2 × 2 Block Representations of the Moore–Penrose Inverse And

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2 Notation

Throughout this paper let m,n,p,q,s,t be positive integers. We denote by Cm×n the set of all complex m × n matrices and by Cn := Cn×1 the set of all column vectors with n complex m×n n×n entries. We write 0m×n for the zero matrix in C and In for the identity matrix in C . n Let U and V be linear subspaces of C . If U ∩ V = {0n×1}, then we write U ⊕ V for the direct sum of U and V . Let U ⊥ be the orthogonal complement of U . If U ⊆ V , we use the notation V ⊖ U := V ∩ (U ⊥). We write R(M) and N (M) for the column space and the null space of a complex matrix M. Let M ∗ be the conjugate transpose of a complex matrix M. If M is an arbitrary complex m × n matrix, then there exists a unique complex n × m matrix X such that the four equations

(1) MXM = M, (2) XMX = X, (3) (MX)∗ = MX, (4) (XM)∗ = XM (2.1) are fulﬁlled (see [16]). This matrix X is called the Moore–Penrose inverse of M and is desig- nated usually by the notation M †. Following [3, Ch. 1, Sec. 1, Def. 1], for each M ∈ Cm×n we denote by M{j,k,...,ℓ} the set of all X ∈ Cn×m which satisfy equations (j), (k),..., (ℓ) from the equations (1)–(4) in (2.1). Each matrix belonging to M{j,k,...,ℓ} is said to be a {j,k,...,ℓ}-inverse of M. Remark 2.1. Let U be a linear subspace of Cn. Then there exists a unique complex n × n ma- ⊥ n trix PU such that PU x ∈ U and x − PU x ∈ U for all x ∈ C . This matrix PU is called n×n the orthogonal projection matrix onto U . If P ∈ C , then P = PU if and only if the three conditions P 2 = P and P ∗ = P as well as R(P )= U are fulﬁlled. Furthermore, the equation PU ⊥ = In − PU holds true. Our strategy to give a block representation of the Moore–Penrose inverse E† of the block matrix E given in (1.1) consists of three elementary steps:

Step (I) We consider the following factorization problem for orthogonal projection matri- r11 r12 P ces: Find a complex (s + t) × (p + q) matrix R = r21 r22 fulﬁlling R(E) = ER with block r Cs×p r Cs×q r Ct×p r Ct×q entries 11 ∈ , 12 ∈ , 21 ∈ , and 22 ∈ expressible explicitly only using the block entries a, b, c, and d of E. Remark 2.2. Let M ∈ Cm×n and let X ∈ Cn×m. In view of Remark 2.1, then:

2 (a) PR(M) = MX if and only if X ∈ M{1, 3}.

P ∗ (b) R(M ) = XM if and only if X ∈ M{1, 4}. We constructing a suitable {1, 3}-inverse R of E using: Theorem 2.3 (Urquhart [22], see also, e. g. [8, Ch. 1, Sec. 5, Thm. 3]). Let M ∈ Cm×n. (a) Let G := MM ∗ and let G(1) ∈ G{1}. Then M ∗G(1) ∈ M{1, 2, 4}. (b) Let H := M ∗M and let H(1) ∈ H{1}. Then H(1)M ∗ ∈ M{1, 2, 3}.

Applying Theorem 2.3, we will get an explicit block representation of PR(E) = ER in terms of a, b, c, and d.

