Generalization of Gracia's Results

Transcription

1 Electronic Journal of Linear Algebra Volume 30 Volume 30 (015) Article Generalization of Gracia's Results Jun Liao Hubei Institute, Heguo Liu Hubei University, Yulei Wang Henan University of Technology, Zuohui Wu Hubei University, Xingzhong Xu Hubei University, Follow this and additional works at: Recommended Citation Liao, Jun; Liu, Heguo; Wang, Yulei; Wu, Zuohui; and Xu, Xingzhong (015), "Generalization of Gracia's Results", Electronic Journal of Linear Algebra, Volume 30 DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information, please contact

2 Volume 30, pp 43-5, June GENERALIZATION OF GRACIA S RESULTS JUN LIAO, HEGUO LIU, YULEI WANG, ZUOHUI WU, AND XINGZHONG XU Abstract Let α be a linear transformation of the m n-dimensional vector space M m n (C) over the complex field C such that α(x) = AX XB, where A and B are m m and n n complex matrices, respectively In this paper, the dimension formulas for the kernels of the linear transformationsα and α 3 aregiven, which generalizes the workofgracia in[jmgracia Dimension of the solution spaces of the matrix equations [A, [A, X]] = 0 and [A[A, [A, X]]] = 0 Linear and Multilinear Algebra, 9:195 00, 1980] Key words Linear transformation, Dimension formulas, Jordan canonical form AMS subject classifications 15A4, 15A7 1 Introduction The notation used in this paper is standard, see [, 3] for example Let C be the complex field Suppose A M m m (C) and B M n n (C) Let α AB be a linear transformation of M m n (C) defined by α AB (X) = AX XB, for X M m n (C) IfA=B, then wewill writeα A insteadofα AA forbrevity Inthe caseofnoconfusion, we write α = α AB for short The well known dimension formula of the kernel kerα A is due to Frobenius [3, Theorem VII1] Then Gracia has obtained the dimension formulas of kerα A and kerα 3 A in [1] It is obvious that the kernels of the liner transformations α and α 3 are the solutions of the matrix equations A(AX XB) (AX XB)B = 0 and A[A(AX XB) (AX XB)B] [A(AX XB) (AX XB)B]B = 0, respectively In this paper, we obtain the dimensions of kerα and kerα 3, which generalizes Gracia s results Received by the editors on September 18, 014 Accepted for publication on May 1, 015 Handling Editor: Joao Filipe Queiro School of Mathematics and Statistics, Hubei University, Wuhan, 43006, PR China (jliao@hubueducn, ghliu@hubueducn, zuohuiwoo@163com, xuxingzhong407@16com) Supported by the National Natural Science Foundation of China (grant no and ) School of Mathematics, Henan University of Technology, Zhengzhou, , PR China (yulwang@163com) Supported by the National Natural Science Foundation of China (grant no ) 43

3 Volume 30, pp 43-5, June Jun Liao, Heguo Liu, Yulei Wang, Zuohui Wu, and Xingzhong Xu For convenience, we introduce the following notations Suppose that the elementary divisors of A and B are (λ λ 1 ) p1, (λ λ ) p,,(λ λ u ) pu and (λ µ 1 ) q1, (λ µ ) q,,(λ µ v ) qv, respectively Let E n be the unit matrix of size n and let N n = be the square matrix of size n in which all the elements of the first superdiagonal are 1 and all the other elements are 0 Let λ 1 E p1 +N p1 λ E p +N p 0 J A = 0 λ u E pu +N pu and J B = µ 1 E q1 +N q1 µ E q +N q 0 0 µ v E qv +N qv be respectively the Jordan normal forms of A and B For 1 α u and 1 β v Let F () αβ and F(3) αβ be defined by the following: F () αβ = 0 if λ α µ β ; min(p α, q β ) 1 if λ α = µ β and p α = q β ; min(p α, q β ) if λ α = µ β and p α q β and F (3) αβ = 0 if λ α µ β ; 3min(p α, q β ) if λ α = µ β and p α = q β ; 3min(p α, q β ) 1 if λ α = µ β and p α q β = 1; 3min(p α, q β ) if λ α = µ β and p α q β The main results are the following: Theorem 11 Let α : M m n (C) M m n (C) be a linear transformation such that α(x) = AX XB, where A and B are m m and n n complex matrices,

