The intrinsic Toeplitz structure and its applications in algebraic Riccati equations

Abstract

In this paper, we derive a Toeplitz-structured closed form of the unique positive semi-definite stabilizing solution for the discrete-time algebraic Riccati equations, especially for the case that the state matrix is not stable. Based on the found form and fast Fourier transform, we propose a new algorithm for solving both discrete-time and continuous-time large-scale algebraic Riccati equations with low-rank structure. It works without unnecessary assumptions, complicated shift selection strategies, or matrix calculations of the cubic order with respect to the problem scale. Numerical examples are given to illustrate its features. Besides, we show that it is theoretically equivalent to several algorithms existing in the literature in the sense that they all produce the same sequence under the same parameter setting.

Appendix. Displacement rank and Toeplitz matrix

In order to prove Lemma 1, we first give a few results to the displacement rank and Toeplitz matrices, and interested readers are referred to the review paper [24] and the references therein.

For any matrix $R\in \mathbb {R}^{pn\times pn}$, its (±)-displacement rank α_±(R,p) with respect to block size p × p, is defined by

$$ \alpha_{+}(R,p):=\text{rank}_{p}(R-Z_{n,p}RZ_{n,p}^{\text{T}}),\qquad \alpha_{-}(R,p):=\text{rank}_{p}(R-Z_{n,p}^{\text{T}}RZ_{n,p}),\qquad $$

where $Z_{n,p}=\begin {bmatrix} 0 & 0 \\ I_{(n-1)p} & 0 \end {bmatrix}_{pn\times pn}$, and rank_p(⋅) is considered as the rank of the linear transformation on the module $\mathbb {R}^{np\times p}$ over the ring $\mathbb {R}^{p\times p}$. For the case p = 1, rank_p(⋅) = rank(⋅), the ordinary rank of matrices in $\mathbb {R}^{n\times n}$.

The definition is based on the following result, namely Lemma 13.

Lemma 13

[21] Given $R_1,R_2\in \mathbb {R}^{pn\times p}$ and $R\in \mathbb {R}^{pn\times pn}$, then

$$ \begin{array}{@{}rcl@{}} R-Z_{n,p}RZ_{n,p}^{\text{T}} = R_{1}R_{2}^{\text{T}} &\iff R=\mathcal{L}_{p\times p}(R_{1})\mathcal{L}_{p\times p}(R_{2})^{\text{T}}, \\ R-Z_{n,p}^{\text{T}}RZ_{n,p} = R_{1}R_{2}^{\text{T}} &\iff R=\mathcal{U}_{p\times p}(R_{1})\mathcal{U}_{p\times p}(R_{2})^{\text{T}}. \end{array} $$

Lemma 13 implies that for a matrix its displacement rank is related to how it can be expressed as a sum of products of block-Toeplitz matrices, as is shown in Lemma 14.

Lemma 14

[21, 23, 24] Given a matrix $R\in \mathbb {R}^{pn\times pn}$.

• 1. Its (+)-displacement rank α₊(R,p) is the smallest integer β such that R can be written in the form

  $$ R=\sum\limits_{i=1}^{\beta}\mathcal{L}_{p\times p}(R_{i})\mathcal{U}_{p\times p}(\widetilde R_{i}), $$

  (A2a)

  where $R_{i},\widetilde R_{i}\in \mathbb {R}^{pn\times p}$.

• 2. Its (−)-displacement rank α₋(R,p) is the smallest integer β such that R can be written in the form

  $$ R=\sum\limits_{i=1}^{\beta}\mathcal{U}_{p\times p}(R_{i})\mathcal{L}_{p\times p}(\widetilde R_{i}), $$

  (A2b)

  where $R_i,\widetilde R_i\in \mathbb {R}^{pn\times p}$.

• 3. If R is symmetric and positive semi-definite, (A2a) and (A2b) can be replaced respectively by

  $$ R=\sum\limits_{i=1}^{\beta}\mathcal{L}_{p\times p}(R_{i})\mathcal{L}_{p\times p}(R_{i})^{\text{T}},\qquad \text{and} \qquad R=\sum\limits_{i=1}^{\beta}\mathcal{U}_{p\times p}(R_{i})\mathcal{U}_{p\times p}(R_{i})^{\text{T}}. $$

• 4. If R is nonsingular, then α₊(R,p) = α₋(R^− 1,p),α₋(R,p) = α₊(R^− 1,p).

Lemma 14 demonstrates the relation between the displacement ranks of a matrix and its inverse, which is actually the theoretical foundation of the fast and superfast algorithms on Toeplitz matrices.

The following result, namely Lemma 15, gives an expression of the inverse related to the displacement rank.

Lemma 15

[25] Given $R\in \mathbb {R}^{pn\times pn}$, suppose

• 1. R is nonsingular, and $R^{-1}=\begin {bmatrix} Q_{1,t} & Q^L_1\\ Q_1 & *\\ \end {bmatrix}=\begin {bmatrix} * & Q_2 \\ Q^L_2 & Q_{2,b}\\ \end {bmatrix}$ where $Q_{1,t},Q_{2,b}\in \mathbb {R}^{p\times p}$ are nonsingular;

• 2. $R-Z_{n,p}RZ_{n,p}^{\text {T}}=\begin {bmatrix} * & * \\ * & D_1{\Sigma } D_2^{\text {T}} \end {bmatrix}$ where $D_1,D_2\in \mathbb {R}^{p(n-1)\times p\alpha }$ and Σ is a diagonal matrix whose diagonal entries are ± 1;

• 3. writing $R=\begin {bmatrix} * & * \\ * & R_s\\ \end {bmatrix}$ where $R_s\in \mathbb {R}^{p(n-1)\times p(n-1)}$, there exist $Q_3,Q^L_3\in \mathbb {R}^{p(n-1)\times p\alpha }$ such that $R_sQ_3=D_1,Q^L_3R_s=D_2^{\text {T}}$.