Step (II) Analogous to Step (I) we construct a suitable complex (s + t) × (p + q) matrix ℓ11 ℓ12 L L E P ∗ LE = ℓ21 ℓ22 fulfilling ∈ {1, 4} and hence R(E ) = . Step (III) With the matrices L and R we apply: Theorem 2.4 (Urquhart [22], see e. g. [8, Ch. 1, Sec. 5, Thm. 4]). If M ∈ Cm×n, then M (1,4)MM (1,3) = M † for every choice of M (1,3) ∈ M{1, 3} and M (1,4) ∈ M{1, 4}. Regarding Remark 2.2, Theorem 2.4 admits the following reformulation: m×n n×m n×m Remark 2.5. Let M ∈ C and let L ∈ C and R ∈ C be such that LM = PR(M ∗) P † and MR = R(M). Then M = LMR. Consider an additive decomposition M = U + V of an m × n matrix M with two m × n ma- ∗ trices U and V , fulfilling UV = 0m×m. In this situation, a result of Cline [5] is applica- ble to obtain a non-trivial representation of M † as a sum of U † and a further matrix. By a 0 0 c Hung/Markham [12] a decomposition E = U + V with U = c 0 and V = 0 d is used in E† this way to derive a block representation of involving only Moore–Penrose inverses of block size matrices. P ∗ Regarding Remarks 2.2 and 2.1 as well as (2.1), the orthogonal projection matrix Q := R(U ) † ∗ ∗ † ∗ ∗ fulfills UQ = UU U = U and VQ = (QV ) = (U UV ) = 0m×n. Consequently, MQ = U m×n and M(In − Q) = V . Conversely, given M ∈ C and an orthogonal projection matrix n×n Q ∈ C , then it is readily checked that U := MQ and V := M(In − Q) fulfill M = U + V ∗ ∗ and UV = 0m×m. Thus, every decomposition M = U + V with UV = 0m×m can be written n×n as M = MQ + M(In − Q) with some orthogonal projection matrix Q ∈ C occurring on the right-hand side and vice versa. Although not explicitly using Cline’s theorem, our investigations involve an analogous decomposition, namely E = PS E + (Ip+q − PS )E with the orthogonal projection matrix PS onto the linear subspace S spanned by the first s columns of E occurring on the left-hand side (see Lemma 3.2).

3 Main results

We consider a complex (p + q) × (s + t) matrix E. Let (1.1) be the block representation of E with p × s block a. Setting a b Y := [a, b], Z := [c, d], S := , T := (3.1) "c# "d#

3 then

Y E = , E = [S, T ]. (3.2) "Z# Let

µ := aa∗ + bb∗, σ := a∗a + c∗c, (3.3) ζ := cc∗ + dd∗, τ := b∗b + d∗d, ρ := ca∗ + db∗, λ := a∗b + c∗d. (3.4)

In view of (3.1), then

µ = YY ∗, σ = S∗S, (3.5) ζ = ZZ∗, τ = T ∗T, (3.6) ρ = ZY ∗, λ = S∗T. (3.7)

Choose µ(1) ∈ µ{1} and σ(1) ∈ σ{1}. Regarding (3.5), then Theorem 2.3 shows that

Y (1,2,4) := Y ∗µ(1), S(1,2,3) := σ(1)S∗ (3.8) fulﬁll

Y (1,2,4) ∈ Y {1, 2, 4}, S(1,2,3) ∈ S{1, 2, 3}. (3.9)

Let

φ := c − (ca∗ + db∗)µ(1)a, ψ := d − (ca∗ + db∗)µ(1)b, (3.10) η := b − aσ(1)(a∗b + c∗d), θ := d − cσ(1)(a∗b + c∗d). (3.11)

Because of (3.8), (3.7), (3.1), (3.4), (3.10), and (3.11), then

η V := [φ, ψ] and W := (3.12) "θ# admit the representations

(1,2,4) ∗ (1) (1) Z(Is+t − Y Y )= Z − ZY µ Y = Z − ρµ Y (3.13) = [c, d] − ρµ(1)[a, b] = [φ, ψ]= V and

(1,2,3) (1) ∗ (1) (Ip+q − SS )T = T − Sσ S T = T − Sσ λ b a η (3.14) = − σ(1)λ = = W. "d# "c# "θ#

Using (3.13), (3.14), (3.9), Remarks 2.2 and 2.1, and [R(Y ∗)]⊥ = N (Y ), we can infer P P V = Z N (Y ), W = [R(S)]⊥ T. (3.15)

4 Let

ν := φφ∗ + ψψ∗, ω := η∗η + θ∗θ. (3.16)

In view of (3.12), (3.15), and Remark 2.1 then ∗ P ∗ ∗ ∗ ∗P ∗ ν = V V = Z N (Y )Z = V Z , ω = W W = T [R(S)]⊥ T = T W. (3.17)

Choose ν(1) ∈ ν{1} and ω(1) ∈ ω{1}. Regarding (3.17), then Theorem 2.3 shows that

V (1,2,4) := V ∗ν(1) and W (1,2,3) := ω(1)W ∗ (3.18) fulﬁll

V (1,2,4) ∈ V {1, 2, 4} and W (1,2,3) ∈ W {1, 2, 3}. (3.19)

