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IJRRAS 8 (3) ● September 2011 www.arpapress.com/Volumes/Vol8Issue3/IJRRAS_8_3_07.pdf THE NULLSPACE OF NESTED MAGIC SQUARES Ayoub ElShokry1 & Saleem Al-Ashhab2 University of Khartoum ,Faculty of Education ,Department of Mathematics , Omdurman ,Sudan Depatement of Mathematics, Al Al-Bayt University, Mafrq, Jordan Email: ayou1975@ maktoob.com; [email protected] ABSTRACT In this paper we formulate the general structure of the nullspace and subspaces of the nullspace for nested magic squares, where we consider two different types of centre. Further, we study the properties of these spaces. AMS classification number: 15A15. Key Words: Nullspace, Magic Squares, Mathematical induction. 1. INTRODUCTION We consider magic squares here as matrices and study the algebraic properties for them. Hence, a semi magic square is a n by n matrix such that the sum of the entries in each row and columns is the same. The common value is called the magic constant. If, in addition, the sum of all entries in each left-broken diagonal and each right-broken diagonal is the magic constant, then we call the matrix a pandiagonal magic square. It is well-known that the following structure A B C 2s–A–B–C E 2s– A–B–E A+E – C B+C – E s–C A+B+C – s s–A s–B s–A–E+C s–B–C+E s–E A+B+E–s is a general structure of the pandiagonal magic square 4 by 4. Here, the magic constant is 2s. When we want to obtain pandiagonal magic squares 6 by 6, having a similar structure to that of the pandiagonal magic squares 4 by 4, we consider the matrix a1 a7 a13 s – a4 s – a10 s – a16 a2 a8 a14 s – a5 s – a11 s – a17 a3 a9 a15 s – a6 s – a12 s – a18 a4 a10 a16 s – a1 s – a7 s – a13 a5 a11 a17 s – a2 s – a8 s – a14 a6 a12 a18 s – a3 s – a9 s – a15 We then require the following conditions: a1 a2 a3 a4 a5 a6 3s a7 a8 a9 a1 0 a1 1 a1 2 3s a1 3 a1 4 a1 5 a1 6 a1 7 a1 8 3s a1 a4 a7 a1 0 a1 3 a1 6 0 a2 a5 a8 a1 1 a1 4 a1 7 0 a3 a6 a9 a1 2 a1 5 a1 8 0 311 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares These conditions suffice to ensure that the resulting square is pandiagonal. We have as a possible form of these squares: J–L –K+J+I+H+E+D+C – 3/2 s L–J+G– E+B K–I+F– D+A –L–K–H–G–F–C–B–A+ 9/2 s L K –J–I–H–G–F+3s –E–D–C–B–A+3s L+K+H+G+F+C+B+A–7/2 s s–H s– C J E s–L I D s–K s–G s–B s–F s–A H C L+K–J–I–H–E–D–C+ 5/2s J+I+H+G+F–2s E+D+C+B+A –2s G B s–L+J– G+E–B s–J s–E F A s–K+I– F+D–A s–I s–D We call this kind of squares the pandiagonal magic square 6 by 6 of the special structure. Note that the magic constant is now 3s. The nullspace of a matrix A is the solution set of the homogenous system Ax = 0. The pandiagonal magic square 4 by 4 possesses a nontrivial null space. Using the previous notation for a pandiagonal magic square 4 by 4 we can easily prove (see [1]) that this nullspace can be written in the following form A 2 B C 2s C A {z : z R} 2s A 2 B C AC Note that the sum of the entries is zero. Furthermore, the multiplication of the square as a matrix with the vector A 2 B C 2s 1 q 1 C A z 2 s A 2 B C 2 s 1 AC 1 , q, z R yields the vector q q q q This is due to definition of the magic square. In [1] we find that the pandiagonal magic square 6 by 6 of the special structure possesses also a nontrivial nullspace, which can be written in the following form: (DJ - DG + AG + BI - EI + EF - AJ - BF) (3Is - BI - 2AI - 2IC - EI - 2AH + 2DH + 3As - 3Ds - 3Fs + EF + BF + DJ + 2DF + 2CF - AG - AJ + DG) (BI - EI - 2EH + 2BH - 3Bs - 3Js + 3Gs + 3Es - AG - EF - 2CG + BF - 2EG + 2BJ + 2JC + DJ - DG + AJ) : z R} {z ( DJ - DG + AG + BI - EI + EF - AJ - BF) (3Is - BI - 2AI - 2IC - EI - 2AH + 2DH + 3As - 3Ds - 3Fs + EF + BF + DJ + 2DF + 2CF - AG - AJ + DG) ( BI - EI - 2EH + 2BH - 3Bs - 3Js + 3Gs + 3Es - AG - EF - 2CG + BF - 2EG + 2BJ + 2JC + DJ - DG + AJ) We will therefore denote the vectors in this nullspace as follows: x1 x2 x 3 x1 x2 x 3 In this case the multiplication of the square with the vector 312 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares x1 1 x 1 2 x3 q 1 z x1 3s 1 1 x2 1 x 3 , q, z R yields the vector q q q q q q Nested magic squares By nested squares we mean matrices having certain properties, which make them and some of their submatrices semi magic squares or pandiagonal magic squares. By a nested semi magic square 6 by 6 with a pandiagonal magic square 4 by 4 we mean a matrix of the form r11 p11 P21 P31 P41 e11 where we require b11 A E s– C s–A–E+C s – b11 b12 B 2s–A–B–E A+B+C–s S–B–C+E s – b12 b13 C A+E–C s–A s–E s – b13 b14 2s–A–B–C B +C – E s–B A+B+E–s s – b14 d11 s – p11 s – p21 s – p31 s – p41 f11 d 11 3s r11 b11 b12 b13 b14 e11 3s r11 p11 p 21 p 31 p 41 .......... (1) f 11 b11 b12 b13 b14 e11 s This ensures that the matrix as a whole is a semi magic square with 3s as a magic constant. In the centre we have a pandiagonal magic square. We introduce also the concept of the multi-nested semi magic square 2n by 2n with a pandiagonal magic square 4 by 4. This will be the following matrix 313 IJRRAS 8 (3) ● September 2011 rmm r(m-1)m rm(m-1) . . . . ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares rm1 . bm1 . bm2 . bm3 . bm4 . dm1 . . . . . . b22 b12 . . b23 b13 . . b24 b14 . . d21 d11 s– p11 s– P21 d22 s– r12 s– p12 s– P22 s– P31 s– P41 f11 . . dm(m-1) . dmm s –r(m1)m . . . r22 r12 r21 r11 . . B21 B11 r2m r1m . . . . . . . p1m . . . p12 p11 A B C 2s–A–B–C P2m . . . P22 P21 E 2s–A– B–E A+ E– C B +C – E P3m . . . P32 P31 s–B P4m . . . P42 P41 e1m . . . e12 e11 e2m . . . e22 s– A s– E s– b13 s– b23 . . . . . . . e(m-1)m . emm s– . rm(m-1) . . s–C –s + A +B+C s–A– E+C s – b11 s–B– C+E s – b12 s– r21 . . . s – b21 s – b22 . . . . . . s– rm1 s –bm1 s – bm2 s– bm3 A+B+E–s s – b14 s – b24 . . . s – bm4 s– d21 . . . s– dm1 . . . . . . . . . . s –r2m s –r1m . . . s –p1m . . . s– P2m s– P32 . . . s– P3m s– P42 s– e12 f22 . . . . . . s– P4m s –e1m . . . s –e2m . . . s– s– e(m-1)m fmm . . . . . dm(m-1) where m= n – 2 and d mm ns rmm rm( m1) ... rm1 bm1 bm2 bm3 bm4 d m1 ... d m( m1) emm ns rmm r( m1) m ... r1m p1m p2m p3m p4m e1m ... e( m1) m f mm rm( m1) ... rm1 bm1 bm 2 bm3 bm4 d m1 ... d m( m1) emm (n 2)s Note that we obtain semi magic square each time we remove the outer frame of the square. By a multi-nested semi magic square 2n by 2n with a pandiagonal magic square 6 by 6 we mean the following matrix: 314 IJRRAS 8 (3) ● September 2011 pmm pm(m-1) P(m- … ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares Pm1 Vm1 Vm2 Vm3 Vm4 Vm5 Vm6 rm1 . . . . . . . . . . . . p21 . . v21 . . v22 . . v23 . . v24 . . v25 . . v26 . . r21 p11 v11 v12 v13 v14 v15 v16 r11 L–J +G– E+B K– I+F– D+A L K J I –L –K –H–G –F–C – B–A+ 9/2s H E D C s–L s –K s–G s–F L+K– J – I –H –E– D– C+ 5/2s J + I +H+G +F–2s E+D+ C+B+ A –2s s – v14 ... rm(m-1) rmm . s – p(m-1)m . . . … s –p1m s – u11 s – p12 s – u12 … s – u1m F s – u21 s – u22 … s – u2m B A s – u31 s – u32 … s –u3m s–L –J– G+E –B s– K + I –F +D – A s – u41 s – u42 … s –u4m s–J s–I s – u51 s – u52 … s –u5m s – E s–D s – u61 s – u62 … s –u6m s – v15 s – v25 . . . s – v16 s – v26 . . . t11 s – w12 t22 … s – w1m . 