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Why does the following hold?: $A$ circulant matrix iff it has a representation of the form $F^{-1}DF$ where $D$ is a diagonal matrix and $F$ is a discrete Fourier transformation. I get that $F^{-1}DF$ is circulant but what about the other direction?

Rodrigo de Azevedo
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vandrfl
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2 Answers2

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I think the fastest way to see this is to decompose the circulant matrix into a linear combination of powers of the permutation matrix associated with long permutation, ie. $(n\,n-1\,\ldots\,1)$ (This is basically the definition of a circulant matrix). This permutation matrix obviously has eigenvectors $(\omega^k,\omega^{2\cdot k},\ldots,\omega^{(n-1)\cdot k} )$, so we can diagonalize the permutation matrix (and hence linear combinations of powers of this matrix) by conjugating by a matrix with these eigenvectors as columns, which is the discrete fourier matrix.

There might be a more elegant way to express this, but all my attempts basically boil down to definitions that expand the above.

Callus - Reinstate Monica
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    the OP asks about the opposite direction, this refers to the direct direction – Nikos M. Sep 08 '15 at 23:08
  • @NikosM. I don't think so. I read the OP's statement "I get that $F^{-1}DF$ is circulant" to mean that s/he knows how to show that any diagonal matrix conjugated by the discrete Fourier matrix is circulant. What I showed was that you can take any circulant matrix and decompose it in to sums of powers of the matrix representing the long permutation, and then point out that the discrete Fourier is the matrix of eigenvectors, ie. conjugating by discrete Fourier gives you a diagonal matrix. So that gives you the $D$ matrices, then conjugate again to get the representation the OP is looking for. – Callus - Reinstate Monica Sep 08 '15 at 23:19
  • can you see this for a minute? – Nikos M. Sep 08 '15 at 23:30
  • @NikosM. checked it, I agree with you on the direction I proved. I think your statement about the form of the eigenvalues comes from the wikipedia page, where they are expressed in a certain way, but that expression does not prohibit any set of $n$ numbers from being a set of eigenvectors, which I think is what your proof that the implication was strictly in one direction relied on. Actually, looking at timestamps on comments, it looks like you agree as well, and pterojacktyl gave a quick proof of the opposite direction. – Callus - Reinstate Monica Sep 09 '15 at 03:49
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In case anyone's looking for a detailed proof of diagonalization of circulant matrices and their eigenvalues and eigenvectors, the below PowerPoint has a great explanation between slides 37-45:

(Link)

108_mk
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user3842413
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