Cp×p Cs×s Cq×q Ct×t Cq×q Ct×t Cq×p Obviously, we have µ ∈ ≥ , σ ∈ ≥ , ζ ∈ ≥ , τ ∈ ≥ , ν ∈ ≥ , ω ∈ ≥ , ρ ∈ , and Cs×t Cn×n λ ∈ , where ≥ denotes the set of all non-negative Hermitian complex n × n matrices. Remark 3.1. Let

(1,2,3) (1,2,3) (1,2,4) (1,2,4) (1,2,4) S (Ip+q − TW ) L := (Is+t − V Z)Y , V , R := (1,2,3) . (3.20) " W # h i Regarding (3.20), (3.18), (3.8), (3.7), (3.1), and (3.12), then

∗ (1) ∗ (1) ∗ (1) ∗ ∗ (1) ∗ (1) ∗ (1) L = (Is+t − V ν Z)Y µ , V ν = (Y − V ν ZY )µ , V ν h ∗ ∗ (1) (1) ∗ (1) i ∗ h(1) ∗ (1) (1) i∗ (1) = (Y − V ν ρ)µ , V ν = [Y µ , 0(s+t)×q] + [−V ν ρµ , V ν ] h ∗ (1) ∗ (1) i (1) = Y µ [Ip, 0p×q]+ V ν [−ρµ ,Iq] ∗ ∗ a (1) φ (1) (1) ℓ11 ℓ12 = ∗ µ [Ip, 0p×q]+ ∗ ν [−ρµ ,Iq]= , "b # "ψ # "ℓ21 ℓ22# where

∗ (1) ∗ (1) ∗ (1) ∗ (1) ℓ12 := φ ν , ℓ11 := (a − ℓ12ρ)µ , ℓ22 := ψ ν , ℓ21 := (b − ℓ22ρ)µ , (3.21) and

(1) ∗ (1) ∗ (1) ∗ ∗ (1) ∗ σ S (Ip+q − Tω W ) σ (S − S Tω W ) R = (1) ∗ = (1) ∗ " ω W # " ω W # σ(1)(S∗ − λω(1)W ∗) σ(1)S∗ −σ(1)λω(1)W ∗ = (1) ∗ = + (1) ∗ " ω W # "0t×(p+q)# " ω W # I −σ(1)λ = s σ(1)S∗ + ω(1)W ∗ "0t×s# " It # I −σ(1)λ r r = s σ(1)[a∗, c∗]+ ω(1)[η∗,θ∗]= 11 12 , "0t×s# " It # "r21 r22# where

(1) ∗ (1) ∗ (1) ∗ (1) ∗ r21 := ω η , r11 := σ (a − λr21), r22 := ω θ , r12 := σ (c − λr22). (3.22)

5 Lemma 3.2. Let E := R(E), let S := R(S), and let W := R(W ). Then W = E ⊖ S and PE = PS + PW = ER.

Proof. We ﬁrst check W = E ∩ (S ⊥). Because of (3.19), (3.9), and Remark 2.2(a), we have (1,2,3) (1,2,3) (1,2,3) PW = WW and PS = SS . According to Remark 2.1, hence Ip+q − SS = PS ⊥ . ⊥ By virtue of (3.14), then W = PS ⊥ T follows. From Remark 2.1 we know R(PS ⊥ ) = S . Consequently, W ⊆ S ⊥. Regarding (3.2) and (3.14), we have S ⊆ E and

(1,2,3) (1,2,3) (1,2,3) (1,2,3) −S T −S T W = (Ip+q − SS )T = T − SS T = [S, T ] = E , " It # " It # implying W ⊆ E . Thus, W ⊆ E ∩(S ⊥) is proved. Now we consider an arbitrary w ∈ E ∩(S ⊥). s+t x Then w ∈ E ; so there exists some v ∈ C with w = Ev. Let v = y be the block v x Cs w Sx Ty w S ⊥ representation of with ∈ . Regarding (3.2), then = + . In view of ∈ and Sx ∈ S , furthermore PS ⊥ w = w and PS ⊥ Sx = 0(p+q)×1. Taking additionally into account W = PS ⊥ T , we obtain then

w = PS ⊥ w = PS ⊥ (Sx + Ty)= PS ⊥ Ty = W y, implying w ∈ W . Thus, we have also shown E ∩ (S ⊥) ⊆ W . Therefore, W = E ∩ (S ⊥) holds true. Since S ⊆ E , hence W = E ⊖ S . Consequently, PW = PE − PS follows (see, , , e. g. [23, Thm. 4.30(c)]). Thus, PE = PS +PW . Taking additionally into account PS = SS(1 2 3) , , and PW = WW (1 2 3) as well as (3.14), (3.2), and (3.20), then we can conclude

