Let A={A0,A1,…,Ak-1}⊂Cd1×d2,ζ=e-2πi/k,Fl=∑m=0k-1ζlmAm,0⩽l⩽k-1, and FA=⊕l=0k-1Fl.All operations in indices are modulo k.It is well known that if d1=d2=1 then [As-r]r,s=0k-1=ΦFAΦ∗, where Φ=1k[ζlm]l,m=0k-1.However, to our knowledge it has not been emphasized that FA plays a fundamental role in connection with all the matrices [As-αr]r,s=0k-1,0⩽α⩽k-1, with d1,d2 arbitrary.We begin by adapting a theorem of Ablow and Jenner with d1=d2=1 to the case where d1 and d2 are arbitrary.We show that A=[As-αr]r,s=0k-1 if and only if A=UαFAP∗ where Uα and P are related to Φ,P is unitary, and Uα is invertible (in fact, unitary) if and only if gcd(α,k)=1, in which case we say that A is a proper circulant.We prove the following for proper circulants A=[As-αr]r,s=0k-1:(i) A†=[Br-αs]r,s=0k-1 with Bm=1k∑l=0k-1ζlmFl†,0⩽m⩽k-1.(ii) Solving Az=w reduces to solving Flul=vαl,0⩽l⩽k-1, where v0,v1,…,vk-1 depend only on w.(iii) A singular value decomposition of A can be obtained from singular value decompositions ofF0,F1,…,Fk-1. (iv) The least squares problem for A reduces to independent least squares problems for F0,F1,…,Fk-1. (v) If d1=d2=d, the eigenvalues of [As-r]r,s=0k-1 are the eigenvalues of F0,F1,…,Fk-1, and the corresponding eigenvectors of A are easily obtainable from those of F0,F1,…,Fk-1. (vi) If d1=d2=d and α>1 then the eigenvalue problem for [As-αr]r,s=0k-1 reduces to eigenvalue problems for d×d matrices related to F0,F1,…,Fk-1 in a manner depending upon α.
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