4 Volume 30, pp 43-5, June Generalization of Gracia s Results 45 respectively Then the dimension formula for kerα is u v dim(kerα ) = F () αβ (11) α=1β=1 Theorem 1 Let α : M m n (C) M m n (C) be a linear transformation such that α(x) = AX XB, where A and B are m m and n n complex matrices, respectively Then the dimension formula for kerα 3 is dim(kerα 3 ) = u v α=1β=1 F (3) αβ (1) The proof of Theorem 11 Before proving the theorem, we first give a lemma in the following: Lemma 1 For a matrix M M l l (C), let Λ(M) be the set of its different eigenvalues If Λ(A) Λ(B) =, then kerα k = 0 for k = 1,,3, Proof It is well known that Λ(A) Λ(B) = if and only if the unique solution of the matrix equation AX XB = 0 is X = 0 Thus, when Λ(A) Λ(B) =, we can prove by induction on k that if Λ(A) Λ(B) =, the equality α k (X) = 0 implies X = 0 In fact, if α k (X) = 0 then α(α k 1 (X)) = 0 and Aα k 1 (X) α k 1 (X)B = 0 Since Λ(A) Λ(B) =, it follows that α k 1 (X) = 0 By hypothesis of the induction the equality α k 1 (X) = 0 implies that X = 0 So that, Λ(A) Λ(B) = implies that kerα k = 0 for k = 1,,3, Proof It is obvious that there are invertible matrices U and V such that A = UJ A U 1 andb = VJ B V 1 AssumeX kerα, thena(ax XB) (AX XB)B = 0 Hence, Thus, UJ A U 1 (UJ A U 1 X XVJ B V 1 ) = (UJ A U 1 X XVJ B V 1 )VJ B V 1 J A (J A U 1 XV U 1 XVJ B ) = (J A U 1 XV U 1 XVJ B )J B Let X = U 1 XV Then, the equation is J A (J A X XJ B ) = (J A X XJ B )J B (1) Now we partition X into blocks (X αβ ) where X αβ = (ε ik ) pα q β is a p α q β matrix for 1 α u and 1 β v Then we get uv matrix equations from (3): (λ α E pα +N pα )[(λ α E pα +N pα )X αβ X αβ (µ β E qβ +N qβ )] = [(λ α E pα +N pα )X αβ X αβ (µ β E qβ +N qβ )](µ β E qβ +N qβ )

5 Volume 30, pp 43-5, June Jun Liao, Heguo Liu, Yulei Wang, Zuohui Wu, and Xingzhong Xu Write P α := N pα and Q β := N qβ An easy calculation gives (µ β λ α ) X αβ = (µ β λ α )(P α X αβ X αβ Q β ) +P α (X αβ Q β P α X αβ ) (X αβ Q β P α X αβ )Q β () If µ β λ α, then X αβ = 0 by Lemma 1 Next we assume that µ β = λ α In this case, we have Case 1 p α = q β P α (P α X αβ X αβ Q β ) = (P α X αβ X αβ Q β )Q β (3) If p α = 1, then it is obvious that X αβ = (ε 11 ) ( ) ε11 ε 1 If p α =, an easy computation gives X αβ = 0 ε If p α 3, then ε 31 ε 3 ε 1 ε 33 ε ε 3, qβ ε, qβ 1 ε 41 ε 4 ε 31 ε 43 ε 3 ε 4, qβ ε 3, qβ 1 = ε pα 1 ε pα ε pα 1, 1 ε pα 3 ε pα 1, ε pα qβ ε pα 1, qβ 1 0 ε pα 1 ε pα ε pα, qβ ε 1 ε ε 11 ε, qβ 1 ε 1, qβ 0 ε 31 ε 3 ε 1 ε 3, qβ 1 ε, qβ 0 ε pα 1, 1 ε pα 1, ε pα, 1 ε pα 1, qβ 1 ε pα, qβ 0 ε pα 1 ε pα ε pα 1, 1 ε pα, qβ 1 ε pα 1, qβ 0 0 ε pα 1 ε pα, qβ This leads to the following equations: ε s1 = ε pα t = 0, ε pα 1, i = ε pα, i+1, ε h = ε h 1, 1, ε jk = ε j 1, k 1 ε j, k, where 3 s p α, 1 t q β, i q β, 3 j p α, 3 k q β, 1 k j q β 3 and 3 h p α According to these equations, we have ε 3 = ε 1 ε 43 = ε 3 ε 1 ε pα, q β 1 = ε pα 1, q β ε pα, q β 3 ε pα, q β 1 = ε pα 1, q β

