# Inverse Eigenvalue Problem and Least-Squares Problem for Skew-Hermitian {P,K + 1}-Reflexive Matrices

Authors: Chang-Zhou Dong; Hao-Xue Li
Journal: Journal of Mathematics
 (2022)

---

## Abstract

This paper involves related inverse eigenvalue problem and least-squares problem of skew-Hermitian {P,k + 1}-reflexive(antireflexive) matrices and their optimal approximation problems. The above problems are studied by converting them into two simpler cases: k = 1 and k = 2. Firstly, with some special properties of skew-Hermitian {P,k + 1}-reflexive(antireflexive) matrices, the necessary and sufficient conditions for the solvability and the general solution are presented, and the solution of corresponding optimal approximation problems also given, respectively. Then, we give the least-squares solution of AX=B satisfying the special condition by the singular value decomposition. Finally, we give an algorithm and an example to illustrate our results.


---

## Body

## 1. Introduction

Throughout this paper,Cn×m and SHCn×n stand for the sets of all n×m complex matrices and n×n skew-Hermitian matrices, respectively. The symbols AH, A, rA, and A† stand for the conjugate transpose, the Frobenius norm, the rank, and the Moore-Penrose inverse of a matrix A∈Cn×m, respectively. Let Ik be identity matrix of size k, and Jk cross-identity matrix of size k having the elements 1 along the southwest-northeast diagonal and the remaining elements being zeros. Symbol Om×n is the m×n matrix of all zeros entries (if no confusion occurs, we will omit the subscript). For two matrices A,B..., the inner product is defined by A,B:=trBHA. Then, it is apparent that Cn×m is a Hilbert inner product space and the norm that is generated by this inner product is the Frobenius norm.A matrixP∈Cn×n is called Hermitian and {k + 1}-potent matrix if Pk+1=P=PH.Definition 1.
LetP∈Cn×nQ∈Cm×m be two Hermitian and {k + 1}-potent matrices, we say that matrix A∈Cn×m is {P,Q,k + 1}-reflexive(antireflexive) if A=PAQA=−PAQ.Definition 2.
LetP∈Cn×n be a Hermitian and {k + 1}-potent matrix, we say that matrix A∈Cn×n is skew-Hermitian {P,k + 1}-reflexive(antireflexive) if A∈SHCn×n, A=PAPA=−PAP.The (skew-)Hermitian {P,k + 1}-reflexive(antireflexive) matrices, {P,k + 1}-reflexive(antireflexive) matrices can be seen as a particular case of {P,Q,k + 1}-reflexive(antireflexive) matrices. These kind of matrices have many special properties and have been discussed widely. In [1], Li and Wang obtained the (P,Q,k + 1)-reflexive solutions to a system of matrix equations. In [2], Liang et al. discussed the {P,Q,k + 1}-reflexive solution of matrix equation AXB=C, the general solution was given by using Moore–Penrose inverse and the associated optimal approximation problem was considered. In [3, 4], Cao et al. considered the {P, Q,k + 1}-reflexive solutions and {P,k + 1}-reflexive solutions of AX=C,XB=D, respectively. In [5, 6], Shen and Yu et al. studied Hermitian {P,k + 1} (anti)-reflexive solutions of a linear matrix equation of AX=B.There are still some other properties worth exploring about skew-Hermitian {P,k + 1}-reflexive(antireflexive) matrices, in this paper, we will discuss the following three problems:Problem 1 (inverse eigenvalue problem).
GivenX=x1,x2,…,xn∈Cn×m and Λ=diagλ1,λ2,…,λm∈Cm×m, find skew-Hermitian {P,k + 1}-reflexive(antireflexive) matrix A such that(1)AX=XΛ.Problem 2 (optimal approximation problem).
GivenA∗∈Cn×n, find A°∈ΨA such that(2)A°−A∗=minA∈ΨAA−A∗,where ΨA is the solution set of Problem 1.
In the first problem, it is clear thatΛH=−Λ. The inverse problem and the approximation problem have been discussed extensively. For reflexive and antireflexive matrices, bisymmetric matrices, symmetric or ortho-symmetric matrices, skew-Hermitian reflexive and antireflexive matrices, and the inverse eigenvalue problem were discussed in [7–13], respectively. As for the left and right inverse eigenvalue of reflexive and antireflexive matrices, it was also considered in [14]. But the kinds of skew-Hermitian {P,k + 1}-reflexive(antireflexive) matrices have not been discussed, which is the motivation of this work.Problem 3 (least-squares problem).
GivenA,B∈Cm×n, find skew-Hermitian {P,k + 1}-reflexive(antireflexive) matrix X such that(3)AX−B=min.
For the matrix equationAX=B, the Hermitian reflexive, antireflexive, generalized reflexive, and antireflexive solutions have been discussed in [15, 16]. And different solution sets for other equations have also been discussed in [17–20]. Besides, the optimization problems of different matrices have also been studied in [21–24]. To our knowledge, so far, there are relatively few research on the least-squares skew-Hermitian {P,k + 1}-reflexive and antireflexive solutions of AX=B, in this paper, we will explore this problem.


