A few days ago I wrote that circulant matrices all have the same eigenvectors. This post will show that it follows that circulant matrices commute with each other.
Recall that a circulant matrix is a square matrix in which the rows are cyclic permutations of each other. If we number the rows from 0, then the kth row takes the last k rows of the top row and moves them to the front.
We can generate two random circulant matrices and confirm that they commute by modifying the Python code from the earlier post.
np.random.seed(42) N = 6 row = np.random.random(N) A = np.array([np.roll(row, i) for i in range(N)]) row = np.random.random(N) B = np.array([np.roll(row, i) for i in range(N)]) AB = np.matmul(A, B) BA = np.matmul(B, A) print(np.absolute(AB - BA).max())
This computes two random 6 by 6 circulant matrices A and B and shows that the difference between AB and BA is on the order of the limit of floating point precision. Specifically, this prints 4.44 × 10-16.
Now for a proof. Let A and B be two matrices of size n that have the same set of independent eigenvectors e1 through en. In the case of circulant matrices we can take ek to be the kth column of the FFT matrix, but the proof here applies to any matrices with common eigenvectors.
Let αk be the eigenvalue for ek as an eigenvector of A and let βk be the eigenvalue for ek as an eigenvector of B. Then
and so multiplying by AB or BA has the same effect on ek since the complex numbers αk and βk commute. Since the eigenvalues form a basis, and multiplication by AB and BA has the same effect on each basis vector, AB = BA.
In short, matrices that share eigenvectors commute because eigenvalues commute.