SCIENTIFIC PUBLICATIONS OF THE STATE UNIVERSITY OF NOVI PAZAR SER A: APPL MATH INFORM AND MECH vol 6, 1 (214), 45-55 A Note on Solutions of the Matrix Equation AXB C I V Jovović, B J Malešević Abstract: This paper deals with necessary and sufficient condition for consistency of the matrix equation AXB C We will be concerned with the minimal number of free parameters in Penrose s formula X A (1) CB (1) +Y A (1) AY BB (1) for obtaining the general solution of the matrix equation and we will establish the relation between the minimal number of free parameters and the ranks of the matrices A and B The solution is described in the terms of Rohde s general form of the {1}-inverse of the matrices A and B We will also use Kronecker product to transform the matrix equation AXB C into the linear system (B T A) X C Keywords: Generalized inverses, Kronecker product, matrix equations, linear systems 1 Introduction In this paper we consider matrix equation AXB C, (1) where X is an n k matrix of unknowns, A is an m n matrix of rank a, B is a k l matrix of rank b, and C is an m l matrix, all over C The set of all m n matrices over the complex field C will be denoted by C m n, m,n N The set of all m n matrices over the complex field C of rank a will be denoted by C m n a We will write A i (A j ) for the i th row (the j th column) of the matrix A C m n and A will denote an ordered stock of columns of A, ie A A 1 A 2 Using the Kronecker product of the matrices B T and A we can transform the matrix equation (1) into linear system A n (B T A) X C (2) Manuscript received September 3, 213 ; accepted November 17, 213 I V Jovović and B J Malešević are with School of Electrical Engineering, University of Belgrade, Serbia 45
46 I V Jovović, B J Malešević For the proof we refer the reader to A Graham 2 Necessary and sufficient condition for consistency of the linear system Ax c, as well as the minimal number of free parameters in Penrose s formula x A (1) c + (I AA (1) )y has been considered in the paper B Malešević, I Jovović, M Makragić and B Radičić 3 We will here briefly sketch this results in the case of the linear system (2) Any matrix X satisfying the equality AXA A is called {1}-inverse of A and is denoted by A (1) For each matrix A C m n a there are regular matrices P C n n and Q C m m such that Ia QAP E A, (3) where I a is a a identity matrix It can be easily seen that every {1}-inverse of the matrix A can be represented in the Rohde s form A (1) Ia U P V W Q, (4) where U u i j, V v i j and W w i j are arbitrary matrices of corresponding dimensions a (m a), (n a) a and (n a) (m a) with mutually independent entries, see C Rohde 1 and V Perić 8 We will explore the minimal numbers of free parameters in Penrose s formula X A (1) CB (1) +Y A (1) AY BB (1) for obtaining the general solution of the matrix equation (1) Some similar considerations can be found in papers B Malešević and B Radičić 4, 5, 6 and 9 2 Matrix equation AXBC and the Kronecker product of the matrices B T and A The Kronecker product of matrices A a i j m n C m n and B b i j k l C k l, denoted by A B, is defined as block matrix a 11 B a 12 B a 1n B a 21 B a 22 B a 2n B A B a m1 B a m2 B a mn B The matrix A B is mk nl matrix with mn blocks a i j B of order k l Here we will mention some properties and rules for the Kronecker product Let A C m n, B C k l, C C n r and D C l s Then the following propositions holds: A T B T (A B) T ; rank(a B) rank(a) rank(b);