## 2. Properties

In this section, we will give some properties of skew-Hermitian {P,k + 1}-reflexive(antireflexive) matrices.Lemma 1 (Lemma 2.1 in [2]).
LetP∈Cn×n be the Hermitian, then P is {k + 1}-potent matrix if and only if P is idempotent (i.e., P2=P) when k is odd, or tripotent (i.e., P3=P) when k is even. Moreover, there exist U∈Un×n such that
Ifk is odd,(4)P=UIpOUH,or if k is even,(5)P=UIr−Ip−rOUH,where p=rP.Throughout this paper, we always assume the Hermitian and {k + 1}-potent matrix P is fixed, which is given by (4) and (5).From Lemma1, we can see that the general {k + 1}-potent matrix P can be reduced to only two simpler cases: P2=P (i.e., k = 1) and P3=P (i.e., k = 2). Consequently, we will discuss our problems through {P, 2}-reflexive(antireflexive) and {P, 3}-reflexive(antireflexive) constraints.Lemma 2.
X∈Cn×n is skew-Hermitian {P, 2}-reflexive if and only if(6)X=UX11OOOUH,where X11∈SHCp×p.Proof.
IfX is skew-Hermitian {P,2}-reflexive, by Definition 1 and (4), it follows that(7)X=PXP=UIpOUHXUIpOUH.
Let(8)UHXU=X11X12X21X22,then (7) becomes(9)UHXU=UHUIpOUHXUIpOUHU,i.e.,(10)X11X12X21X22=IpOX11X12X21X22IpO,yielding X12=0, X21=0, and X22=0. So, (7) becomes (6).
Conversely, ifX can be expressed as the form (6), then XH=−Xand PXP=X, so X is skew-Hermitian {P,2}-reflexive matrix.Lemma 3.
X∈Cn×n is skew-Hermitian {P,3}-reflexive if and only if(11)X=UX11X22OUH,where X11∈SHCr×r and X22∈SHCp−r×p−r.The proof of it is similar to Lemma2.Assume thatP is idempotent (i.e., k = 1) and PXP=−X, then by premultiplying and postmultiplying it by P, we get PXP=−PXP, which implies that PXP=O and thus, X=O. In this case, the matrix equation AX=XΛ is meaningless. So, we only consider skew-Hermitian {P,3}-antireflexive matrix.Lemma 4.
X∈Cn×n is skew-Hermitian {P,3}-antireflexive if and only if(12)X=UOX12O−X12HOOOOOUH,where X12∈Cr×p−r.Proof.
IfX is skew-Hermitian {P,3}-antireflexive, by Definition 1 and (5), it follows that(13)−X=PXP=UIr−Ip−rOUHXUIr−Ip−rOUH
Let(14)UHXU=X11X12X13X21X22X23X31X32X33,then (13) becomes(15)−UHXU=UHUIr−Ip−rOUHXUIr−Ip−rOUHU,i.e.,(16)−UHXU=Ir−Ip−rOUHXUIr−Ip−rO,yielding X11=X13=X22=X23=X31=X32=X33=0, and X21=U2HXU1=−U1HXU2H=−X12H, so (13) becomes (12).
Conversely, ifX can be expressed as the form (12), then XH=−X and PXP=−X, so X is skew-Hermitian {P,3}-antireflexive matrix.Lemma 5 (Lemma 2.5 in [12]).
GivenX,B∈Cn×n, then AX=B has a solution A∈SHCn×n if and only if(17)BX†X=B,XHB=−BHX,and its general solution can be expressed as(18)A=BX†−BX†HIn−XX†+In−XX†GIn−XX†,where G is arbitrary.Lemma 6 (Lemma 12 in [17]).
IfA∈Cm×n, B∈Cm×k, C∈Ck×l, and D∈Cn×l be known; and X∈Cn×k be variable. Then, the following equation is as follows:(19)AX=BXC=D,has a common solution if and only if(20)AA†B=B,DC†C=D,BC=AD.Also, the general solution(21)X=A†B+In−A†ADC†+In−A†AZIn−CC†,where Z is arbitrary.


