When dealing with an undamped MDOF dynamic system, one can find the natural frequency (eigenvalue) and corresponding modeshape-vector (eigenvector) for mode $n$ by solving the eigenvalue problem:

$$ [K] \{ \phi \}_n = \omega_n^2 [M] \{ \phi \}_n $$

Furthermore, it is stated in a lot of litterature, that for any two modes, $n$ and $r$, their eigenvectors are orthogonal, because they can be shown to fulfill the following:

$$ \{ \phi \}_n [M] \{ \phi \}_r = 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \{ \phi \}_n [K] \{ \phi \}_r = 0 $$

Two vectors are, however generally, defined to be orthogonal if: $\;\;\; \{ \phi \}_n^T \{ \phi \}_r = 0 \;\;\;$, and this is actually not true for the modeshape-vectors you find by solving the eigenvalue problem for an undamped dynamic system!

Just to be completely clear, i will give an example using a 2DOF system:

2DOF Dynamic system

For such a system, the stiffness and mass matrices will be the following:

$$K = \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; M = \begin{bmatrix} m_1 & 0\\ 0 & m_2 \end{bmatrix} $$

I will just select some arbitrary values:

$$m_1 = 50 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; m_2 = 20 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; k_1 = 90 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; k_2 = 50$$

which gives:

$$K = \begin{bmatrix} 140 & -50 \\ -50 & 50 \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; M = \begin{bmatrix} 50 & 0\\ 0 & 20 \end{bmatrix} $$

By solving the eigenvalue problem using these matrices, i get the following eigenvectors:

$$ \{ \phi \}_1 = \begin{Bmatrix} 0.575\\ 1.000 \end{Bmatrix} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \{ \phi \}_2 = \begin{Bmatrix} -0.695\\ 1.000 \end{Bmatrix} $$

which gives:

$$ \{ \phi \}_1^T \{ \phi \}_2 = 0.6 \neq 0 $$

By using the mass or stiffness matrix, sure enough we get:

$$ \{ \phi \}_1 [M] \{ \phi \}_2 = 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \{ \phi \}_1 [K] \{ \phi \}_2 = 0 $$

but still $\{ \phi \}_1^T \{ \phi \}_2 \neq 0$ so they are not orthogonal! So why do we call them orthogonal?

  • 2
    $\begingroup$ They are mutualy orthogonal with repect to the inner product defined by the $M$ matrix. $\endgroup$
    – mike stone
    Mar 26, 2021 at 12:37
  • $\begingroup$ The eigenvalue problem is $[K] \{ \phi \}_n = \omega_n^2 [M] \{ \phi \}_n$. You forgot the ^2 in the post. $\endgroup$
    – JAlex
    Mar 26, 2021 at 20:15
  • $\begingroup$ JAlex. Sorry, i fixed it $\endgroup$
    – MMU_SDU
    Mar 27, 2021 at 3:02

1 Answer 1


No reference I encountered on the topic so far so has called the eigenmodes orthogonal with respect to the standard scalar product, for the obvious reason you point out: They are not orthogonal in general.

In any instance something like orthogonality is mentioned, it means (as the comments to your question state) orthogonality with respect to the inner product that is induced by the mass- or the stiffness-matrix.

One might think that the modes are orthogonal in general, because they are solutions to an equation that is generally referred to as "eigenvalue problem", but it's not an eigenvalue problem in the usual sense. Even if it were, this would not in general mean orthogonality.

There is actually only one special case when the modes are orthogonal, and that is for the case that mass and stiffness matrix commute - in that case, both matrices share the same set of eigenvectors, and those are then also a solution to finding the eigenmodes. Those eigenvectors will then be orthogonal, because mass- and stiffness matrix are orthogonal.

If the modes were orthogonal in general, this would mean that one could diagonalize two abitrary symmetric matrices at the same time (see this question of mine on stack exchange) - That can't be!


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