(1,2,3) (1,2,3) (1,2,3) (1,2,3) (1,2,3) PE = SS + WW = SS + (Ip+q − SS )TW (1,2,3) (1,2,3) (1,2,3) (1,2,3) (1,2,3) S (Ip+q − TW ) = SS (Ip+q − TW )+ TW = [S, T ] (1,2,3) = ER. " W # The following result can be proved analogously. We omit the details.

Lemma 3.3. Let E˜ := R(E∗), let Y˜ := R(Y ∗), and let V˜ := R(V ∗). Then V˜ = E˜ ⊖ Y˜ and PE˜ = PY˜ + PV˜ = LE. Remark 3.4. From Lemma 3.2 and Remark 2.2(a) we can infer R ∈ E{1, 3}, whereas Lemma 3.3 and Remark 2.2(b) yield L ∈ E{1, 4}. Now we obtain the announced block representations of orthogonal projection matrices.

Proposition 3.5. Let E be a complex (p + q) × (s + t) matrix and let (1.1) be the block representation of E with p × s block a. Let S be given by (3.1). Let σ and λ be given by (3.3) and (3.4). Let σ(1) ∈ σ{1}. Let η,θ and W be given by (3.11) and (3.12). Let ω be given by (3.16) and let ω(1) ∈ ω{1}. Then

aσ(1)a∗ + ηω(1)η∗ aσ(1)c∗ + ηω(1)θ∗ P (1) ∗ (1) ∗ R(E) = Sσ S + Wω W = .  cσ(1)a∗ + θω(1)η∗ cσ(1)c∗ + θω(1)θ∗    P (1,2,3) (1,2,3) Proof. In the proof of Lemma 3.2, we have already shown R(E) = SS + WW . Taking additionally into account (3.8), (3.18), (3.1), and (3.12), the assertions follow.

Now we are able to prove a 2 × 2 block representation of the Moore–Penrose inverse.

6 Theorem 3.6. Let E be a complex (p + q) × (s + t) matrix and let (1.1) be the block representation of E with p × s block a. Let Y,S be given by (3.1). Let µ,σ and ρ, λ be given by (3.3) and (3.4). Let µ(1) ∈ µ{1} and let σ(1) ∈ σ{1}. Let φ, ψ and η,θ be given by (3.10) and (3.11). Let V and W be given by (3.12). Let ν and ω be given by (3.16). Let ν(1) ∈ ν{1} and let ω(1) ∈ ω{1}. Then

α β E† = LER = = (Y ∗ −V ∗ν(1)ρ)µ(1)aσ(1)(S∗ −λω(1)W ∗)+(Y ∗ −V ∗ν(1)ρ)µ(1)bω(1)W ∗ "γ δ# + V ∗ν(1)cσ(1)(S∗ − λω(1)W ∗)+ V ∗ν(1)dω(1)W ∗ with

α := ℓ11ar11 + ℓ11br21 + ℓ12cr11 + ℓ12dr21 β := ℓ11ar12 + ℓ11br22 + ℓ12cr12 + ℓ12dr22

γ := ℓ21ar11 + ℓ21br21 + ℓ22cr11 + ℓ22dr21 δ := ℓ21ar12 + ℓ21br22 + ℓ22cr12 + ℓ22dr22 where, for each j, k ∈ {1, 2}, the matrices ℓjk and rjk are given by (3.21) and (3.22), resp.