Then

$$ \begin{array}{@{}rcl@{}} &&{\kern-3.5pc} R^{-1}= -\mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{1}\\0_{p\times p}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{1,t})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{1}}&0_{p\times p} \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\ &&\quad +\mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{2}\\Q_{2,b}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{2,b})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{2}}&Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\ &&\quad +\mathcal{U}_{p\times p\alpha }\left( \begin{bmatrix} Q_{3}\\0_{p\times p\alpha}\\ \end{bmatrix}\right)(I_{n}\otimes W)^{-1}\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0_{p\alpha\times p}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}, \end{array} $$

(A3a)

or alternatively,

$$ \begin{array}{@{}rcl@{}} &&{\kern-3.5pc} R^{-1}= \mathcal{L}_{p\times p}\left( \begin{bmatrix} Q_{1,t}\\Q_{1}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{1,t})^{-1}\mathcal{L}_{p\times p}\left( \begin{bmatrix} Q_{1,t}&{Q^{L}_{1}} \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\ &&\quad -\mathcal{L}_{p\times p}\left( \begin{bmatrix} 0_{p\times p}\\Q_{2}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{2,b})^{-1}\mathcal{L}_{p\times p}\left( \begin{bmatrix} 0_{p\times p}&{Q^{L}_{2}}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\&&\quad -\mathcal{L}_{p\times p\alpha}\left( \begin{bmatrix} 0_{p\times p\alpha}\\Q_{3}\\ \end{bmatrix}\right)(I_{n}\otimes W)^{-1}\mathcal{L}_{p\times p\alpha}\left( \begin{bmatrix} 0_{p\alpha\times p}&{Q^{L}_{3}}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} , \end{array} $$

(A3b)

where $W={\Sigma }-Q^L_3D_1$.

Moreover, if R is symmetric, then there exists a factorization to make D₁ = D₂; for that case, (15) can be rewritten by $Q^L_1=Q_1^{\text {T}},Q^L_2=Q_2^{\text {T}},Q^L_3=Q_3^{\text {T}}$.

Remark 2

Item 2 of Lemma 15 implies that $R-Z_{n,p}^{\text {T}}RZ_{n,p}=\begin {bmatrix} -D_1{\Sigma } D_2^{\text {T}} & *\\ * &* \end {bmatrix}$.

Note that (15) presents a sum of α + 2 products of block-Toeplitz matrices, in which the number of terms may not be the smallest one, namely α_∓(R,p).

In the following, we will derive a sum of the α₊(R,p) = α₋(R^− 1,p) terms, called a shortest sum, to coincide with Lemma 14. Using the same way a sum of α₋(R,p) = α₊(R^− 1,p) terms can also be derived, so we omit the details.

Write $R=\begin {bmatrix} R_{11} & R_{12} \\ R_{21} & R_s\\ \end {bmatrix}$, and then $R-Z_{n,p}RZ_{n,p}^{\text {T}}=\begin {bmatrix} R_{11} & R_{12}\\ R_{21} & D_1{\Sigma } D_2^{\text {T}}\\ \end {bmatrix}$. Thus, α ≤ α₊(R,p) ≤ α + 2, provided that $\text {rank}(D_1{\Sigma } D_2^T)=\text {rank}({\Sigma })=p\alpha$.

On the other hand, by (3), under sufficient nonsingular conditions, it is easy to have

$$ \begin{array}{@{}rcl@{}} &&{\kern-2.1pc} \begin{bmatrix} R_{11} & R_{12} \\ R_{21} & R_{s} \\ \end{bmatrix}^{-1} \\&=& \begin{bmatrix} R_{11}^{-1}+R_{11}^{-1}R_{12}(R_{s}-R_{21}R_{11}^{-1}R_{12})^{-1}R_{21}R_{11}^{-1} & -R_{11}^{-1}R_{12}(R_{s}-R_{21}R_{11}^{-1}R_{12})^{-1}\\ -(R_{s}-R_{21}R_{11}^{-1}R_{12})^{-1}R_{21}R_{11}^{-1} &(R_{s}-R_{21}R_{11}^{-1}R_{12})^{-1} \\ \end{bmatrix} \\&=&\begin{bmatrix} (R_{11}-R_{12}R_{s}^{-1}R_{21})^{-1} &-(R_{11}-R_{12}R_{s}^{-1}R_{21})^{-1}R_{12}R_{s}^{-1}\\ -R_{s}^{-1}R_{21}(R_{11}-R_{12}R_{s}^{-1}R_{21})^{-1} & R_{s}^{-1}+R_{s}^{-1}R_{21}(R_{11}-R_{12}R_{s}^{-1}R_{21})^{-1}R_{12}R_{s}^{-1}\\ \end{bmatrix} . \end{array} $$

Compared with the conditions,

$$ \begin{array}{@{}rcl@{}} Q_{1} = -R_{s}^{-1}R_{21}Q_{1,t},\quad {Q^{L}_{1}} = -Q_{1,t}R_{12}R_{s}^{-1},\quad Q_{1,t} = (R_{11}-R_{12}R_{s}^{-1}R_{21})^{-1}. \end{array} $$

If α₊(R,p) = α, then it has to hold that $R-Z_{n,p}RZ_{n,p}^{\text {T}}=\begin {bmatrix} S_1^{\text {T}}{\Sigma }^{-1}S_2 & S_1^{\text {T}} D_2^{\text {T}}\\ D_1S_2 & D_1{\Sigma } D_2^{\text {T}}\\ \end {bmatrix}$ for some $S_1,S_2\in \mathbb {R}^{p\alpha \times p}$. Clearly S₁,S₂ are of full column rank for R is nonsingular. Noticing Σ^− 1 = Σ, we have

$$ \begin{array}{@{}rcl@{}} Q_{1} &=& -R_{s}^{-1}D_{1}S_{2}Q_{1,t}=-Q_{3}S_{2}Q_{1,t},\quad\\ {Q^{L}_{1}} &=& -Q_{1,t}S_{1}^{\text{T}}D_{2}^{\text{T}}R_{s}^{-1}=-Q_{1,t}S_{1}^{\text{T}}{Q^{L}_{3}},\quad\\ Q_{1,t} &=& (S_{1}^{\text{T}}{\Sigma}^{-1}S_{2}-S_{1}^{\text{T}}D_{2}^{\text{T}}R_{s}^{-1}D_{1}S_{2})^{-1} = (S_{1}^{\text{T}}{\Sigma}^{-1}S_{2}-S_{1}^{\text{T}}{Q^{L}_{3}}D_{1}S_{2})^{-1}=(S_{1}^{\text{T}}WS_{2})^{-1}. \end{array} $$