1)m . . . P1m … p22 p12 u1m … u12 u11 u2m … u22 u21 u3m … u32 u31 u4m … u42 u41 u5m … u52 u51 –L – K+J+I +H+E+ D+C– 3/2s –J–I – H–G– F + 3s –E–D – C–B A+3s L+K+ H+G+ F+C+B +A– 7/2 s s–H u6m … u62 u61 s–C s–B s–A w1m ... w12 w11 s – v11 w22 s – p21 . . . s – v21 s – v12 s – v22 . . . s – v13 s – v23 . . . . . . w(m- . . . . s – v24 . . . G s – r21 r22 . . . . 1)m . . s – w(m1)m wmm s –pm(m-1) . s – pm1 s –vm1 s –vm2 s –vm3 s –vm4 s – vm5 s –vm6 s –rm1 . s –rm(m- tmm 1) where m = n – 2, rmm ns pmm pm( m1) ... pm1 vm1 vm 2 vm3 vm4 vm5 vm6 rm1 ... rm( m1) , wmm ns pmm p( m1) m ... p1m u1m u 2m u3m u 4m u5m u 6m w1m ... w( m1) m , t mm pm( m1) ... pm1 vm1 vm2 vm3 vm 4 vm5 vm6 rm1 ... rm( m1) wmm (n 2)s. For example, we define the nested semi magic square 8 by 8 with a 6 by 6 pandiagonal square as the following square: 315 IJRRAS 8 (3) ● September 2011 p11 u11 v11 – L –K + J +I +H + E + D + C – 3/2 s u21 –J –I –H –G– F+ 3s u31 u41 u51 where ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares v12 L–J+G –E+B v13 K–I– D+F+A v14 –L –K –H –G –F –C –B –A + 9/2 s v15 L v16 K J I H G F –E –D –C –B– A+ 3s L+K+H+G +F+C+B+ A – 7/2 s E s–L D s –K B s – L +J – G+ E –B A s–K – F +D – A +I s–H s–G s–F s–J s–I s–B s–A C L+K–J – I –H –E –D – C + 5/2s J +I+ H + G +F – 2s E+D+C + B+A –2s s – v12 s – v13 s–v14 s–v15 u61 s –C w11 s – v11 r11 s – u11 s – u21 s – u31 s–E s–D s – v16 s – u41 s – u51 s – u61 t11 r11 4 s p11 v11 v12 v13 v14 v15 v16 w11 4 s p11 u11 u 21 u 31 u 41 u 51 u 61 t11 v11 v12 v13 v14 v15 v16 w11 2 s 2. MAIN RESULTS We will prove several statements about the nullspace of the mentioned nested magic squares. By doing this we will use mathematical induction. Hence, some results will be used for proving others. We start with Proposition 1: The nested semi magic square 6 by 6 with a pandiagonal magic square 4 by 4 has the following space 1 A 2 B C 2 s 1 2 C A 1 2 {z : z R},1 R 1 2s A 2 B C 2 1 AC 2 1 as a subspace of the nullspace of this square. Proof: We search for real numbers a1 ,... , a6 such that a1 C1 ... a6 C6 0 where the capital letters denote the columns of the matrix. We set a1 a6 and q sa1 . We set 316 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares a2 A 2 B C 2s 1 C A q a3 1 a z 2s A 2 B C 2s 1 ....................... (2) 4 AC 1 a 5 where z is a unknown real number. Now, the expression a1 C1 ... a6 C6 is the result of matrix multiplication of the square with the vector a1 a2 a 3 a4 a5 a 6 The result of this operation will be according to our choice a vector which has the following middle entries p11a1 p 21a1 p 31a1 p 41a1 sa1 sa1 sa1 sa1 sa1 sa1 sa1 sa1 p11a1 p 21a1 p 31a1 p 41a1 Hence, the middle entries are zero and we have to equate the first and last entry of the vector to zero. We will use this requirement to determine the values of a1 and z. Thus, we solve the following two equations: r 11a1 b11a2 b12a3 b13a4 b14a5 d11a1 0 