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HiPPO LegS, LegT, and LagT matrices have normal-plus-low-rank form after specified similarities

Statement

Index rows and columns by 0,…,N−1. Define the real LegS coefficient matrix GS by GnkS=−(2n+1)(2k+1) when n>k, GnnS=−(n+1), and GnkS=0 when n<k. In the LegS coefficient ODE for t>0 this matrix is multiplied by the time-dependent scalar 1/t; the matrix assertion below does not make that ODE time-invariant. For θ>0, define the LegT state generator by GnkT=−(2n+1)(−1)n−k/θ when n≥k and GnkT=−(2n+1)/θ when n<k. Define the original HiPPO LagT state generator by GnkL=−1 when n≥k and GnkL=0 when n<k. Here X∗ is conjugate transpose, a complex matrix H is normal when H∗H=HH∗, and a square matrix V is unitary when V∗V=VV∗=I. For LegS and LagT use the identity similarity; for LegT use the invertible diagonal similarity S0=diag⁡(2n+1)n=0N−1diag⁡((−1)n)n=0N−1. Each transformed matrix is a difference H−UV∗ with H normal and rank⁡(UV∗)≤1 for LegS and LagT and rank⁡(UV∗)≤2 for LegT. Consequently the equivalent finite-dimensional complex state-space matrix has a normal-plus-low-rank representation, and a unitary change of state coordinates gives diagonal-plus-low-rank form. Under any state change x=Sx~, the input and output vectors change to S−1B and S∗C, respectively.

Facts & Assumptions

Given: The finite matrices in the Statement, the parameter θ>0 for LegT, and finite-dimensional complex state coordinates.

[F1]

DPLR structured state-space parameterization: a normal-plus-low-rank matrix has form A=VΛV∗−PQ∗ with unitary V and diagonal Λ; under the state change x~=V∗x, the input and output vectors become V∗B and V∗C.

[F2]

Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely: every normal endomorphism of a finite-dimensional complex inner product space has an orthonormal basis of eigenvectors.

Proof

1.1F1algebra

Put wn=2n+1 and define the real skew-symmetric matrix S by Snk=−wnwk/2 for n>k, Snk=wnwk/2 for n<k, and Snn=0. The transpose is −S, so over C its conjugate transpose is also −S; hence S is normal. With v=w/2, entrywise comparison gives GS=(S−12I)−vv∗: below the diagonal the entries are −wnwk, above they cancel to zero, and on the diagonal they are −12−(2n+1)/2=−(n+1). Since a scalar shift of a normal matrix is normal and vv∗ has rank at most one, this is the required LegS decomposition. This assertion concerns the finite coefficient matrix only; the factor 1/t in the LegS ODE remains time-dependent.

1.2F1algebra

Define S with entries −1/2 below the diagonal, +1/2 above it, and zero on the diagonal. It is real skew-symmetric and therefore normal. For e=(1,…,1)T, entrywise comparison gives GL=(S−12I)−12ee∗: below the diagonal the two terms sum to −1, above they sum to zero, and on the diagonal they sum to −1. Thus the original LagT generator is a normal matrix minus a rank-at-most-one matrix.

1.3F1algebra

For LegT let D=diag⁡(2n+1)n=0N−1 and E=diag⁡((−1)n)n=0N−1. Define Tnk=1 for n≥k, and Tnk=(−1)k−n for n<k. Then direct entry evaluation gives GT=−DETE/θ: for n≥k, (−1)n(−1)k=(−1)n−k, while for n<k, (−1)n(−1)k−n(−1)k=1. Write T=S+R, where Rnk=1 when n and k have the same parity and is zero otherwise. The matrix R is the sum of the outer products of the even-index and odd-index indicator vectors, so its rank is at most two. For opposite parities, Snk=1 below the diagonal and Snk=−1 above it; for equal parities and on the diagonal, Snk=0. Thus ST=−S. The invertible diagonal matrix S0=D1/2E gives S0−1GTS0=−θ−1D1/2SD1/2−θ−1D1/2RD1/2. The first term is real skew-symmetric and hence normal; the second has rank at most two. The diagonal similarity is essential: this proves the claim for an equivalent state-space matrix, not unitary similarity of the raw LegT matrix.

2.1F1F2step 1.1step 1.2step 1.3algebra∎

In each case write the resulting generator as H−UV∗ with H normal: for LegS and LagT the factors are the displayed real vectors divided by 2, and for LegT use the even and odd indicator columns multiplied by D1/2/θ as both factors. By [F2], choose a unitary matrix V0 with V0∗HV0=Λ diagonal. Then V0∗(H−UV∗)V0=Λ−(V0∗U)(V0∗V)∗, preserving the low-rank bound. If the original coordinates were changed by S0, the total state change is x=S0V0z; substitution into x′=Gx+Bu, y=C∗x gives z′=(V0∗S0−1GS0V0)z+V0∗S0−1Bu and output vector V0∗S0∗C. These are exactly the state-coordinate transformations required for an equivalent DPLR model.

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