## 3. The Solution of Problem1

Based on the analysis in previous section, this section solves the inverse eigenvalue problems of skew-Hermitian {P,2}-reflexive, skew-Hermitian {P,3}-(anti-)reflexive matrices, respectively. And give the general solutions by the Moore-Penrose inverse.Theorem 1.
GivenX=x1,x2,…,xn∈Cn×m, Λ=diagλ1,λ2,…,λm, andP is {k + 1}-potent matrix, let(22)UHX=X1X2,where X1∈Cp×m and X2∈Cn−p×m. Then, AX=XΛ is consistent for {P,2}-reflexive matrix A if and only if X1ΛX1†X1=X1Λ and X1HX1Λ=ΛX1HX1, and the general solution can be expressed as(23)A=UX1ΛX1†−X1ΛX1†HIp−X1X1†+Ip−X1X1†GIp−X1X1†OOOUH,where G∈SHCp×p is arbitrary.Proof.
AssumeA is skew-Hermitian {P,2}-reflexive matrix is the solution of AX=XΛ. By Lemma 2, there exist A11∈SHCp×p such that(24)A=UA11OOOUH,and AX=XΛ. Thus, combining with (22), the above matrix equation AX=XΛ can be expressed as(25)UA11OOOUHX=XΛ,i.e.,(26)A11OOOX1X2=X1X2Λ.
So, it is equivalent toA11X1=X1Λ. By Lemma 5, this equation is consistent if and only if(27)X1ΛX1†X1=X1Λ,X1HX1Λ=−X1ΛHX1=ΛX1HX1.
Moreover, the general solution isA11=X1ΛX1†−X1ΛX1†HIp−X1X1†+Ip−X1X1†GIp−X1X1† and G∈SHCp×p is arbitrary, which implies that (23) holds.
Assume (23) holds. For any G∈SHCp×p, we know that implies A11∈SHCp×p, thus A∈SHCn×n and PAP=A, so A is skew-Hermitian {P,2}-reflexive matrix. From Lemma 5, there exist unique A11 such that A11X1=X1Λ, which is equivalent to(28)A11OOOX1X2=X1ΛX2Λ.
From (22), AX=XΛ. Therefore, A is a solution of Problem 1.Remark 1.
The skew-Hermitian {P,3}-reflexive solution of Problem1 can be reduced similar to Theorem 1, the conclusion is omitted here.Theorem 2.
GivenX=x1,x2,…,xn∈Cn×m,Λ=diagλ1,λ2,…,λm, and Pis {k + 1}-potent matrix, let(29)UHX=X1X2X3.Then,AX=XΛ is consistent for {P,3}-antireflexive matrix A if and only if(30)X1HX1H†ΛX2H=ΛX2H,X1ΛX2†X2=X1Λ,X1HX1Λ=ΛX2HX2,and the general solution can be expressed as(31)A=UOA12O−A12HOOOOOUH,where A12=X1H†ΛX2H+I−X1H†X1HX1ΛX2+I−X1H†X1HUI−X2X2† and U∈Cr×p−r is arbitrary.Proof.
AssumeA is skew-Hermitian {P,3}-antireflexive matrix is the solution of AX=XΛ. By Lemma 4, there exist A12∈Cr×p−r such that(32)A=UOA12O−A12HOOOOOUH,and AX=XΛ. Thus, combining with (29), the above matrix equation is equivalent to(33)A12X2=X1Λ,−A2HX1=X2Λ,.i.e.,(34)X1HA12=ΛX2H,A12X2=X1Λ,.
By Lemma6, this equation is consistent if and only if(35)X1HX1H†ΛX2H=ΛX2H,X1ΛX2†X2=X1Λ,X1HX1Λ=ΛX2HX2.
Moreover, the general solution is(36)A12=X1H†ΛX2H+I−X1H†X1HX1ΛX2+I−X1H†X1HUI−X2X2†,and U∈Cp×p is arbitrary, which implies that (31) holds.
Assume (31) holds. We can obtain AH=−A and PAP=−A. So A is skew-Hermitian {P,3}-antireflexive matrix. From Lemma 6, there exist unique A12 such that X1HA12=−X2ΛH and A12X2=X1Λ, which is equivalent to(37)OA121O−A12HOOOOOX1X2X3=X1ΛX2ΛX3Λ,i.e., AX=XΛ. Therefore, A is a solution of Problem 1.