Proof. According to Lemmas 3.3 and 3.2, we have PR(E∗) = LE and PR(E) = ER. Thus, we can apply Remark 2.5 to obtain E† = LER. Using Remark 3.1 and (1.1), we furthermore obtain

(1) ∗ (1) ∗ ∗ ∗ (1) (1) ∗ (1) a b σ (S − λω W ) LER = (Y − V ν ρ)µ , V ν (1) ∗ "c d# " ω W # h i = (Y ∗ − V ∗ν(1)ρ)µ(1)aσ(1)(S∗ − λω(1)W ∗) + (Y ∗ − V ∗ν(1)ρ)µ(1)bω(1)W ∗ + V ∗ν(1)cσ(1)(S∗ − λω(1)W ∗)+ V ∗ν(1)dω(1)W ∗ as well as ℓ ℓ a b r r α β LER = 11 12 11 12 = . "ℓ21 ℓ22# "c d# "r21 r22# "γ δ#

4 Examples for consequences

In this section, we give some examples of applications of the block representations of orthogonal projection matrices and Moore–Penrose inverses given in Proposition 3.5 and Theorem 3.6. In order to avoid lengthy formulas, we give only hints for computations. Example 4.1. Let N be a positive integer and let U and V be two complementary linear subspaces of CN , i. e., the subspaces U and V fulﬁll U ⊕ V = CN . Then there exists a unique N complex N × N matrix PU ,V such that PU ,V x = u for all x ∈ C , where x = u + v is the unique representation of x with u ∈ U and v ∈ V . This matrix PU ,V is called the oblique projection matrix on U along V and admits the representations † † PU ,V = (PV ⊥ PU ) = [(IN − PV )PU ] , (4.1) see [8] or also [3, Ch. 2, Sec. 7, Ex. 60, formula (80)]. Assume that N = p + q and that U = R(E) and V = R(F) for two matrices E ∈ C(p+q)×(s+t) and F ∈ C(p+q)×(m+n) with block e f representations (1.1) and F = g h , where a is a p × s block and e is a p × m block. Then (4.1) together with Proposition 3.5 and Theorem 3.6 could be used to obtain a block representation of PU ,V in terms of a, b, c, d and e,f,g,h.

7 Example 4.2. Because U ⊕ (U ⊥) = Cn holds true for every linear subspace U of Cn, the Moore–Penrose inverse of matrices is a special case of the uniquely determined {1, 2}-inverse with simultaneously prescribed column space and null space. More precisely, if M ∈ Cm×n and linear subspaces U of Cm and V of Cn with R(M) ⊕ U = Cm and N (M) ⊕ V = Cn are given, then there exists a unique complex n × m matrix X such that the four conditions

MXM = M, XMX = X, R(X)= V , and N (X)= U

(1,2) are fulﬁlled. This matrix X is denoted by MU ,V and admits the representations

(1,2) (1) P P † (1) P P † MV ,U = PV ,N (M)M PR(M),U = ( V ⊥ N (M)) M ( [R(M)]⊥ U ) (4.2) with every M (1) ∈ M{1} (see, e. g. [3, Ch. 2, Sec. 6, Thm. 12]). In particular, (4.2) is valid (1) † † (1,2) for M = M . Furthermore, M = M[N (M)]⊥,[R(M)]⊥ . Example 4.1 could be used to obtain a (1,2) U V block representation of MV ,U , if the subspaces and are given as column spaces of certain matrices, partitioned accordingly to a block representation of M, and if a matrix M (1) ∈ M{1} the corresponding block representation of which is known. (In particular, for M (1) = M † one can use Theorem 3.6.)

5 An alternative approach

In this ﬁnal section, we give alternative representations of the matrices L and R occurring in Theorem 3.6 and Lemmas 3.3 and 3.2. Utilizing these representations, further block representations of the Moore–Penrose inverse E† could possibly be obtained, in particular, in the case of E satisfying additional conditions. We will not pursue this direction any further here. We continue to use the notations given above. C(p+q)×(p+q) Lemma 5.1 (Rohde [17], see, e. g. [8, Ch. 5, Sec. 2, Ex. 10(a)]). Let M ∈ ≥ and m11 m12 (1) let M = m21 m22 be the block representation of M with p × p block m11. Let m11 ∈ m11{1} (1) (1) and let ς := m22 − m21m11 m12. Let ς ∈ ς{1}. Then

(1) (1) (1) (1) (1) (1) m + m m12ς m21m −m m12ς M (1) := 11 11 11 11  (1) (1) (1)  −ς m21m11 ς     belongs to M (1) ∈ M{1}. Remark 5.2. Regarding (3.17), (3.13), (3.14), (3.8), (3.6), and, (3.7), we can infer

∗ (1,2,4) ∗ ∗ (1) ∗ (1) ∗ ν = V Z = Z(Is+t − Y Y )Z = Z(Is+t − Y µ Y )Z = ζ − ρµ ρ (5.1) and