Thus

$$ \begin{array}{@{}rcl@{}} &&{\kern-2.5pc}\mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{1}\\0\\ \end{bmatrix}\right)(I_{n}\otimes Q_{1,t})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{1}}&0 \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\&=&\!\mathcal{U}_{p\times p}\!\left( \begin{bmatrix} -Q_{3}S_{2}Q_{1,t}\\0\\ \end{bmatrix}\right)(I_{n}\otimes Q_{1,t})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} -Q_{1,t}S_{1}^{\text{T}}{Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\&=&\!\mathcal{U}_{p\times p\alpha}\!\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)\!(I_{n}\! \otimes\! S_{2}Q_{1,t})(I_{n}\otimes Q_{1,t})^{-1}\!(I_{n}\!\otimes\! Q_{1,t}S_{1}^{\text{T}})\mathcal{U}_{p\times p\alpha}\!\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\!\!\!\!\!\!\!\!\!\\ &=&\!\mathcal{U}_{p\times p\alpha}\!\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes S_{2}Q_{1,t}S_{1}^{\text{T}})\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\&=&\!\mathcal{U}_{p\times p\alpha}\!\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)\left( I_{n}\otimes S_{2}(S_{1}^{\text{T}}WS_{2})^{-1}S_{1}^{\text{T}}\right)\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} . \end{array} $$

Note that

$$ \begin{array}{@{}rcl@{}} \left[W^{-1}-S_{2}\left( S_{1}^{\text{T}}WS_{2}\right)^{-1}S_{1}^{\text{T}}\right]WS_{2} =0 . \end{array} $$

Complement S₂ to a nonsingular matrix $\begin {bmatrix} S_2 & S_2^{c} \end {bmatrix}$, and then

$$ \begin{array}{@{}rcl@{}} \left[\!W^{-1}\! -\! S_{2}\!\left( \!S_{1}^{\text{T}}WS_{2}\!\right)^{-1}\!S_{1}^{\text{T}}\!\right]\! W\! \begin{bmatrix} S_{2} & {S_{2}^{c}} \end{bmatrix} \! &=&\! \begin{bmatrix} S_{2} & {S_{2}^{c}} \end{bmatrix}\begin{bmatrix} 0_{p\times p} & -\left( S_{1}^{\text{T}}WS_{2}\right)^{-1}S_{1}^{\text{T}}W{S_{2}^{c}}\\ 0 & I_{p(\alpha-1)}\\ \end{bmatrix} \\ & = &\! \begin{bmatrix} S_{2} & {S_{2}^{c}} \end{bmatrix}\!\begin{bmatrix} \! -\!\left( S_{1}^{\text{T}}WS_{2}\right)^{-1}S_{1}^{\text{T}}W{S_{2}^{c}}\\ I_{p(\alpha-1)}\\ \end{bmatrix}\!\begin{bmatrix} 0 & \!I_{p(\alpha-1)} \end{bmatrix}\!, \end{array} $$

whose rank is p(α − 1). Write

$$ \begin{array}{@{}rcl@{}} W_{1}&=&\begin{bmatrix} S_{2} & {S_{2}^{c}} \end{bmatrix}\begin{bmatrix} -\left( S_{1}^{\text{T}}WS_{2}\right)^{-1}S_{1}^{\text{T}}W{S_{2}^{c}}\\ I_{p(\alpha-1)} \end{bmatrix}\in \mathbb{R}^{p\alpha\times p(\alpha-1)}, \qquad \\ {W^{L}_{1}}&=&\begin{bmatrix} 0 & I_{p(\alpha-1)} \end{bmatrix}\begin{bmatrix} S_{2} & {S_{2}^{c}} \end{bmatrix}^{-1}W^{-1}\in \mathbb{R}^{p(\alpha-1)\times p\alpha}, \end{array} $$

and then $W^{-1}-S_2\left (S_1^{\text {T}}WS_2\right )^{-1}S_1^{\text {T}}=W_1W^L_1$. Hence

$$ \begin{array}{@{}rcl@{}} R^{-1} &= & \mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{2}\\Q_{2,b}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{2,b})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{2}}&Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\\ &&+~\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes W_{1}{W^{L}_{1}})\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\\ &=& \mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{2}\\Q_{2,b}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{2,b})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{2}}&Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\\ && +~\mathcal{U}_{p\times p(\alpha-1)}\left( \begin{bmatrix} Q_{3}W_{1}\\0\\ \end{bmatrix}\right)\mathcal{U}_{p\times p(\alpha-1)}\left( \begin{bmatrix} {W^{L}_{1}}{Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} . \end{array} $$

(A4)

If α₊(R,p) = α + 1, then it holds that

$$ \begin{array}{@{}rcl@{}} R-Z_{n,p}RZ_{n,p}^{\text{T}} &=\begin{bmatrix} S_{1}^{\text{T}}{\Sigma}^{-1}S_{2} & S_{1}^{\text{T}} D_{2}^{\text{T}}\\ D_{1}S_{2} & D_{1}{\Sigma} D_{2}^{\text{T}}\\ \end{bmatrix}+\begin{bmatrix} S_{3} & D_{3}^{\text{T}}\\0 & 0 \end{bmatrix} \end{array} $$

(A5a)

$$ \begin{array}{@{}rcl@{}} &\quad\text{or}\quad \begin{bmatrix} S_{1}^{\text{T}}{\Sigma}^{-1}S_{2} & S_{1}^{\text{T}} D_{2}^{\text{T}}\\ D_{1}S_{2} & D_{1}{\Sigma} D_{2}^{\text{T}}\\ \end{bmatrix}+\begin{bmatrix} S_{3} & 0\\ D_{3} & 0 \end{bmatrix} \end{array} $$

(A5b)

for some $S_1,S_2\in \mathbb {R}^{p\alpha \times p}$, $S_3\in \mathbb {R}^{p\times p}$ and $D_3\in \mathbb {R}^{p(n-1)\times p}$.