6 Volume 30, pp 43-5, June Generalization of Gracia s Results 47 Then, ε 1 = ε 3 = = ε pα, q β 1 = 0 We also have ε 33 = ε ε 11 ε pα, q β = ε pα 1, q β 1 ε pα, q β By induction on the subscript, we obtain ε ii = (i 1)ε (i )ε 11, where 3 i p α Note that ε jk = ε j 1, k 1 ε j, k, where 3 j p α, 3 k q β and 1 k j q β 3 By induction, we get ε rh = (r 1)ε, h r+ (r )ε 1, h r+1 where 3 r p α, 0 h r q β 3 and 3 h q β Thus, we conclude that ε 11 ε 1 ε 13 ε 1, qβ 1 ε 1 qβ 0 ε ε 3 ε qβ X αβ = 0 0 ε ε 11 ε, qβ 1 ε 1, qβ (p α )ε (p α 3)ε 11 (p α )ε 3 (p α 3)ε (p α 1)ε (p α )ε 11 Let D 1 = diag{1,0, 1,, (p α )} and D = diag{0,1,,p α 1} be diagonal matrices of size p α It is obvious that X αβ = q β (ε 1iD 1 +ε,i+1 D )Np i 1 α Then the number of arbitrary parameters in X αβ is p α 1 Case p α q β We can assume that p α < q β, since the case p α > q β is analogous 0 If p α = 1, then X αβ = (0,,0,ε 1, qβ 1,ε 1, qβ ) In this case, q β columns are If p α =, then X αβ has the following form: ( ) 0 0 ε, qβ 1 ε 1, qβ 1 ε 1, qβ X αβ =, ε, qβ 1 ε, qβ and q β 3 columns are 0 If p α 3, then X αβ has the following form: 0 0 ε 1, qβ pα ε1, q β pα+1 ε1, q β pα+ ε1, q β ε1, q β 1 ε1, q β ε, qβ pα+1 ε, qβ pα+ ε, qβ ε, qβ 1 ε, qβ ε 3, qβ pα+ ε 3, qβ ε 3, qβ 1 ε 3, qβ ε pα,qβ ε pα,qβ 1 ε pα,qβ ε pα 1,qβ ε pα 1,qβ 1 ε pα 1,qβ ε pα,qβ 1 ε pα,qβ where ε i, qβ p α 1+i = (p α +1 i)ε pα, q β 1,ε rh = (r 1)ε, h r+ (r )ε 1, h r+1 and ε jk = 0,p α q β + j k p α 1, and 3 r p α, 0 h r q β 3, q β p α +3 h q β, 1 i p α In this case, q β p α 1 columns are 0,

7 Volume 30, pp 43-5, June Jun Liao, Heguo Liu, Yulei Wang, Zuohui Wu, and Xingzhong Xu Therefore, there are n = u v α=1 β=1 F() αβ linearly independent solutions X j of (3) For each X, there are k 1,k,,k n, such that X = n j=1 k jx j Note that X = U 1 XV It is straightforward to show that every solution of A(AX XB) (AX XB)B = 0 is a linear combination of n linearly independent solutions Let us illustrate Theorem 11 with an example Example Suppose that the elementary divisors of A and B are (λ λ 1 ) 4, (λ λ 1 ) 3, (λ λ ), (λ λ )and(λ λ 1 ) 5, (λ λ 1 ) 3, (λ λ ) 3, (λ λ ), (λ λ 3 ), respectively, where λ 1 λ λ 3, then dim(kerα ) = 4 5 α=1β=1 F () αβ = = 36 Let A be an n n complex matrix with distinct eigenvalues a 1,, a r the elementary divisors of A are (λ a i ) n ik and Segre characteristic }{{} j ik times [ ] (n j11 11,nj1,,nj1m 1 1 ),,(n jr1 r1,njr,,njrmr r ), where0 < n i1 < n i < < n imi for1 i r; herewewriten j ik ik forn ik,n ik,,n }{{ ik } j ik times Then we can show the following corollaries Corollary 3 ([1]) Let α A : M n n (C) M n n (C) be a linear transformation such that α A (X) = AX XA, where A is an n n complex matrix Then r m i m i 1 m i dim(kerα A) = (n ik 1)jik +4 n ik j ik β=k+1 j iβ Proof Let the elementary divisors of A are (λ a i ) n ik, where 1 i r and }{{} j ik times 1 k m i Using Theorem 11, we have dim(kerα A ) = u u α=1 α=1 F() αα, where u = r mi j ik The result follows from that u u α=1 α=1 r αα = m i m i 1 (n ik 1)jik +4 n ik j ik F () This completes the proof m i β=k+1 Corollary 4 Let α A : M n n (C) M n n (C) be a linear transformation such that α A (X) = AX XA, where A is an n n complex matrix Then n j iβ