## 4. The Solution of Problem2

In this section, we will obtain the unique optimal approximation solution for Problem2.If the solution setΨA of Problem 1 is nonempty, it is easy to verify that ΨA is a closed convex set. Thus, there exist unique solution of Problem 2.Lemma 7 (Lemma 4 in [3]).
IfA∈Cq×m, B∈Cq×q, C∈Cm×n, B2=B=BH, andC2=C=CH. Then, A−BAC=minG∈Cq×mA−BGC if and only if(38)BA−GC=0.In that case,(39)minG∈Cq×mA−BGC=BAC−A.Theorem 3.
IfX∈Cn×m, Λ=diagλ1,λ2,…,λm∈Cm×m. Let ΨA be the set of all skew-Hermitian {P,2}-reflexive solutions to Problem 1 and A∗∈Cn×n is a given matrix. Then, A°−A∗=minA∈ΨAA−A∗ has a unique solution A°, which can be expressed as(40)A°=UX1ΛX1†−X1ΛX1†HIp−X1X1†+Ip−X1X1†GIp−X1X1†OOOUH,where G=X1ΛX1†+X1ΛX1†HIp−X1X1†−A11∗ and A11∗=U1HA∗U1∈Cp×p.Proof.
It is easy to verify that the solution setΨA is a closed and convex set in matrix space Cn×n under the Frobenius norm, so Problem 2 has a unique solution. Let(41)UHA∗U=A11∗A12∗A21∗A22∗.
From Theorem1 and the unitary invariance of the Frobenius norm, we have(42)A−A∗2=UA11OOOUH−A∗2=A11OOO−UHA∗U2=A11OOO−A11∗A12∗A21∗A22∗2=A11−A11∗2+A12∗2+A21∗2+A22∗2,where A11=X1ΛX1†−X1ΛX1†HIp−X1X1†+Ip−X1X1†GIp−X1X1†. Hence, minΨAA−A∗2 is equivalent to the problem(43)minX1ΛX1†−X1ΛX1†HIp−X1X1†+Ip−X1X1†GIp−X1X1†−A11∗2,i.e.,(44)minIn−X1X1†GIn−X1X1†−X1ΛX1†+X1ΛX1†HIp−X1X1†−A11∗2.
From Lemma7, it is equivalent to the problem(45)In−X1X1†G−X1ΛX1†+X1ΛX1†HIp−X1X1†−A11∗In−X1X1†=0.
Thus,G=X1ΛX1†+X1ΛX1†HIp−X1X1†−A11∗.The following two theorems obtain the skew-Hermitian {P,3}-reflexive constraint and skew-Hermitian {P,3}-antireflexive constraint of Problem1 and the proof of them is similar to that of Theorem 2.Theorem 4.
IfX∈Cn×m and Λ=diagλ1,λ2,…,λm∈Cm×m. Let ΨA be the set of all skew-Hermitian {P,3}-reflexive solutions to Problem 1 and A∗∈Cn×n is a given matrix. Then, A°−A∗=minA∈ΨAA−A∗ has a unique solution A°, which can be expressed as(46)A°=UA11°OOOA22°OOOOUH,where(47)A11°=X1ΛX1†−X1ΛX1†HIr−X1X1†+Ir−X1X1†G1Ir−X1X1†,A22°=X2ΛX2†−X1ΛX1†HIp−r−X1X1†+Ip−r−X2X2†G2Ip−r−X2X2†,G1=X1ΛX1†−X1ΛX1†HIr−X1X1†−A11∗,G2=X2ΛX2†−X1ΛX1†HIp−r−X1X1†−A22∗,A11∗=U1HA∗U1,A22∗=U2HA∗U2.Theorem 5.
IfX∈Cn×m and Λ=diagλ1,λ2,…,λm∈Cm×m. Let ΨA be the set of all skew-Hermitian {P,3}-antireflexive solutions to Problem 1 and A∗∈Cn×n is a given matrix. Then, A°−A∗=minA∈ΨAA−A∗ has a unique solution A°, which can be expressed as(48)A°=UOA12°O−A12°HOOOOOUH,where(49)A12°=−X1H†X2ΛH+X1ΛX2†+I−X1H†X1HX1ΛX2+I−X1H†X1HUI−X2X2†,U=−X1H†X2ΛH+X1ΛX2†+I−X1H†X1HX1ΛX2−A12∗,A12∗=U1HA∗U2.


