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A Toeplitz matrix has lower-shift displacement rank at most two

Statement

Let T=(ti−j)0≤i,j<n be any square Toeplitz matrix, and let Z=Zn be the lower shift of Lower-shift displacement rank and generator pairs. With e0=(1,0,…,0)T and t0=T00, T−ZTZ∗=e0(e0∗T−t0e0∗)+(Te0)e0∗. Consequently, its displacement rank is at most two. A width-two generator is G=[ e0  Te0 ],H∗=[e0∗T−t0e0∗e0∗].

Facts & Assumptions

Given: A square Toeplitz matrix T=(ti−j), its first diagonal entry t0, and the lower shift Z.

[F1]

The lower-shift displacement is ∇Z(T)=T−ZTZ∗, and a generator is a factorization ∇Z(T)=GH∗. Lower-shift displacement rank and generator pairs

[F2]

Toeplitz entries satisfy Tij=ti−j for zero-based indices. Toeplitz, circulant, and finite-index conventions for structured solvers

[F3]

Proof

1.1F1F2F3algebra

For i,j≥1, (ZTZ∗)ij=Ti−1,j−1=t(i−1)−(j−1)=ti−j=Tij by [F2]. If i=0 or j=0, the corresponding entry of ZTZ∗ is zero because the first row of Z and the first column of Z∗ are zero. Thus T−ZTZ∗ agrees with T on its first row and first column and is zero in the interior. The displayed expression has exactly these entries: its first row is e0∗T, its first column is Te0, and its (0,0) entry is t0.

2.1F1step 1.1algebra∎

The factorization in the statement multiplies to the displayed expression from step 1.1. It has two columns, so the rank of T−ZTZ∗ is at most two by the definition of rank and a product through C2.

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