Suppose we are given that A=\begin{bmatrix}4&11&20&30\\ 0&4&11&20\\ 0&0&4&11\\ 0&0&0&4\end{bmatrix} I need to show that if F is a Fourier matrix then $$F^{-1}AF=D$$ where D is a diagonal matrix containing eigen values of A. I am doubtful before starting solving this problem. Clearly the matrix A is not diagonalizable. Is this scenario possible?
2026-04-05 21:45:42.1775425542
Question related to diagonalization of a matrix.
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Circulant matrices are ensured to be diagonalizable by the Fourier matrix.
The matrix you have is not circulant, but it is Toeplitz.
You can follow this answer for how to construct a circulant matrix from a Toeplitz one.
https://math.stackexchange.com/a/3415198/213607
Depending on what you want to do, this might help you.