A note on solutions of the matrix equation AXB C 47 (A B) (C D) (A C) (B D) (mixed product rule); if A and B are regular n n and k k matrices, then (A B) 1 A 1 B 1 The proof of these facts can be found in A Graham 2 and A Ben Israel and TNE Greville 1 Matrix A (1) B (1) is {1}-inverse of A B Using mixed product rule we have (A B)(A (1) B (1) )(A B) (A A (1) A) (B B (1) B) A B Let R C k k and S C l l be regular matrices such that Ib RBS E B (5) An {1}-inverse of the matrix B can be represented in the Rohde s form B (1) Ib M S R, (6) N K where M m i j, N n i j and K k i j are arbitrary matrices of corresponding dimensions b (k b), (l b) b and (l b) (k b) with mutually independent entries From now on, we will look more closely at the linear system (2) Firstly, by mixed product rule we obtain (S T Q) (B T A) (R T P) (S T B T R T ) (Q A P) E B T E A Unfortunately, the matrix Ib E E B T E A A E A E A E A I a I a I a is not of the needed form E B T A Equality E B T E A E B T A holds for b 1 Swapping the rows and the columns corresponding to blocks I a and to zero diagonal blocks we get required matrix E B T A If matrices D and G are the elementary matrices obtained by swapping rows and columns corresponding to mentioned blocks of the identity matrices, then D (E B T E A ) G E B T A Thus, we have (D (S T Q)) (B T A) ((R T P) G) E B T A (7) and so an {1}-inverse of the matrix B T A can be represented in the Rohde s form (B T A) (1) (R T Iab F P) G D (S T Q), (8) H L
48 I V Jovović, B J Malešević where F f i j, H h i j and L l i j are arbitrary matrices of corresponding dimensions ab (ml ab), (nk ab) ab and (nk ab) (ml ab) with mutually independent entries If the matrices A and B are square matrices, then D G T For the simplicity of notation, we will write c a (c a ) for the submatrix corresponding to the first (the last) a coordinates of the vector c Now, we can rephrase Lemma 21 and Theorem 22 from the paper B Malešević, I Jovović, M Makragić and B Radičić 3 in the case of the linear system (2) Let C be given by C QCS Then C (S T Q) C Lemma 21 The linear system (2) has a solution if and only if the last ml ab coordinates of the vector c D C are zeros, where D is elementary matrix such that (7) holds Theorem 22 The vector X (B T A) (1) C + (I (B T A) (1) (B T A))y, (9) y C nk 1 is an arbitrary column, is the general solution of the system (2), if and only if the {1}-inverse (B T A) (1) of the system matrix B T A has the form (8) for arbitrary matrices F and L and the rows of the matrix H(c ab y ab ) + y nk ab are free parameters, c where D (S T Q) C c ab and G 1 ((R 1 ) T P 1 )y y y ab y nk ab In the paper B Malešević, I Jovović, M Makragić and B Radičić 3 we have seen that general solution (9) can be presented in the form X (R T c P) G ab H(c ab y ab ) + y (1) nk ab We illustrate this formula in the next example Example 23 We consider the matrix equation where A 1 2 1 1 1, X x11 x 12 x 13 x 21 x 22 x 23 AXB C,, B 1 1 1 1 1 1 and C 3 1 4 4 Using the Kronecker product the matrix equation may be considered in the form of the equivalent linear system (B T A) X C, ie 1 2 1 2 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 x 11 x 21 x 12 x 22 x 13 x 23 3 1 4 4
1 Matrices Q 1 and P 1 1 1 1 E A 1 and matrices R matrices such that RBS E B A note on solutions of the matrix equation AXB C 49 1 1 1 1 1 1 1 1 1 are regular matrices such that QAP and S Therefore, 1 1 1 1 are regular E B T E A 1 1 1 1 and for matrix D 1 1 1 1 1 1 1 1 1 we have E B T A D (E B T E A ) 1 1 1 1 Let us remark that E B T A (D (S T Q)) (B T A) (R T P), and hence according to the Theorem 22 X (R T c 4 P) H(c 4 y 4 ) + y is the general solution of the linear 2 system iff elements of the column H(c 4 y 4 )+y 2 are two mutually independent parameters
