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I believe that a (finite) square matrix is diagonalised by the DFT basis $F$ iff it is circulant

$$ A \text{ is circulant} \quad \Leftrightarrow \quad F^{-1} A F \text{ is diagonal} $$

as described in the question Diagonalization of circulant matrices. However, I do not have a reference for this. Is it true? Which source can I reference for this fact?

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pterojacktyl
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  • why isnt the answer on that question enough? actually it stems from the definition of both a circulant matrix (Toeplitz matrix) and the properties of DFT. https://en.wikipedia.org/wiki/Circulant_matrix – Nikos M. Sep 08 '15 at 21:56
  • I would like to have a book/chapter/article/similar that I can cite. – pterojacktyl Sep 08 '15 at 22:46

1 Answers1

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From the wikipedia article

In numerical analysis, circulant matrices are important because they are diagonalized by a discrete Fourier transform, and hence linear equations that contain them may be quickly solved using a fast Fourier transform.[1]

[1]: Davis, Philip J., Circulant Matrices, Wiley, New York, 1970 ISBN 0471057711

http://www-ee.stanford.edu/~gray/toeplitz.pdf

In simple terms, why this holds is that:

  1. A circulant matrix has all its rows being cyclic permutations (cyclic shifts) of the same row.
  2. The fourier transform (DFT) is circular, meaning its basis is polynomials on the roots of unity which are invariant under cyclic shifts (on the unit circle).
  3. Thus if expressed on the DFT basis, each row of the matrix is only a shift away from the reference (original) row. But a shift, in DFT terms, is simple multiplication by a root of unity, thus only the diagonal elements need be non-zero, to describe the appropriate shift (think of a clock, and shifting as changing the angle of the pointer, i.e simple multiplication by a root of unity)
Nikos M.
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  • Thanks! Theorem 3.1 in [1] states that $A$ is circulant $\Rightarrow$ $F^{-1} A F$ is diagonal and $A = F \operatorname{diag}(a) F^{-1}$ $\Rightarrow$ $A$ is circulant, however I can't see the proof for the latter? – pterojacktyl Sep 08 '15 at 22:41
  • @pterojacktyl, well one can do the opposite direction, or use the definition. where is the problem exactly? – Nikos M. Sep 08 '15 at 22:57
  • @pterojacktyl, if you check the wikipedia article you will see the that the diagonal vector has a specific form and is related to the DFT (dont have access to the Davis book myself), i dont think the opposite direction applies in general for arbitrary diagonal vector(??) – Nikos M. Sep 08 '15 at 23:05
  • oh sorry! I said [1] above where I meant [Gray 2006] (the linked PDF) – pterojacktyl Sep 08 '15 at 23:10
  • @pterojacktyl, ok, still i think the opposite direction does not hold for an arbitray diagonal vector $a$, the eigenvalues of a cicrulant matrix have a specific form (similar to roots of unity), so if the diagonal vector does not exhibit this form it will not be a circulant matrix, updating answer with this comment. Opposite direction dows not hold excep;t for special cases of diagonal vectors – Nikos M. Sep 08 '15 at 23:17
  • I think the previous question that I referenced (http://math.stackexchange.com/questions/601168/diagonalization-of-circulant-matrices) and Theorem 3.1 in [Gray 2006] both seem to state that it does hold for an arbitrary vector on the diagonal. – pterojacktyl Sep 08 '15 at 23:17
  • no i checked it again, the answer there talks about the direct direction, not the opposite, it starts from a circulant matrix and how this is diagonalised – Nikos M. Sep 08 '15 at 23:18
  • I tried this in Matlab n = 8; F = fft(eye(n)); a = randn(n,1) + 1i*randn(n,1); A = F * diag(a) * inv(F); norm(A - toeplitz(A(:,1), A(1,:))) and got about 1e-15. I know it's not a proof though :) – pterojacktyl Sep 08 '15 at 23:21
  • @pterojacktyl, still you see the eigenvalues (on which diagonalisation depends) of any circulant matrix are not arbitrary, how can this hold for arbitrary vectors? i think not, (which is also not a proof, but close) – Nikos M. Sep 08 '15 at 23:24
  • hmm ok.. if the opposite direction does not hold, can you show me a vector $a$ such that $A = F \operatorname{diag}(a) F^{-1}$ is not circulant? – pterojacktyl Sep 08 '15 at 23:30
  • If $a$ is the first row of the circulant matrix and $\hat{a} = F a$ is its Fourier transform then $A = F \operatorname{diag}(\hat{a}) F^{-1}$. Since the Fourier transform is invertible, there exists a unique $a$ for every $\hat{a}$ and a unique $\hat{a}$ for every $a$. – pterojacktyl Sep 08 '15 at 23:41
  • yes, that pretty much seems to settle it, still thinking about it though – Nikos M. Sep 09 '15 at 00:00
  • do you think that it is sufficient to cite Gray for the original statement? – pterojacktyl Sep 09 '15 at 00:10
  • i dont know why you need this so much? obviously you gave a clear proof that this is so. i would cite it, along with a brief explanation why it holds – Nikos M. Sep 09 '15 at 00:14