## 5. The Solution of Problem3

In this section, by the singular value decomposition, we will consider the least-squares skew-Hermitian {P,2}-reflexive and skew-Hermitian {P,3}-(anti-)reflexive solution of matrix equationAX=B and obtain the general expression.Lemma 8 (Lemma 11 in [4]).
GivenE,F∈Cm×n, Ω1=diaga1,…,am, Ω2=diagb1,…,bn, ai>0i=1,2,…,m, and bj>0j=1,2,…,n. Then, there exist unique matrix S∗∈Cm×n such that(50)Ω1S−E2+SΩ2−F2=min,and S∗ can be expressed as(51)S∗=Φ∗Ω1E+FΩ2,whereΦ=1/ai2+bj2.Lemma 9 (Lemma 2.7 in [6]).
SupposeG∈Cs×s and Σ0=diagσ1,…,σs, σi>0, and i=1,2,…,s. Then, there exists a unique S∗∈SHCn×n such that(52)minΣ0S−G=Σ0S∗−G,and S∗=Φ∗GHΣ0−Σ0G, where(53)Φ=φij∈HCn×n,φ=1σi2+σj2,1≤i,j≤s.Theorem 6.
LetA,B∈Cm×n, and AU=A1,A2, BU=B1,B2. Assume that the singular value decomposition of A1 is as follows:(54)A1=WΣ1OOOVH,where W=W1,W2, and V=V1,V2 are unitary matrices and Σ1=diagσ1,…,σr1, σi>0i=1,2,…,r1, and r1=rankA1. Then, the least-square skew-Hermitian {P,2}-reflexive solution can be expressed as(55)X=UVΦ∗W1HB1V1HΣ1−Σ1W1HB1V1Σ1W1HB1V2−Σ1W1HB1V2HY4VHOOOUH,where Y4 is arbitrary.Proof.
IfX is skew-Hermitian {P,2}-reflexive matrix, from Lemma 2 and (54), we have(56)AX−B2=AUX11OOOUH−B2=AUX11OOO−BU2=A1A2X11OOO−B1B22=A1X11O−B1B22=A1X11−B12+B22.
Thus,AX−B=min is equivalent to A1X11−B12=min. It yielding from (54) that(57)A1X11−B12=WΣ1OOOVHX11−B12=Σ1OOOVHX11V−WHB1V2.
Let(58)VHX11V=Y1Y2−Y2HY4,WHB1V=W1HB1V1W1HB1V2W2HB1V1W2HB1V2,then (57) becomes(59)Σ1OOOY1Y2−Y2HY4−W1HB1V1W1HB1V2W2HB1V1W2HB1V22=Σ1Y1−W1HB1V12+Σ1Y2−W1HB1V22+W2HB1V12+W2HB1V22.
Thus,AX−B=min is solvable if and only if there exist Y1,Y2 such that(60)Σ1Y1−W1HB1V12=min,Σ1Y2−W1HB1V22=min.
By Lemma9,(61)Y1=Φ∗W1HB1V1HΣ1−Σ1W1HB1V1,Φ=1σi2+σj2,Y2=Σ1W1HB1V2.
SubstitutingY1,Y2, and Φ into (58) and X can be expressed as (55).Remark 2.
The skew-Hermitian {P,3}-reflexive constraint least squares problem can be reduced similar to Theorem 5.1, the conclusion is omitted here.Theorem 7.
LetA,B∈Cm×n, AU=A1,A2,A3, and BU=B1,B2,B3. Assume that the singular value decomposition of A1,A2 are as follows:(62)A1=WΣ1OOOVH,A2=MΣ2OOONH,where W=W1,W2, V=V1,V2, M=M1,M2, and N=N1,N2 are unitary matrices; Σ1=diagσ1,…,σr1, σi>0i=1,2,…,r1, r1=rankA1, Σ2=diagδ1,…,δr2, δi>0i=1,2,…,r2, and r2=rankA2. Then, the least-square skew-Hermitian {P,3}-antireflexive solution can be expressed as(63)X=UOX12O−X12HOOOOOUH,where(64)X12=VΦ∗Σ1W1HB2N1−V1HB1HM1Σ2Σ1W1HB2N2−Σ2V2HB1HM1Z4NH.Proof.
IfX is skew-Hermitian {P,3}-antireflexive matrix, then from Lemma 4 and (62), we have(65)AX−B2=AUOX12O−X12HOOOOOUH−B2=AUOX12O−X12HOOOOO−BU2=A1X12−B22+−A2X12H−B12+B32=A1X12−B22+−X12A2H−B1H2+B32.
Thus,AX−B=min is equivalent to A1X12−B22+−X12A2H−B1H2=min. It yielding from (62) that(66)A1X12−B22+−X12A2H−B1H2=Σ1OOOVHX11N−WHB2N2+−VHX12NΣ2OOO−VHB1HM2.
Let(67)VHX12N=Z1Z2Z3Z4,(68)WHB2N=W1HB2N1W1HB2N2W2HB2N1W2HB2N2,VHB1HM=V1HB1HM1V1HB1HM2V2HB1HM1V2HB1HM2,then (66) becomes(69)Σ1Z1−W1HB2N12+−Z1Σ2−V1HB1HM12+Σ1Z2−W1HB2N22+−Z3Σ2−V2HB1HM12+W2HB2N12+W2HB2N22+V1HB1HM22+V2HB1HM22=Σ1Z1−W1HB2N12+Z1Σ2−−V1HB1HM12+Σ1Z2−W1HB2N22+Z3Σ2−−V2HB1HM12+W2HB2N12+W2HB2N22+V1HB1HM22+V2HB1HM22.
Thus,AX−B=min is solvable if and only if there exist Z1,Z2,Z3 such that(70)Σ1Z1−W1HB2N12+Z1Σ2−−V1HB1HM12=min,Σ1Z2−W1HB2N22=min,Z3Σ2−−V2HB1HM12=min.
By Lemma8,(71)Z1=Φ∗Σ1W1HB2N1−V1HB1HM1Σ2,Φ=1σi2+δj2,Z2=Σ1W1HB2N2,Z3=−Σ2V2HB1HM1.
SubstitutingZ1,Z2,Z3Φ into (67), X can be expressed as (63).