e11a1 (s b11 )a2 (s b12 )a3 (s b13 )a4 (s b14 )a5 f11a1 0 This system can be simplified to (r11d11)a1 b11a2 b12a3 b13a4 b14a5 0 .......... ..... (3) (e11 f11)a1 s(a2 a3 a4 a5 ) b11a2 b12a3 b13a4 b14a5 0 According to the definition of the variables (see (1) and (2)) we have r11 d11 3s b11 b12 b13 b14 e11 f11 b11 b12 b13 b14 s 2q 2a1 s Note that we used the sign properties of the vectors belonging to the nullspace of the 4 by 4 square. Upon using the last relations we convert the system (3) into a 2 a3 a 4 a5 (3s b11 b12 b13 b14 )a1 b11a 2 b12 a3 b13a4 b14a5 0 (b11 b12 b13 b14 3s)a1 b11a2 b12a3 b13a4 b14a5 0 We recognize that one equation is redundant. We substitute the values of a 2 , a 3 , a 4 and a 5 . Then, we obtain from the last equation 3 (2s b11 b12 b13 b14 )a1 (b11( A 2B C 2s) b12 (C A) b13 (2s A 2B C ) b14 ( A C )) z 0 2 If the coefficient of a1 is not zero, then we conclude that a1 1 z Upon substituting this value for that a1 we are done with the proof. If the coefficient of a1 is zero, then we conclude a1 can be any number, while z might be zero. In this case we can say a1 1u, 1 1, u R 317 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares and the proof is done with u instead of z. Remark: 1) We note that the sum of all entries of any vector in the subspace is zero. 2) When we consider the following numerical example for a nested semi magic square 6 by 6 9 5 2 0 3 –16 7 2 6 1 –7 –6 4 3 –9 4 4 –3 1 0 8 –1 –5 0 8 –3 –3 –2 10 –7 –26 –4 –1 1 –2 35 We obtain the following reduced row echelon form 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 35 11 19 11 46 11 8 11 0 Hence, the nullity of the square is one. Therefore, the described subspace in proposition 1 is actually the nullspace of the square. Proposition 2: The multi-nested semi magic square 2n by 2n with a pandiagonal magic square 4 by 4 has the following space n 2 n 2 n 3 n 1 n 3 n 2 n 4 n 2 n 1 . . . 1 2 3 ... n 2 3 4 n 1 n 2 1 A 2 B C 2 s ... 2 n 1 n 2 1 ... C A 2 n 1 : z R} {z n 2 1 2 s A 2 B C ... 2 n 1 n 2 1 A C 2 ... n 1 n 2 2 3 1 3 4 ... n 1 . . . n 4 n 3 n 2 n2 n 1 n 3 n 2 n 1 n2 for all n 3 as a subspace of the nullspace of this square. Proof: We will use mathematical induction in our proof. We start with 318 IJRRAS 8 (3) ● September 2011 Basis step: when n as claimed. We continue now with 3 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares we obtain the semi magic square 6 by 6, which has according to proposition 1 a subspace Induction step: we suppose that the given form of the nullspace is true for the nullspace of any multi-nested square following structure 2k by 2 k n k , i. e. there exists a subspace of with a pandiagonal magic square 4 by 4, which has the k 2 k 2 k 3 k 1 k 3 k 2 k 4 k 2 k 1 . . . x k 2 2 3 1 ... 1 x 3 4 k 1 2 k 2 1 . A 2 B C 2s ... 2 k 1 . . C A 1 ... k 2 2 k 1 , xk z k 2 1 2s A 2 B C ... xk 1 2 k 1 . A C 1 ... k 2 . 2 k 1 3 k 2 . 2 ... x 1 3 4 k 1 2k . . . k 3 k 2 k 4 k 2 k 1 k 2 k 3 k 1 k 2 We construct now a subspace of the nullspace of the multi-nested square zR 2k 2 by 2 k 2 . We search for a1 ,... , a2 k 2 such that the following relation holds a1 C1 ... a2k 2 C2k 2 0 where the capital letters denote the columns of the square. We choose a2 a 3 a 4 . . . a 2k a1 a2 k 2 , q sa1 . We choose further x 1 x 1 2 . 1 1 . . . q . .......( 4) z xk ks . x k 1 1 . 