To set up my question, I will briefly recap the derivation that is usually given for eigenmodes of a system:

If there is no force present, and a body is described by a vector $u$, then the equations of motion can be given (with $M$ the mass matrix, and $K$ the stiffness matrix):

\begin{align} M \ddot{u}(t) - K u(t) = 0 \end{align}

We did neglect the damping matrix at this point. If we want to know which motions will behave in a very simple oscillating manner, we use \begin{align} u(t) = e^{i \omega t} u_0 \end{align} And then the equations simplify to:

\begin{align} -\omega^2 M u_0 - K u_0 = (-\omega^2 M + K) u_0 = 0 \end{align} or equivalently \begin{align} \omega^2 u_0 = - M^{-1} K u_0 \end{align} or equivalently \begin{align} \frac{1}{\omega^2} u_0 = - K^{-1} M u_0 \end{align} Solving this for $u_0$ leads to the solutions $u_i$, which are equivalently:

  • Eigenvectors of $K - \omega^2 M$ with the Eigenvalue 0
  • Eigenvectors of $ - M^{-1} K$ with Eigenvalue $\omega^2$
  • Eigenvectors of $ - K^{-1} M$ with Eigenvalue $\frac{1}{\omega^2}$

In either way, these vectors will form a complete orthogonal basis, meaning that linear combinations of them will span the entire space of possible configurations of the system, and that none of these eigenmodes can be modelled by a linear combination of the other ones.

After the Eigenvectors and Eigenvalues have been found, many texts that I encountered (for example this text on smart FEM, p. 3.8 or this script of the university munich) proposes normalizations for the found eigenvectors in the form \begin{align} u_j^T M u_i = \delta_{ij} \end{align} Or \begin{align} u_j^T K u_i = \omega_i^2 \delta_{ij} \end{align}

This is supposedly called "orthogonality property of normal modes". I do however fail to understand why those equations can be achieved by scaling the eigenvectors $u_i$. What those equations basically tell us is that the $u_i$ are eigenvectors of $M$ and $K$ as well - but why would they be? From what we calculated, they are only Eigenvectors of $- M^{-1} K$ or $- K^{-1} M$.

Of course, if the $u_i$ were Eigenvectors of both $M$ and $K$ simultaneously, then they'd also be solutions to our initial problem. But the reverse implication is not true, unless there are some additional assumptions at hand, that are not talked about.

Are there any other assumptions to add to this topic?

I could imagine that for example one just sets the mass matrix to a constant times unity, that seems far fetched.

  • $\begingroup$ Consider posting at Math SE if you don't receive and answer here. $\endgroup$
    – AJN
    May 11, 2023 at 15:32
  • $\begingroup$ @AJN I will consider this, but ultimately I feel that from mathematics perspective, there isn't anything else to be said than what is already contained in the question. $\endgroup$ May 11, 2023 at 17:01
  • $\begingroup$ The first equation is not "achieved" by scaling the eigenvectors. It's a property of the eigenvectors that u^T_j M u_i =\alpha \delta_{ij}. Scaling the eigenvectors only allows you to get \alpha to be equal to 1. This in turn is used in another property of the eigenvectors, namely that \frac{u^T_i K u_i}{u^T_i M u_i} = \omega_i^2. Combining the first normalization with the second equation gets you the second equation. Finally (and I think this is the crux of your question), the eigenvalues are NOT orthogonal between themselves, only through M or K. $\endgroup$
    – Zegpi
    Oct 24, 2023 at 1:16

1 Answer 1


I think your question stems from a simple misunderstanding. The equation $$ u_j^T M u_i = \delta_{j} $$ would imply $u_i$ is an eigenvalue of $M$ only if $u_i^T u_j=0$, i.e. the eigenvalues of $(M,K)$ were orthogonal. This is not the case, I'll show an example below to illustrate this.

Assume you have the following system: Problem set-up The mass and stiffness matrices for this problem are $$ \mathbf{M}= \begin{bmatrix} 2 & 0 \\ 0 & 1 \\ \end{bmatrix}m. \qquad \qquad \mathbf{K}= \begin{bmatrix} \phantom{-}3 & -1 \\ -1 & \phantom{-}1 \\ \end{bmatrix}k. $$

For simplicity, assume $k=m=1$. The eigenvalues and eigenvectors for the $(M,K)$ pair are: $$ \lambda_1=\frac{1}{2} \qquad \qquad \lambda_2=2 \\ u_1=\left[\begin{matrix} 1 \\ 2 \end{matrix}\right] \qquad \qquad u_2=\left[\begin{matrix} \phantom{-}1 \\ -1 \end{matrix}\right] $$

Note that I "normalized" the vectors by setting the first component to $1$.

So, first thing to notice, $u_1^T u_2 \neq0 $. However, $$ u_2^T M u_1 =0 \qquad \qquad u_2^T K u_1 =0, $$ so the property $u_j^T M u_i =0$ comes automatically.

And $$ u_1^T M u_1 =6 \qquad \qquad u_2^T M u_2 =3. $$ So, by scaling $u_1$ by $\frac{1}{\sqrt{6}}$ and $u_2$ by $\frac{1}{\sqrt{3}}$ you can get the property $$ u_i^T M u_i =1. $$

I hope this clears things somewhat.


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