## 6. An Algorithm and Numerical Example

We propose the following algorithm to compute the optimal approximate solution of Problem2 over skew-Hermitian {P,2}-reflexive matrix.Example 1.
SupposeX∈C4×3, P∈C4×4, A∗∈C4×4, Λ∈C3×3, and Λ=diag−4.3i,−4.3i,−4.3i,(72)X=2.1+i−3.7i0.623.4i2.3−0.5i−2.7ii2−1.8i2.34.2−i−2i−6.8i,P=0000010000100001,A∗=2.253.01+i2.7i−0.25i1.7+2i3−4i0.25−2i2+0.5i−3.1i0.28i2.70.6i4.12.8+3i3.041.5.
Algorithm 1: Compute the optimal approximate solution of Problem 2.
(1)
InputP∈Cn×n, X∈Cn×m, A∗∈Cn×n, and Λ∈Cm×m.(2)
ComputeU according to (4) and X1 according to (22).(3)
If the conditionsX1ΛX1†X1=X1Λ and X1HX1Λ=ΛX1HX1 hold, then continue. Otherwise, stop.(4)
ComputeA by Theorem 1.(5)
ComputeA11∗ according to (41).(6)
ComputeA° according to (40).
From Algorithm 1, we can obtain(73)U=0001001001001000,X1=4.2−i−2i−6.8ii2−1.8i2.33.4i2.3−0.5i−2.7i,A11∗=1.53.042.8+3i0.6i2.70.28i2+0.5i0.25−2i3−4i.
So, the conditionsX1ΛX1†X1=X1Λ, X1HX1Λ=ΛX1HX1 is hold. By Theorem 1 and problem (40), we can obtain(74)A°=U−4.3i0000−4.3i0000−4.3i00000UH.



---

*Source: 2940377-2022-07-21.xml*