Consider (A5a). Then,

$$ \begin{array}{@{}rcl@{}} Q_{1} &= &-R_{s}^{-1}D_{1}S_{2}Q_{1,t}=-Q_{3}S_{2}Q_{1,t},\quad\\ {Q^{L}_{1}} &=& -Q_{1,t}(S_{1}^{\text{T}}D_{2}^{\text{T}}+D_{3}^{\text{T}})R_{s}^{-1}=-Q_{1,t}S_{1}^{\text{T}}{Q^{L}_{3}}-Q_{1,t}D_{3}^{\text{T}}R_{s}^{-1},\quad\\ Q_{1,t} &= &(S_{1}^{\text{T}}{\Sigma}^{-1}S_{2}+S_{3}-(S_{1}^{\text{T}}D_{2}^{\text{T}}+D_{3}^{\text{T}})R_{s}^{-1}D_{1}S_{2})^{-1} \\&=& (S_{1}^{\text{T}}{\Sigma}^{-1}S_{2}+S_{3}-D_{3}^{\text{T}}Q_{3}S_{2}-S_{1}^{\text{T}}{Q^{L}_{3}}D_{1}S_{2})^{-1} \\&=&(S_{3}-D_{3}^{\text{T}}Q_{3}S_{2}+S_{1}^{\text{T}}WS_{2})^{-1}. \end{array} $$

Thus,

$$ \begin{array}{@{}rcl@{}} &&{\kern-2.6pc}\mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{1}\\0\\ \end{bmatrix}\right)(I_{n}\otimes Q_{1,t})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{1}}&0 \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\&=&\mathcal{U}_{p\times p}\left( \begin{bmatrix} -Q_{3}S_{2}Q_{1,t}\\0\\ \end{bmatrix}\right)(I_{n}\otimes Q_{1,t})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} -Q_{1,t}S_{1}^{\text{T}}{Q^{L}_{3}}-Q_{1,t}D_{3}^{\text{T}}R_{s}^{-1}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\!\!\!\!\!\!\!\!\!\!\!\\ &=& \mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes S_{2}Q_{1,t})(I_{n}\otimes Q_{1,t})^{-1}(I_{n}\otimes Q_{1,t}S_{1}^{\text{T}})\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\ && + \mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes S_{2}Q_{1,t})(I_{n}\otimes Q_{1,t})^{-1}(I_{n}\otimes Q_{1,t})\mathcal{U}_{p\times p}\left( \begin{bmatrix} D_{3}^{\text{T}}R_{s}^{-1}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\!\!\!\!\!\!\!\!\!\!\! \\&=& \mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes S_{2}Q_{1,t}S_{1}^{\text{T}})\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\ && + \mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes S_{2}Q_{1,t})\mathcal{U}_{p\times p}\left( \begin{bmatrix} D_{3}^{\text{T}}R_{s}^{-1}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\&=& \mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes S_{2}(S_{3}-D_{3}^{\text{T}}Q_{3}S_{2}+S_{1}^{\text{T}}WS_{2})^{-1}S_{1}^{\text{T}})\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\!\!\!\! \\ && + \mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes S_{2}Q_{1,t})\mathcal{U}_{p\times p}\left( \begin{bmatrix} D_{3}^{\text{T}}R_{s}^{-1}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}. \end{array} $$

Since

$$ \begin{array}{@{}rcl@{}} &&{\kern-2.4pc}W^{-1}-S_{2}(S_{3}-D_{3}^{\text{T}}Q_{3}S_{2}+S_{1}^{\text{T}}WS_{2})^{-1}S_{1}^{\text{T}} \\&=&W^{-1}\left( W-WS_{2}(S_{3}-D_{3}^{\text{T}}Q_{3}S_{2}+S_{1}^{\text{T}}WS_{2})^{-1}S_{1}^{\text{T}}W\right)W^{-1} \\&\overset{(3)}{=}& W^{-1}\left( W^{-1}+S_{2}(S_{3}-D_{3}^{\text{T}}Q_{3}S_{2})^{-1}S_{1}^{\text{T}}\right)^{-1}W^{-1} \\&=& \left( W+WS_{2}(S_{3}-D_{3}^{\text{T}}Q_{3}S_{2})^{-1}S_{1}^{\text{T}}W\right)^{-1} =:W_{1}^{-1} , \end{array} $$

we have

$$ \begin{array}{@{}rcl@{}} R^{-1} &=& \mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{2}\\Q_{2,b}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{2,b})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{2}}&Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\ && +~\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes W_{1})^{-1}\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\\ && -~ \mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes S_{2}Q_{1,t})\mathcal{U}_{p\times p}\left( \begin{bmatrix} D_{3}^{\text{T}}R_{s}^{-1}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\\ &=& \mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{2}\\Q_{2,b}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{2,b})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{2}}&Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\ &&+~\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{n}\otimes W_{1})^{-1}\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}-W_{1}S_{2}Q_{1,t}D_{3}^{\text{T}}R_{s}^{-1}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} .\\ \end{array} $$

(A6)

Similarly, for (A5b),

$$ \begin{array}{@{}rcl@{}} R^{-1} &=& \mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{2}\\Q_{2,b}\\ \end{bmatrix}\right)(I_{n}\otimes Q_{2,b})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{2}}&Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\\ && +~\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} Q_{3}-R_{s}^{-1}D_{3}Q_{1,t}S_{1}^{\text{T}}W_{1}\\0\\ \end{bmatrix}\right)(I_{n}\otimes W_{1})^{-1}\mathcal{U}_{p\times p\alpha}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}},\\ \end{array} $$

(A7)

where $W_1=W+WS_2(S_3-S_1^{\text {T}}Q^L_3D_3)^{-1}S_1^{\text {T}}W$.

To sum up, we have Lemma 16.

Lemma 16

Given $R\in \mathbb {R}^{pn\times pn}$, suppose the conditions in Lemma 15 hold. Then α ≤ α₊(R,p) ≤ α + 2, and the following statements hold.

• 1. if α₊(R,p) = α, then (A4) is a shortest sum.

• 2. if α₊(R,p) = α + 1, then (A6) or (A7) is a shortest sum.

• 3. if α₊(R,p) = α + 2, then (A3a) is a shortest sum.

Moreover, if R is symmetric, then there exists a factorization to make D₁ = D₂; for that case, (A3a), (A4), (A6) and (A7) can be rewritten by $Q^L_1=Q_1^{\text {T}},Q^L_2=Q_2^{\text {T}},Q^L_3=Q_3^{\text {T}},S_1=S_2,D_3=0$.

Then we devote Lemma 17.