5 I V Jovović, B J Malešević α 1 i α 2 for the vector c c 4 (D (S T Q)) C 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 1 4 4 3 1 Finally, the general solution of the linear system is X (R T P) 3 1 α 1 α 2 1 1 2 1 1 1 1 2 1 1 1 1 3 1 α 1 α 2 1 α 1 + 2α 2 1 α 2 α 1 + 2α 2 α 2 2α 3 α 4 α 2 α 1 + 2α 2 α 2 3 Matrix equation AXBC and the {1} inverses of the matrices A and B In this section we indicate how technique of an {1}-inverse may be used to obtain the necessary and sufficient condition for an existence of a general solution of the matrix equation (1) without using Kronecker product We will use the symbols C a,b, C a,b, C a,b and C a,b for the submatrices of the matrix C corresponding to the first a rows and b columns, the last a rows and the first b columns, the first a rows and the last b columns, the last a rows and b columns, respectively Lemma 31 The matrix equation (1) has a solution if and only if the last m a rows and l b columns of the matrix C QCS are zeros, where Q C m m and S C l l are regular matrices such that (3) and (5) hold Proof: The matrix equation (1) has a solution if and only if C AA (1) CB (1) B, see R Penrose 7 Since A (1) and B (1) are described by the equations (4) and (6), it follows that AA (1) Ia U AP V W Q Q 1 Ia U Q
A note on solutions of the matrix equation AXB C 51 and B (1) Ib M B S N K Ib RB S N S 1 Hence, since Q and S are regular matrices we have the following equivalences C AA (1) CB (1) B QCSQAA (1) CB (1) BS C QCS C Ia U C a,b C a,l b C m a,b C m a,l b C a,b C a,l b C m a,b C m a,l b for C C a,b C a,l b C m a,b C m a,l b Ia U C a,b C a,l b C m a,b C m a,l b C Ib N Ib N C a,b +U C m a,b +C a,l b N +U C n a,l b N Therefore, we conclude, C AA (1) CB (1) B C a,l b C m a,b C m a,l b As we have seen in the Lemma 21 the matrix equation (1) has a solution if and only if the last ml ab coordinates of the column c D C are zeros, where D is elementary matrix such that (7) holds Here we obtain the same result without using Kronecker product The last m(l b) elements of the column C are zeros and there are b blocks of m a zeros Multiplying by the left column C with elementary matrix D switches the rows corresponding to this zeros blocks under the blocks C a,b i, 1 i b Hence, the last m(l b) + (m a)b ml ab entries of the column c are zeros Furthermore, we give a new form of the general solution of the matrix equation (1) using {1}-inverses of the matrices A and B Theorem 32 The matrix X A (1) CB (1) +Y A (1) AY BB (1), (11) Y C n k is an arbitrary matrix, is the general solution of the matrix equation (1) if and only if the {1}-inverses A (1) and B (1) of the matrices A and B have the forms (4) and (6) for arbitrary matrices U, W, N and K and the entries of the matrices V ( C Y ) + Y, ( C Y ) M + Y, V ( C Y ) M + Y a,b a,b n a,b a,b a,b a,k b a,b a,b n a,k b (12) are mutually independent free parameters, where QCS C C a,b and P 1 Y R 1 Y Y a,b Y a,k b Y n a,b Y n a,k b (13)
