# Use symmetry to simplify eigenvalue problem

This is a mathematics question originating from a problem in physics, so I will give the physics problem, too, to clarify what I mean.

Physics problem

Consider N identical masses joined by identical springs to form a circle. This problem is clearly symmetric under rotation by $\frac{2 \pi}{N}$, because all springs and masses are identical. One can write the equations of motion for each mass as a vector ODE $$\ddot{x} = A \cdot x .$$ The solution to this is a superposition of oscillations with frequencies being the square root of the eigenvalues of the matrix.

Mathematics problem

We want to find the eigenvalues of A using the symmetry described above. Let $$S(x_1, ..., x_N) = (x_2, ..., x_{N+1} = x_1)$$ be the matrix describing the symmetry. Because it is a symmetry, we have $S \cdot A = A \cdot S$ and therefore $$S \cdot x = \lambda x \implies S \cdot A \cdot x = \lambda \cdot A \cdot x$$

Up to this point, this makes sense to me. However, the book now states:

Therfore, it is sufficient to solve the eigenvalue problem of A in every eigenspace of S (reduction of dimension)

This, I don't understand. What is meant by "solving the eigenvalue problem of a matrix in the eigenspace of another matrix" and how does this simplify the initial problem of finding the eigenvalues of A?

• See math.stackexchange.com/q/2016330/115115, in this case because of the shift invariance the transformation will be the Discrete Fourier Transform. – LutzL Nov 19 '16 at 9:55
• Thanks for your comment! Unfortunately, I don't quite see how this answer relates to using the symmetry of the problem... Could you perhaps elaborate? – tomet Nov 19 '16 at 10:12

Because of the symmetry you can assemble the coordinates of $x$ as a polynomial $$x(t,z)=x_0(t)+x_1(t)z+x_2(t)z^2…+x_{n-1}(t)z^{n-1}$$ and the first row of the matrix $$a(z)=a_{00}+a_{0,1}z^{n-1}+a_{0,2}z^{n-2}+…+a_{0,n-1}z$$ so that your equation then reads $$\ddot x(t,z)=a(z)·x(z)\pmod{z^n-1}$$ Evaluated at the roots $z=\xi_k=\exp(i\frac{2\pi k}n)$ of $0=z^n-1$ this reduces to the scalar equations $$\ddot x(t,\xi_k)=a(\xi_k)·x(\xi_k)$$ This evaluation at unit roots corresponds to the inverse Discrete Fourier Transform. To get the coefficients of $x$ back you then only need to apply the forward transform to the solutions, $$x_j(t)=\frac1n\sum_{k=0}^{n-1}{\bar \xi_k}^j·x(\xi_k)$$