Lemma 17

Given $Y\in \mathbb {R}^{p_1\times p_2}$, $Y^L\in \mathbb {R}^{p_2\times p_1}$, $D_{t-1}\in \mathbb {R}^{p_1(t-1)\times p_2}$, $D^L_{t-1}\in \mathbb {R}^{p_2\times p_1(t-1)}$, let

$$ \begin{array}{@{}rcl@{}} T_{t} &=\mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix} Y \\ D_{t-1}\\ \end{bmatrix}\right) =\begin{bmatrix} Y & 0\\ D_{t-1} & T_{t-1} \end{bmatrix}\in \mathbb{R}^{p_{1}t\times p_{2}t} ,\quad \\ {T^{L}_{t}} &=\mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix} Y^{L}& D^{L}_{t-1}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} =\begin{bmatrix} Y^{L} & D^{L}_{t-1}\\ 0 & T^{L}_{t-1} \end{bmatrix}\in \mathbb{R}^{p_{2}t\times p_{1}t} . \end{array} $$

If $I_{p_1t}-T_tT^L_t$ is nonsingular, then

$$ \begin{array}{@{}rcl@{}} (I_{p_{1}t}-T_{t}{T^{L}_{t}})^{-1}\!&=&\! \mathcal{U}_{p_{1}\times p_{1}}\left( \begin{bmatrix} Q_{2}\\Q_{2,b}\\ \end{bmatrix}\right)(I_{t}\otimes Q_{2,b})^{-1}\mathcal{U}_{p_{1}\times p_{1}}\left( \begin{bmatrix} {Q^{L}_{2}}~~ Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}\\ && \!+ \mathcal{U}_{p_{1}\times p_{2}}\!\left( \begin{bmatrix} Q_{3}\\0_{p_{1}\times p_{2}}\\ \end{bmatrix}\right)\!(I_{t}\otimes \left[W+WY^{L}Y W\right])^{-1}\mathcal{U}_{p_{1}\times p_{2}}\!\left( \begin{bmatrix} {Q^{L}_{3}}~~0_{p_{2}\times p_{1}}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} ,\!\!\\ \end{array} $$

(A8a)

where the following equations are solvable and Q_2,b,W + WY^LY W are nonsingular:

$$ \begin{array}{@{}rcl@{}} &&{}{Q^{L}_{3}}\left( I_{p_{1}(t-1)}-D_{t-1}D^{L}_{t-1}-T_{t-1}T^{L}_{t-1}\right)=D^{L}_{t-1}, \end{array} $$

(A8b)

$$ \begin{array}{@{}rcl@{}} \left( I_{p_{1}t}-D_{t-1}D^{L}_{t-1}-T_{t-1}T^{L}_{t-1}\right)Q_{3}&=&D_{t-1}, W=-(I_{p_{2}}+{Q^{L}_{3}}D_{t-1}), \end{array} $$

(A8c)

$$ \begin{array}{@{}rcl@{}} \begin{bmatrix} {Q^{L}_{2}}& Q_{2,b} \end{bmatrix}(I_{p_{1}t}-T_{t}{T^{L}_{t}}) &=& \begin{bmatrix} 0 & I_{p_{1}}\\ \end{bmatrix}, \end{array} $$

(A8d)

$$ \begin{array}{@{}rcl@{}} (I_{p_{1}t}-T_{t}{T^{L}_{t}}) \begin{bmatrix} Q_{2}\\ Q_{2,b} \end{bmatrix}&=& \begin{bmatrix} 0 \\ I_{p_{1}}\\ \end{bmatrix}, \qquad Q_{2,b}\in \mathbb{R}^{p_{1}\times p_{1}}. \end{array} $$

(A8e)

Proof

First consider the case p₁ = p₂ = p. Since

$$ I_{pt}-T_{t}{T^{L}_{t}}=\begin{bmatrix} I_{p}-Y Y^{L} & -Y D^{L}_{t-1} \\ -D_{t-1}Y^{L} & I_{p(t-1)}-D_{t-1}D^{L}_{t-1}-T_{t-1}T^{L}_{t-1} \end{bmatrix}, $$

and

$$ \begin{array}{@{}rcl@{}} &&{\kern-2.5pc}(I_{pt}-T_{t}{T^{L}_{t}})-Z_{t,p}(I_{pt}-T_{t}{T^{L}_{t}})Z_{t,p}^{\text{T}} \\&=&I_{pt}-Z_{t,p}Z_{t,p}^{\text{T}}-T_{t}{T^{L}_{t}}+Z_{t,p}T_{t}{T^{L}_{t}}Z_{t,p}^{\text{T}} \\&=&\begin{bmatrix} I_{p} & \\ & 0 \end{bmatrix}-\begin{bmatrix} Y & 0\\ D_{t-1} & T_{t-1} \end{bmatrix}\begin{bmatrix} Y^{L} & D^{L}_{t-1}\\ 0 & T^{L}_{t-1} \end{bmatrix}+\begin{bmatrix} 0 & 0\\ T_{t-1} & 0\\ \end{bmatrix}\begin{bmatrix} 0& T^{L}_{t-1} \\0&0\\ \end{bmatrix} \\& =&\begin{bmatrix} I_{p}-Y Y^{L} & -Y D^{L}_{t-1} \\ -D_{t-1}Y^{L} & -D_{t-1}D^{L}_{t-1} \end{bmatrix} , \end{array} $$

we have $\alpha _+(I_{pt}-T_tT^L_t,p)=2$. By Lemma 16, since α = 1, the case falls in Item 2 with substitutions

$$ D_{3}\leftarrow 0, S_{3}\leftarrow I_{p},D_{1}\leftarrow D_{t-1},{\Sigma}\leftarrow -I_{p(t-1)},D_{2}^{\text{T}}\leftarrow D^{L}_{t-1}, S_{1}^{\text{T}}\leftarrow Y, S_{2}\leftarrow Y^{L}. $$

Then (A6) (or equivalently (A7)) becomes

$$ \begin{array}{@{}rcl@{}} (I_{pt}-T_{t}{T^{L}_{t}})^{-1} \! &= &\! \mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{2}\\Q_{2,b}\\ \end{bmatrix}\right)(I_{t}\otimes Q_{2,b})^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{2}}~~Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\ && +\mathcal{U}_{p\times p}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{t}\otimes [W+WY^{L}YW])^{-1}\mathcal{U}_{p\times p}\left( \begin{bmatrix} {Q^{L}_{3}}~~0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} ,\\ \end{array} $$

(A6)

where $Q_2,Q_{2,b},Q^L_2,Q_3,Q^L_3,W$ is as in (17).