52 I V Jovović, B J Malešević Proof: Since the {1}-inverses A (1) and B (1) of the matrices A and B have the forms (4) and (6), the solution of the system X A (1) CB (1) +Y A (1) AY BB (1) can be represented in the form Ia U X P V W Ib M QCS N K Ia U R + Y P V W QAPP 1 Y R 1 Ib M RBS N K R According to Lemma 31 and from (3) and (5) we have Ia U C X P a,b Ib M R + Y V W N K Ia U Ia P P 1 Y R 1 Ib Ib M V W N K R Furthermore, we obtain C a,b C a,b M I a X P V C a,b V C R + Y P P 1 Y R 1 I b M R a,b M V ( ) C a,b C a,b M P V C a,b V C + Y I a Y I b M R a,b M V ( C a,b C a,b M P V C a,b V C + a,b M ) Y a,b Y a,k b I a Y a,b Y a,k b I b M + Y n a,b Y n a,k b V Y n a,b Y R n a,k b ( ) C a,b C a,b M Y a,b Y a,k b Y a,b Y a,b M P V C a,b V C + a,b M V Y a,b V Y R, a,b M Y n a,b Y n a,k b where Y P 1 Y R 1 We now conclude X P C a,b V ( C a,b Y a,b) + Y n a,b ( C a,b Y a,b)m +Y a,k b V ( C a,b Y a,b)m +Y n a,k b R (14) According to the Theorem 22 the general solution of the equation (1) has nk ab free parameters Therefore, since the matrices P and R are regular we deduce that the solution (14) is the general if and only if the entries of the matrices separately V ( C a,b Y a,b ) + Y n a,b, ( C a,b Y a,b ) M +Y a,k b, V ( C a,b Y a,b ) M +Y n a,k b are nk ab free parameters We can illustrate the Theorem 32 on the following two examples
A note on solutions of the matrix equation AXB C 53 Example 33 Consider again the matrix equation from previous example A B C, where A B 1 2 1 1 1 1 1 1 1 1 1 x11 x, X 12 x 13 x 21 x 22 x 23 and C, 3 1 4 4 We have C QCS and 1 1 1 1 1 3 1 4 4 3 C 2,2 1 The solution of the matrix equation AXB C is 3 α1 X P 1 α 2 R 1 1 Example 34 We now consider the matrix equation 1 1 1 1 3 α1 1 α 2 3 1 1 1 1 1 1 1 α1 + 2α 2 α 1 + 2α 2 α 1 2α 2 AXB C, 1 3 2 x 11 x 12 where A 2 6 4, X x 21 x 22, B 1 3 2 x 31 x 32 1 For regular matrices Q 1 and P 1 1 1 equality QAP E A 1 α 2 α 2 α 2 1 3 6 and C 1 3 1 1 2 4 12 2 the following holds Thus, rank of the matrix A is a 1 There
54 I V Jovović, B J Malešević 1 1 3 1 are regular matrices R and S such that RBS E 1 1 B holds Thus, rank of the matrix B is b 1 Since the ranks of the matrices A and B are a b 1, according to the Lemma 31 all entries of the last column and two rows of the matrix C QCS are zeros, ie we get that the matrix C is of the form C QCS 1 1 1 1 2 4 12 2 1 3 1 2 Applying the Theorem 32, we obtain the general solution of the given matrix equation X P 2 α 1 β 1 γ 1 β 2 γ 2 R 1 3 1 1 2 α β 1 γ 1 β 2 γ 2 1 1 2 2α 3β 1 2β 2 + 6γ 1 + 4γ 2 α 3γ 1 2γ 2 β 1 2γ 1 γ 1 β 2 2γ 2 γ 2 Acknowledgment Research is partially supported by the Ministry of Science and Education of the Republic of Serbia, Grant No17432 References 1 A BEN ISRAEL AND TNE GREVILLE, Generalized Inverses: Theory & Applications, Springer, New York, 23 2 A GRAHAM, Kronecker Products and Matrix Calculus with Applications, Ellis Horwood Lim, Chichester, England, 1981 3 B MALEŠEVIĆ, I JOVOVIĆ, M MAKRAGIĆ AND B RADIČIĆ, A note on solutions of linear systems, ISRN Algebra, Vol 213, Article ID 142124, 6 pages (in the press) 4 B MALEŠEVIĆ AND B RADIČIĆ, Non-reproductive and reproductive solutions of some matrix equations, Proceedings of International Conference Mathematical and Informational Technologies 211, Vrnjačka Banja, 211, pp 246 251 5 B MALEŠEVIĆ AND B RADIČIĆ, Reproductive and non-reproductive solutions of the matrix equation AXB C, Zbornik radova simpozijuma Matematika i primene, Matematički fakultet Beograd, 27 i 28 maj 211, pp 157 163 6 B MALEŠEVIĆ AND B RADIČIĆ, Some considerations of matrix equations using the concept of reproductivity, Kragujevac J Math, Vol 36 (212,) No 1, pp 151 161
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