8 Volume 30, pp 43-5, June Generalization of Gracia s Results 49 dim(kerα A ) n Moreover, dim(kerα A ) = n if and only if A is similar to a diagonal matrix diag{a 1,a,,a n }, where a i a j if i j Proof By Corollary 3, r r dim(kerα A ) [(n i1 1)ji1 ] ji1 ( r j i1) r = n r n and this equality holds if and only if m 1 = 1,n i1 = 1 and j i1 = 1,r = n Clearly kerα A is a subspace of the vector space M n n(c), thus dim(kerα A ) n 3 The proof of Theorem 1 This theorem can be proved by the same method as employed in the previous section Proof For convenience, we still use the same notations as in the proofof Theorem 11 Assume X kerα 3 Then J A (J AX 3XJ B ) = (XJ B 3J A X)J B, where X = U 1 XV Consequently, we get uv matrix equations: (λ α E pα +N pα ) [(λ α E pα +N pα )X αβ 3X αβ (µ β E qβ +N qβ )] = [(λ α E pα +N pα )X αβ 3X αβ (µ β E qβ +N qβ )](µ β E qβ +N qβ ), for 1 α u and 1 β v An easy calculation gives (µ β λ α ) 3 X αβ = 3(µ β λ α ) (P α X αβ X αβ Q β ) 3(µ β λ α )(P α X αβ +X αβ Q β P αx αβ Q β ) +P α(p α X αβ 3X αβ Q β )+(3P α X αβ X αβ Q β )Q β (31) If µ β λ α then X αβ = 0 by Lemma 1 We assume that µ β = λ α In this case, we have P α (P αx αβ 3X αβ Q β ) = (X αβ Q β 3P α X αβ )Q β (3) Case 1 p α = q β If p α < 4, then the number of arbitrary parameters in X αβ is 3p α If p α 4, then we obtain ε ij = 0, ε h = 3ε h 1, 1, ε pα 1, k = 3ε pα, k+1, ε h3 = 3ε h 1, 3ε h, 1, ε pα, k = 3ε pα 1, k+1 3ε pα, k+, ε fg = ε f 3, g 3 3ε f, g +3ε f 1, g 1,

9 Volume 30, pp 43-5, June Jun Liao, Heguo Liu, Yulei Wang, Zuohui Wu, and Xingzhong Xu where 4 i p α, i j 3, 4 h p α, 1 k q β 3, 4 f p α and 4 g q β Hence, we have ε 4 = 3ε 31 ε 53 = 3ε 4 3ε 31 ε 64 = 3ε 53 3ε 4 +ε 31 ε pα, q β = 3ε pα 1, q β 3 3ε pα, q β 4 +ε pα 3, q β 5 ε pα, q β 4 = 3ε pα 1, q β 3 3ε pα, q β ε pα 1, q β 3 = 3ε pα, q β Thus, ε 31 = ε 4 = = ε pα, q β = 0 We also have ε 43 = 3ε 3 3ε 1 ε 54 = 3ε 43 3ε 3 +ε 1 ε pα, q β 1 = 3ε pα 1, q β 3ε pα, q β 3 +ε pα 3, q β 4 ε pα, q β 3 = 3ε pα 1, q β 3ε pα, q β 1 By induction on the subscript, we obtain [ (s 1)(s )(pα ) ε s, s 1 = (p α 1) ] (s ) +1 ε 1, where s p α Note that ε fg = ε f 3, g 3 3ε f, g + 3ε f 1, g 1, where 4 f p α and 4 g q β Continuing by induction, we finally have ε fg = (f )(f 3) ε 1, g f+1 [f(f 4)+3]ε, g f+ + (f 1)(f ) ε 3, g f+3, where 4 f p α, 4 g q β and 0 g f q β 4 Then it is evident to see that the number of arbitrary parameters in X αβ is 3p α Case p α q β We can assume that p α > q β, since the case p α < q β is analogous