Then consider the case p₁ > p₂. Complement Y to a p₁ × p₁ matrix $\widetilde Y=\begin {bmatrix} Y & 0 \end {bmatrix}$ and similarly for $\widetilde D_{t-1}=\begin {bmatrix} D_{t-1} & 0 \end {bmatrix}, \widetilde Y^L=\begin {bmatrix} Y^L\\0 \end {bmatrix}, \widetilde D^L_{t-1}=\begin {bmatrix} D^L_{t-1} \\ 0 \end {bmatrix}$. Immediately we are able to use the result above on the case p₁ = p₂ to obtain $\left [I_{p_1t}-{\mathscr{L}}_{p_1\times p_1}\left (\begin {bmatrix}\widetilde Y\\\widetilde D_{t-1}\end {bmatrix}\right ){\mathscr{L}}_{p_1\times p_1}\left (\begin {bmatrix}\widetilde Y^L&\widetilde D^L_{t-1}\end {bmatrix}^{\text {T}}\right )^{\text {T}}\right ]^{-1}$. Note that

$$ \begin{array}{@{}rcl@{}} \mathcal{L}_{p_{1}\times p_{1}}\!\left( \begin{bmatrix}\widetilde Y\\\widetilde D_{t{\kern-.5pt}-{\kern-.5pt}1}\end{bmatrix}\right)\!\mathcal{L}_{p_{1}{\kern-.5pt}\times{\kern-.5pt} p_{1}}\!\left( \!\begin{bmatrix}\widetilde Y^{L}&\!\widetilde D^{L}_{t{\kern-.5pt}-{\kern-.5pt}1}\end{bmatrix}^{\text{T}}\right)^{\text{T}} \! &=&\!\begin{bmatrix} * & 0 & * & 0 & \cdots\\ {\vdots} & {\vdots} & {\vdots} & {\vdots} & \\ * & 0 & * & 0 & \cdots\\ \end{bmatrix}\begin{bmatrix} * &{\cdots} & *\\ 0 &{\cdots} & 0\\ * &{\cdots} & *\\ 0 &{\cdots} & 0\\ {\vdots} & &{\vdots} \\ \end{bmatrix} \\&=&\!\begin{bmatrix} * & * & \cdots\\ {\vdots} & {\vdots} & \\ * & * & \cdots\\ \end{bmatrix}\begin{bmatrix} * &{\cdots} & *\\ * &{\cdots} & *\\ {\vdots} & &{\vdots} \\ \end{bmatrix} \\& = &\! \mathcal{L}_{p_{1}\times p_{2}}\!\left( \begin{bmatrix} Y\\ \!D_{t{\kern-.5pt}-{\kern-.5pt}1}\end{bmatrix}\!\right)\!\mathcal{L}_{p_{1}\!\times\! p_{2}}\!\left( \!\begin{bmatrix} Y^{L}& \!D^{L}_{t{\kern-.5pt}-{\kern-.5pt}1}\end{bmatrix}^{\text{T}}\!\right)^{\text{T}} \! \!. \end{array} $$

Thus, $\widetilde Q_2=Q_2,\widetilde Q^L_2=Q^L_2,\widetilde Q_{2,b}=Q_{2,b}$, and $\widetilde Q_3=\begin {bmatrix} Q_3 & 0 \end {bmatrix}, \widetilde Q^L_3=\begin {bmatrix} Q^L_3\\0 \end {bmatrix}$. Therefore,

$$ \begin{array}{@{}rcl@{}} \widetilde W&=& -I_{p_{1}} - \begin{bmatrix} {Q^{L}_{3}} \\ 0 \end{bmatrix}\begin{bmatrix} D_{t-1} & 0 \end{bmatrix}=\begin{bmatrix} -I_{p_{2}}-{Q^{L}_{3}}D_{t-1} & 0\\ 0 & -I_{p_{1}-p_{2}}\\ \end{bmatrix}=\begin{bmatrix} W & \\ & -I_{p_{1}-p_{2}} \end{bmatrix}, \\ \widetilde W\widetilde Y^{L}\widetilde Y\widetilde W&=&\begin{bmatrix} W & \\ & -I_{p_{1}-p_{2}} \end{bmatrix}\begin{bmatrix} Y^{L}\\0 \end{bmatrix}\begin{bmatrix} Y& 0 \end{bmatrix}\begin{bmatrix} W & \\ & -I_{p_{1}-p_{2}} \end{bmatrix} =\begin{bmatrix} WY^{L}YW & \\ & 0 \end{bmatrix}. \end{array} $$

Hence

$$ \begin{array}{@{}rcl@{}} &&{\kern-2.7pc}\mathcal{U}_{p_{1}\times p_{1}}\left( \begin{bmatrix} \widetilde Q_{3}\\0\\ \end{bmatrix}\right)(I_{t}\otimes [\widetilde W+\widetilde W\widetilde Y^{L}\widetilde Y\widetilde W])^{-1}\mathcal{U}_{p_{1}\times p_{1}}\left( \begin{bmatrix} \widetilde {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} \\&=&\begin{bmatrix} * & 0 & * & 0 & \cdots\\ {\vdots} & {\vdots} & {\vdots} & {\vdots} & \\ * & 0 & * & 0 & \cdots\\ \end{bmatrix}\begin{bmatrix} * & &&&\\ & -I &&&\\ &&* &&\\ &&& -I &\\ &&&&\ddots\\ \end{bmatrix}\begin{bmatrix} * &{\cdots} & *\\ 0 &{\cdots} & 0\\ * &{\cdots} & *\\ 0 &{\cdots} & 0\\ {\vdots} & &{\vdots} \\ \end{bmatrix} \\&=&\begin{bmatrix} * & * & \cdots\\ {\vdots} & {\vdots} & \\ * & * & \cdots\\ \end{bmatrix}\begin{bmatrix} * &&\\ & *&\\ && {\ddots} \\ \end{bmatrix}\begin{bmatrix} * &{\cdots} & *\\ * &{\cdots} & *\\ {\vdots} & &{\vdots} \\ \end{bmatrix} \\&=& \mathcal{U}_{p_{1}\times p_{2}}\left( \begin{bmatrix} Q_{3}\\0\\ \end{bmatrix}\right)(I_{t}\otimes [ W+ W Y^{L} Y W])^{-1}\mathcal{U}_{p_{1}\times p_{2}}\left( \begin{bmatrix} {Q^{L}_{3}}&0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} . \end{array} $$

