Consider a linear system of $n$ differential equations with constant coefficients corresponding to a physical scenario where I have $n$ coupled oscillators (like $n$ masses attached by springs in series or something like that, also no friction). I was told that if I have $n$ oscillators in my coupled system, then I will always have $n$ distinct normal modes corresponding to this system.

For that to be the case, though, my coefficient matrix would need to have $n$ distinct eigenvalues which correspond to each normal mode. Going further, if my system had $n$ distinct normal modes it would mean that all of my normal modes could be decoupled since my coefficient matrix could be diagonalized by a suitable change of basis, thus suggesting that all linear coupled systems with constant coefficients have normal modes which are independent from one another.

My question pertains to how we could say in general that our matrices will have $n$ distinct eigenvalues as from what I am aware, generally speaking that may not be the case. Any help would be appreciated in understanding this generalization.

  • 2
    $\begingroup$ In displacement coordinates with respect to the equilibrium positions you get a symmetric real matrix of dimension $n$, specifying your problem. This matrix can have degeneracies and complex eigenvalues, so you don't have necessarily $n$ distinct real eigenvalues, but you can always diagonalize such a matrix and pick a set of $n$ orthogonal eigenvectors. Are you simply looking for a proof why real symmetric matrices are diagonalizable? $\endgroup$
    – Hans Wurst
    Feb 4 at 9:27
  • $\begingroup$ I know that part, I just didn’t realize that we will always end up with real symmetric matrices. I can see why that would happen in the case of n coupled springs, but is that true for all linear coupled systems? $\endgroup$ Feb 4 at 17:27
  • $\begingroup$ No, not for all linear coupled systems. Just change an off-diagonal element to differ from the "mirror element" across the diagonal and you have a linear coupled system where this is no longer true. That would then be of course no longer be a system of coupled harmonic oscillators, rendering a normal mode analysis moot. $\endgroup$
    – Hans Wurst
    Feb 4 at 18:07
  • $\begingroup$ So what you are saying is that for a system of coupled harmonic oscillators, I can always expect to get a symmetric matrix, and as such, I can always diagonalize and essentially decouple my normal modes? Also if I have a real symmetric matrix, I know that I will never get complex eigenvalues, but I might have degeneracies which is fine since I will still get enough orthogonal eigenvectors, but I thought that each eigenvalue corresponds to a different normal mode, so what does it mean for my normal modes if I have repeated eigenvalues? Would these just be degenerate normal modes, or the same? $\endgroup$ Feb 4 at 18:19
  • 1
    $\begingroup$ Yes, for harmonic systems you always get a real symmetric matrix. But not every general linearly coupled system has to be derived from a harmonic system, which I pointed out, since you mentioned "all linear coupled systems" in your comment. Repeated eigenvalues usually correspond to degenerate normal modes, unless you have accidental degeneracy, where two eigenvalues values just happen to have the same value. That is also possible, although rarely the case. $\endgroup$
    – Hans Wurst
    Feb 5 at 8:29

1 Answer 1


The easiest way to solve the problem is to take advantage of the Lagrangian formalism. Denoting the $n$ generalized coordinates by $x_1, \ldots x_n$, the most general form of a Lagrangian of the system under consideration is given by $$ L(x_1, \ldots x_n, \dot{x}_1, \ldots \dot{x}_n)= \frac{1}{2}\sum\limits_{i, j} \dot{x}_i M_{ij} \dot{x}_j - \frac{1}{2} \sum_{i,j} x_i V_{i j} x_j = \frac{1}{2} \dot{x}^TM\dot{x}-\frac{1}{2}x^T V x$$ with a strictly positive symmetric $n \times n$ mass matrix $M^T=M \gt 0$ and a nonnegative symmetric matrix $V^T= V \ge 0$.

In a first step, the mass matrix $M$ can be diagonalzed by a real orthogonal transformation $R$, such that $M = R^T \hat{M} R$ with the diagonal matrix $\hat{M} = {\rm diag}(m_1, \ldots, m_n)$ (where $m_i \gt 0$) and $R^T = R^{-1}$. Introducing the new generalized coordinates $y= \hat{M}^{1/2} R \, x \Leftrightarrow x= R^T \hat{M}^{-1/2} y$, the Lagrangian takes the form $$L(y, \dot{y}) = \frac{1}{2} \dot{y}^T \dot{y}- \frac{1}{2}y^T K y$$ with the symmetric non-negative coupling matrix $K= \hat{M}^{-1/2}R V R^T \hat{M}^{-1/2} $.

In the final step, the matrix $K$ is diagonalized, $K =S^T \hat{K} S$, with $\hat{K} = {\rm diag} (\omega_1^2, \ldots, \omega_n^2)$ (with $\omega_\alpha^2 \ge 0$ as $K$ is non-negative) and a real orthogonal matrix $S$. Introducing the normal coordinates $Q= S y = S \hat{M}^{1/2} R \, x $, the Lagrangian takes the final form $$L(Q, \dot{Q})= \frac{1}{2} \dot{Q}^T \dot{Q} -\frac{1}{2} Q^T \hat{K} Q= \frac{1}{2} \sum\limits_{\alpha = 1}^n (\dot{Q}_\alpha^2 - \omega_\alpha^2 \, Q_\alpha^2).$$ Depending on the physical context, it is possible that some of the eigenvalues $\omega_\alpha$ might vanish (so-called zero modes), which occurs e.g. in the situation where two mass points are connected by a spring, which, however, does not affect the linear motion of the center-of-mass coordinates. Likewise, some of the eigenvalues may be degenerate, which allows you to arbitrarily rotate the normal coordinates in the corresponding eigenspace.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.