10 Volume 30, pp 43-5, June If p α q β = 1 and q β 4, then we have ε ij = 0, Generalization of Gracia s Results 51 ε s, s 1 = (s 1)(s ) ε 3 [(s ) 1]ε 1, (f )(f 3) ε fg = ε 1, g f+1 (f 1)(f ) [f(f 4)+3]ε, g f+ + ε 3, g f+3, where i j, 3 i p α, s p α, 4 f p α 1, 4 g q β and 0 g f q β 4 Thus, the number of arbitrary parameters in X αβ is 3q β 1 If p α q β = 1 and q β < 4, then a routine computation gives rise to the result If p α q β and q β 4, then we have (g 1)(g ) ε ij = 0,ε g, g = ε 31, ε s,s 1 = (s 1)(s ) ε 3 [(s ) 1]ε 1, (f )(f 3) ε fg = ε 1,g f+1 [f(f 4)+3]ε, g f+ (f 1)(f ) + ε 3, g f+3, where i j 3, 3 g p α and s p α and 4 f p α 1 and 4 g q β, 0 g f q β 4 Thus, the number of arbitrary parameters in X αβ is 3q β If p α q β and q β < 4, then an obvious computation gives rise to the same result v β=1 F(3) Therefore, there are n 3 = u α=1 αβ linearly independent X j kerα 3 For each X kerα 3, there are k 1,k,,k n3, such that X = n 3 j=1 k jx j Let us illustrate Theorem 1 with an example Example 31 Suppose that the elementary divisors of A and B are (λ λ 1 ) 4, (λ λ 1 ) 3, (λ λ ), (λ λ ), (λ λ 3 )and(λ λ 1 ) 5, (λ λ 1 ) 3, (λ λ ) 3, (λ λ ), respectively, where λ 1 λ λ 3, then dim(kerα 3 ) = 5 4 α=1β=1 F (3) αβ = = 49 Let A be an n n complex matrix with distinct eigenvalues a 1,,a r and the elementary divisors of A are (λ a i ) n ik for 1 k m i and 1 i r The }{{} j ik times

11 Volume 30, pp 43-5, June Jun Liao, Heguo Liu, Yulei Wang, Zuohui Wu, and Xingzhong Xu difference n i,h+1 n ih is called an h-th jump of i for 1 h m i 1 We denote by µ i1,µ i,,µ iρi the places where the jumps equal to 1 and µ i1 < µ i < < µ iρi Then we can show the following corollaries Corollary 3 ([1]) Let α A : M n n (C) M n n (C) be a linear transformation such that α A (X) = AX XA, where A is an n n complex matrix Then r m i m i ρ i dim(kerα 3 A) = (3n ik )jik +6n ik j ik j i,µil j i,µil +1 β=k+1 j iβ l=1 Proof Let the elementary divisors of A are (λ a i ) n ik, where 1 i r, 1 }{{} j ik times k m i and 0 < n i1 < n i < < n imi Using Theorem 1, we have dim(kerα 3 A ) = u u α=1 α=1 F(3) αα, where u = r mi j ik Since u u α=1 α=1 F(3) αα is equal to r m i (3n ik )jik +6n m i ρ i ik j ik j i,µil j i,µil +1, we have the dimension formula β=k+1 Corollary 33 Let α A : M n n (C) M n n (C) be a linear transformation such that α A (X) = AX XA, where A is an n n complex matrix Then n dim(kerα 3 A ) n Moreover, dim(kerα 3 A ) = n if and only if A is similar to a diagonal matrix diag{a 1,a,,a n }, where a i a j if i j Proof Since r m i dim(kerα 3 A ) = (3n ik )jik +6n m i ik j ik j iβ β=k+1 j iβ l=1 ρ i j i,µil j i,µil +1, it follows that dim(kerα 3 A ) r [(3n i1 )j i1 ] r j i1 ( r ji1) r l=1 = n r n andthe equalityholdsifand onlyifm 1 = 1,n i1 = 1andj i1 = 1,r = n Clearly, kerα 3 A is a subspace of the n n-dimensional vector space M n n (C), so dim(kerα 3 A ) n REFERENCES [1] JM Gracia Dimension of the solution spaces of the matrix equations [A, [A, X]] = 0 and [A[A, [A, X]]] = 0 Linear and Multilinear Algebra, 9:195 00, 1980 [] N Jacobson Lectures in Abstract Algebra: II Linear Algebra Springer, New York, 1953 [3] M Wedderburn Lectures on Matrices American Mathematical Society Colloquium Publications, Vol 17, New York, 008