Finally consider the case p₁ < p₂. Complement Y to a p₂ × p₂ matrix $\widetilde Y=\begin {bmatrix} Y \\ 0 \end {bmatrix}$ and similarly for $\widetilde D_{t-1}^{\text {T}}=\begin {bmatrix} * & 0 & * & 0&\cdots \end {bmatrix}$ where $D_{t-1}^{\text {T}}=\begin {bmatrix} * & * & \cdots \end {bmatrix}$, and $\widetilde Y^L=\begin {bmatrix} Y^L&0 \end {bmatrix}, \widetilde D^L_{t-1}=\begin {bmatrix} * & 0 & * & 0&\cdots \end {bmatrix}$ where $D^L_{t-1}=\begin {bmatrix} * & * & \cdots \end {bmatrix}$. To make things clear, two permutations P,P_s are used to make $P\begin {bmatrix} \widetilde Y\\ \widetilde D_{t-1} \end {bmatrix}=\begin {bmatrix} Y\\ D_{t-1} \\ 0 \end {bmatrix}, P_s\widetilde D_{t-1}=\begin {bmatrix} D_{t-1} \\ 0 \end {bmatrix}$. So $\begin {bmatrix} \widetilde Y^L & \widetilde D^L_{t-1} \end {bmatrix}P^{\text {T}}=\begin {bmatrix} Y^L & D^L_{t-1} & 0 \end {bmatrix}, \widetilde D^L_{t-1} P_s^{\text {T}}=\begin {bmatrix} D^L_{t-1} & 0 \end {bmatrix}$, and

$$ \begin{array}{@{}rcl@{}} P\mathcal{L}_{p_{2}\times p_{2}}\left( \begin{bmatrix}\widetilde Y\\\widetilde D_{t-1}\end{bmatrix}\right) &=&\begin{bmatrix} \mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix} Y\\ D_{t-1} \end{bmatrix}\right)\\ 0\\ \end{bmatrix} , \\ \mathcal{L}_{p_{2}\times p_{2}}\left( \begin{bmatrix}\widetilde Y^{L}&\widetilde D^{L}_{t-1}\end{bmatrix}^{\text{T}}\right)^{\text{T}}P^{\text{T}} &=&\begin{bmatrix} \mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix} Y^{L} & D^{L}_{t-1} \end{bmatrix}^{\text{T}}\right)^{\text{T}}& 0\\ \end{bmatrix} . \end{array} $$

Then we use the result above on the case p₁ = p₂ to obtain $\left [\vphantom {\left (\begin {bmatrix}\widetilde Y\\\widetilde D_{t-1}\end {bmatrix}\right )}I_{p_2t}-{\mathscr{L}}_{p_2\times p_2}\right .$ $\left .\left (\begin {bmatrix}\widetilde Y\\\widetilde D_{t-1}\end {bmatrix}\right ){\mathscr{L}}_{p_2\times p_2}\left (\begin {bmatrix}\widetilde Y^L&\widetilde D^L_{t-1}\end {bmatrix}^{\text {T}}\right )^{\text {T}}\right ]^{-1}$. Note that

$$ \begin{array}{@{}rcl@{}} &&{\kern-2.5pc} P\left[I_{p_{2}t}-\mathcal{L}_{p_{2}\times p_{2}}\left( \begin{bmatrix}\widetilde Y\\\widetilde D_{t-1}\end{bmatrix}\right)\mathcal{L}_{p_{2}\times p_{2}}\left( \begin{bmatrix}\widetilde Y^{L}&\widetilde D^{L}_{t-1}\end{bmatrix}^{\text{T}}\right)^{\text{T}}\right]P^{\text{T}} \\ && =\begin{bmatrix} I_{p_{1}t}-\mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix}Y\\ D_{t-1}\end{bmatrix}\right)\mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix}Y^{L}& D^{L}_{t-1}\end{bmatrix}^{\text{T}}\right)^{\text{T}} & \\ & I_{(p_{2}-p_{1})t} \end{bmatrix}. \end{array} $$

(A7)

Thus,

$$ \begin{array}{@{}rcl@{}} P\begin{bmatrix} \widetilde Q_{2}\\ \widetilde Q_{2,b}\\ \end{bmatrix} &=&P\left[I_{p_{2}t}-\mathcal{L}_{p_{2}\times p_{2}}\left( \begin{bmatrix}\widetilde Y\\\widetilde D_{t-1}\end{bmatrix}\right)\mathcal{L}_{p_{2}\times p_{2}}\left( \begin{bmatrix}\widetilde Y^{L}&\widetilde D^{L}_{t-1}\end{bmatrix}^{\text{T}}\right)^{\text{T}}\right]^{-1}P^{\text{T}}P\begin{bmatrix} 0\\ I_{p_{2}}\\ \end{bmatrix} \\&=&\begin{bmatrix} \left[ I_{p_{1}t}-\mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix}Y\\ D_{t-1}\end{bmatrix}\right)\mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix}Y^{L}& D^{L}_{t-1}\end{bmatrix}^{\text{T}}\right)^{\text{T}} \right]^{-1} & \\ & I_{(p_{2}-p_{1})t} \end{bmatrix} \begin{bmatrix} 0\\\begin{bmatrix} I_{p_{1}} & 0 \end{bmatrix}\\0\\\begin{bmatrix} 0 & I_{p_{2}-p_{1}} \end{bmatrix} \end{bmatrix} \\&=&\begin{bmatrix} \left[ I_{p_{1}t}-\mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix}Y\\ D_{t-1}\end{bmatrix}\right)\mathcal{L}_{p_{1}\times p_{2}}\left( \begin{bmatrix}Y^{L}& D^{L}_{t-1}\end{bmatrix}^{\text{T}}\right)^{\text{T}} \right]^{-1}\begin{bmatrix} 0 \\ I_{p_{1}} \end{bmatrix} & 0\\ 0 &\begin{bmatrix} 0 \\ I_{p_{2}-p_{1}} \end{bmatrix} \end{bmatrix} \\& =&\begin{bmatrix} \begin{bmatrix} Q_{2}\\ Q_{2,b} \end{bmatrix} & 0\\ 0 &\begin{bmatrix} 0 \\ I_{p_{2}-p_{1}} \end{bmatrix} \end{bmatrix} , \end{array} $$

