3-dimentional space, most likely you will have three Eigenvalues for the mass matrix. The corresponding Eigenvectors will help you form a kind of quadratic solid shape (like a ellipsoid) whose direction and radial distance to the center tells you on which direction you can get how much of the mass observed.

4. By observing the mass effect, it is really case-to-case interpreted. You should be able to interpret it for your case by yourself.

Hope this solve your problem in some extent.

• Are you talking about the moment of inertia tensor? – Brian Moths Oct 5 '15 at 22:40
• I think it is a similar idea, but your case can be interpreted differently as I said. – Xiaodong Qi Oct 5 '15 at 22:52