∗ ∗ (1,2,3) ∗ (1) ∗ ∗ (1) ω = T W = T (Ip+q − SS )T = T (Ip+q − Sσ S )T = τ − λ σ λ. (5.2)

Lemma 5.3. Let

G := EE∗ and H := E∗E. (5.3)

8 Let µ(1) ∈ µ{1} and ν(1) ∈ ν{1} and let σ(1) ∈ σ{1} and ω(1) ∈ ω{1}. Then the matrices µ(1) + µ(1)ρ∗ν(1)ρµ(1) −µ(1)ρ∗ν(1) G(1) := (5.4)  −ν(1)ρµ(1) ν(1)  and   σ(1) + σ(1)λω(1)λ∗σ(1) −σ(1)λω(1) H(1) := (5.5) " −ω(1)λ∗σ(1) ω(1) # fulﬁll G(1) ∈ G{1} and H(1) ∈ H{1}. C(p+q)×(p+q) C(s+t)×(s+t) Proof. Clearly, G ∈ ≥ and H ∈ ≥ . Regarding (3.2), (3.5), and (3.6), we ∗ Y ∗ ∗ YY ∗ YZ∗ µ ρ S∗ S∗S S∗T σ λ have G = Z [Y , Z ] = ZY ∗ ZZ∗ = ρ ζ and H = T ∗ [S, T ] = T ∗S T ∗T = λ∗ τ . Taking additionally into account Remark 5.2, thus from Lemm a 5.1 the assertions immediately follow. Finally, in the following result, we not only get new representations for the matrices L and R occurring in Theorem 3.6 and Lemmas 3.3 and 3.2, but also obtain their belonging to the set E{1, 2, 4} and E{1, 2, 3}, resp., thereby improving Remark 3.4. Lemma 5.4. The matrices L and R admit the representations L = E∗G(1) and R = H(1)E∗ and fulﬁll L ∈ E{1, 2, 4} and R ∈ E{1, 2, 3}.

(1,2,4) ∗ P∗ P ∗ Proof. Using (3.9) and Remarks 2.2 and 2.1, we have (Y Y ) = R(Y ∗) = R(Y ) = Y (1,2,4)Y and, analogously, (SS(1,2,3))∗ = SS(1,2,3). Taking additionally into account (3.13), (3.14), (3.8), and (3.7), we thus can conclude ∗ (1,2,4) ∗ ∗ ∗ (1) ∗ ∗ ∗ (1) ∗ V = (Is+t − Y Y )Z = Z − Y µ Y Z = Z − Y µ ρ (5.6) and ∗ ∗ (1,2,3) ∗ ∗ (1) ∗ ∗ ∗ (1) ∗ W = T (Ip+q − SS )= T − T Sσ S = T − λ σ S . (5.7) Regarding (3.2), (5.4), (5.5), (5.1), (5.2), (5.6), (5.7), and Remark 3.1, we obtain (1) ∗ ∗ (1) ∗ ∗ Ip (1) −µ ρ (1) (1) E G = [Y , Z ] µ [Ip, 0p×q]+ ν [−ρµ ,Iq] "0q×p# " Iq # ! (5.8) ∗ (1) ∗ ∗ (1) ∗ (1) (1) = Y µ [Ip, 0p×q] + (Z − Y µ ρ )ν [−ρµ ,Iq]= L and (1) ∗ (1) ∗ Is (1) −σ λ (1) ∗ (1) S H E = σ [Is, 0s×t]+ ω [−λ σ ,It] ∗ "0t×s# " It # ! "T # (5.9) I −σ(1)λ = s σ(1)S∗ + ω(1)(T ∗ − λ∗σ(1)S∗)= R. "0t×s# " It # According to Lemma 5.3, we have G(1) ∈ G{1} and H(1) ∈ H{1}. Taking additionally into account (5.3), (5.8), and (5.9), then L ∈ E{1, 2, 4} and R ∈ E{1, 2, 3} follow from Theorem 2.3.

Observe that Lemma 5.4 in connection with Theorem 3.6 yields a factorization E† = LER with particular matrices L ∈ E{1, 2, 4} and R ∈ E{1, 2, 3}. This gives a special factorization of the kind mentioned in Urquhart’s result (Theorem 2.4), whereby all matrices can be expressed explicitly in terms of the block entries a, b, c, d of the given matrix E.

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