and similarly, $\begin {bmatrix} \widetilde Q^L_2& \widetilde Q_{2,b}\\ \end {bmatrix}P^{\text {T}} =\begin {bmatrix} \begin {bmatrix} Q^L_2 & Q_{2,b} \end {bmatrix} & 0\\ 0 &\begin {bmatrix} 0 & I_{p_2-p_1} \end {bmatrix} \end {bmatrix}$. Therefore, $\widetilde Q_{2,b}=\begin {bmatrix} Q_{2,b} & 0 \\ 0 & I_{p_2-p_1} \end {bmatrix}$ and

$$ \begin{array}{@{}rcl@{}} &&{}P\mathcal{U}_{p_{2}\times p_{2}}\left( \begin{bmatrix} \widetilde Q_{2}\\ \widetilde Q_{2,b}\\ \end{bmatrix}\right)P^{\text{T}}P(I_{t}\otimes \widetilde Q_{2,b})^{-1}P^{\text{T}}P\mathcal{U}_{p_{2}\times p_{2}}\left( \begin{bmatrix} \widetilde {Q^{L}_{2}}& \widetilde Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}P^{\text{T}} \\&=&\! \begin{bmatrix} \mathcal{U}_{p_{1}\times p_{1}}\!\left( \begin{bmatrix} Q_{2}\\ Q_{2,b}\\ \end{bmatrix}\right) & 0\\ 0& I_{(p_{2}-p_{1})t} \end{bmatrix}\!\begin{bmatrix} (I_{t}\otimes Q_{2,b})^{-1} & \\ & I_{(p_{2}-p_{1})t} \end{bmatrix}\!\begin{bmatrix} \!\mathcal{U}_{p_{1}\times p_{1}}\!\left( \begin{bmatrix} {Q^{L}_{2}}& Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right) & 0\\ 0& I_{(p_{2}-p_{1})t} \end{bmatrix}^{\text{T}}\!\!\!\!\!\!\!\! \\&=&\! \begin{bmatrix} \mathcal{U}_{p_{1}\times p_{1}}\left( \begin{bmatrix} Q_{2}\\ Q_{2,b}\\ \end{bmatrix}\right)(I_{t}\otimes Q_{2,b})^{-1}\mathcal{U}_{p_{1}\times p_{1}}\left( \begin{bmatrix} {Q^{L}_{2}}& Q_{2,b}\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} & \\ & I_{(p_{2}-p_{1})t} \end{bmatrix} . \end{array} $$

Similarly,

$$ \begin{array}{@{}rcl@{}} P_{s}\widetilde Q_{3} &=&P_{s}\left[I_{p_{2}(t-1)}-\widetilde D_{t-1}\widetilde D^{L}_{t-1}-\widetilde T_{t-1}\widetilde T^{L}_{t-1}\right]^{-1}P_{s}^{\text{T}}P_{s}\widetilde D_{t-1} \\& =&\begin{bmatrix} \left[I_{p_{1}(t-1)}- D_{t-1} D^{L}_{t-1}- T_{t-1} T^{L}_{t-1}\right]^{-1} & \\ & I_{(p_{2}-p_{1})(t-1)} \end{bmatrix}\begin{bmatrix} D_{t-1}\\0 \end{bmatrix} =\begin{bmatrix} Q_{3} \\ 0 \end{bmatrix} , \end{array} $$

and similarly, $\widetilde Q^L_3P_s^{\text {T}} =\begin {bmatrix} Q^L_3 & 0 \end {bmatrix} ,$ and then

$$ \begin{array}{@{}rcl@{}} \widetilde W &=& -I_{p_{2}} - \widetilde{Q}_{3}^{L}P_{s}^{\text{T}}P_{s}\widetilde D_{t-1}=-I_{p_{2}}-{Q_{3}^{L}} D_{t-1}=W, \\ \widetilde W\widetilde Y^{L}\widetilde Y\widetilde W& = & W\begin{bmatrix} Y^{L}&0 \end{bmatrix}\begin{bmatrix} Y\\ 0 \end{bmatrix}W = WY^{L}YW. \end{array} $$

Hence

$$ \begin{array}{@{}rcl@{}} &&{\kern-2.5pc}P\mathcal{U}_{p_{2}\times p_{2}}\left( \begin{bmatrix} \widetilde Q_{3}\\ 0\\ \end{bmatrix}\right)(I_{t}\otimes [\widetilde W+\widetilde W\widetilde Y^{L}\widetilde Y\widetilde W])^{-1}\mathcal{U}_{p_{2}\times p_{2}}\left( \begin{bmatrix} \widetilde {Q^{L}_{3}}& 0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}}P^{\text{T}} \\&=& \begin{bmatrix} \mathcal{U}_{p_{1}\times p_{2}}\left( \begin{bmatrix} Q_{3}\\ 0\\ \end{bmatrix}\right) \\ 0 \end{bmatrix} (I_{t}\otimes [ W+ W Y^{L} Y W])^{-1} \begin{bmatrix} \mathcal{U}_{p_{1}\times p_{2}}\left( \begin{bmatrix} {Q^{L}_{3}}& 0\\ \end{bmatrix}^{\text{T}}\right) \\ 0 \end{bmatrix}^{\text{T}} \\&=& \begin{bmatrix} \mathcal{U}_{p_{1}\times p_{2}}\left( \begin{bmatrix} Q_{3}\\ 0\\ \end{bmatrix}\right)(I_{t}\otimes [ W+ W Y^{L} Y W])^{-1}\mathcal{U}_{p_{1}\times p_{2}}\left( \begin{bmatrix} {Q^{L}_{3}}& 0\\ \end{bmatrix}^{\text{T}}\right)^{\text{T}} & \\ & 0 \end{bmatrix} . \end{array} $$

□

Finally, Lemma 1 comes out as a corollary.

Proof Proof of Lemma 1

Use Lemma 17 with $Y=-(Y^L)^{\text {T}}, D_{t-1}=-(D^L_{t-1})^{\text {T}}$. Then we take Q_2,b = Q₁ to obtain the result. □