1 . 1 1 . x 2k 319 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares Since the statement of the proposition holds for 2k by 2 k n k square inside the multi-nested square entry is equal to , we conclude that the multiplication of the multi-nested 2k 2 by 2 k 2 by this vector yields a vector, where each q . Hence, the multiplication of the multi-nested square 2k 2 by 2 k 2 by a1 a 2 a3 . . . a 2k 2 yields r sa r a m( m 1) a1 1 m( m 1) 1 sa 1 sa r r a a ( m 1)(m 1) 1 1 ( m 1)(m 1) 1 sa 1 . . . . . . . . . r2( m 1) a1 sa sa1 r2( m 1) a1 1 r1( m 1) a1 sa sa1 r1( m 1) a1 1 p1( m 1) a1 sa1 sa1 p1( m 1) a1 p2( m 1) a1 sa1 sa1 p2( m 1) a1 p3( m 1) a1 sa1 sa1 p3( m 1) a1 sa a p 1 sa1 p4( m 1) a1 4 ( m 1 ) 1 sa 1 sa e e a a 1( m 1) 1 1( m 1) 1 sa 1 1 e sa e a a 1 2( m 1) 1 2( m 1) 1 . . . . . . . . . sa1 em( m 1) a1 sa1 em( m 1) a1 Therefore, all the entries are zero except the first and last entry. We will use this requirement to determine z and . We obtain therefore the following equations r( m1)( m1) a1 r( m1) m a 2 ... r( m1)1a k 1 b( m1)1a k b( m1) 2 a k 1 b( m1)3 a k 2 b( m1) 4 a k 3 d ( m1)1a k 4 ... d ( m1)( m1) a1 0 e( m1)( m1) a1 ( s r( m1) m )a 2 ... ( s r( m1)1 )a k 1 ( s b( m1)1 )a k ( s b( m1) 2 )a k 1 ( s b( m1)3 )a k 2 ( s b( m1) 4 )a k 3 ( s d ( m1)1 )a k 4 ... f ( m1)( m1) a1 0 This linear system can be simplified to (r( m1)( m1) d ( m1)( m1) )a1 r( m1) m a 2 ... r( m1)1a k 1 b( m1)1a k b( m1) 2 a k 1 b( m1)3 a k 2 b( m1) 4 a k 3 d ( m1)1a k 4 ... d ( m1) m a 2 k 1 0 320 a1 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares (e( m1)( m1) f ( m1)( m1) )a1 s(a2 a3 ... a2 k 1 ) r( m1) m a2 ... r( m1)1ak 1 b( m1)1ak b( m1) 2 ak 1 b( m1)3 ak 2 b( m1) 4 ak 3 d ( m1)1ak 4 ... d ( m1) m a2 k 1 0 According to the definition of our variables (see (4)) we have r( m1)( m1) d ( m1)( m1) (k 1) s r( m1) m r( m1)( m1) ... r( m1)1 b( m1)1 b( m1) 2 b( m1)3 b( m1) 4 d ( m1)1 ... d ( m1) m e( m1)( m1) f ( m1)( m1) r( m1) m ... r( m1)1 b( m1)1 b( m1) 2 b( m1)3 b( m1) 4 d ( m1)1 ... d ( m1) m (k 1) s a2 a3 ... a2k 1 2q 2a1 s According to the definition of f ( m1)( m1) and d ( m1)( m1) we obtain the following linear system ((k 1) s r( m1) m ... r( m1)1 b( m1)1 b( m1) 2 b( m1)3 b( m1) 4 d ( m1)1 ... d ( m1) m )a1 r( m1) m a 2 ... r( m1)1 a k 1 b( m1)1 a k b( m1) 2 a k 1 b( m1)3 a k 2 b( m1) 4 a k 3 d ( m1)1 a k 4 ... d ( m1) m a 2 k 1 0 (r( m1) m ... r( m1)1 b( m1)1 b( m1) 2 b( m1)3 b( m1) 4 d ( m1)1 ... d ( m1) m (k 1) s)a1 r( m1) m a 2 ... r( m1)1 a k 1 b( m1)1 a k b( m1) 2 a k 1 b( m1)3 a k 2 b( m1) 4 a k 3 d ( m1)1 a k 4 ... d ( m1) m a 2 k 1 0 We recognize that one equation is redundant. We substitute the values of a 2 , …, a 2 k 1 and obtain the following equation: k 1 (ks r( m1) m ... r( m1)1 b( m1)1 b( m1) 2 b( m1)3 b( m1) 4 d ( m1)1 ... d ( m1) m )a1 k (r( m1) m x1 ... r( m1)1 xk 2 b( m1)1 xk 1 b( m1) 2 xk b( m1)3 xk 1 b( m1) 4 xk 2 d ( m1)1 xk 3 ... d ( m1) m x2 k ) z 0 We can say using a similar argument to the one used in the proof of proposition 1 a1 k 1 z In analogy to that proof we are done. Remark: We note that the sum of all entries of any vector in the subspace is zero. We turn our attention now to the squares, which have a pandiagonal magic square 6 by 6 of the special structure in the centre. We can prove very similar results for this type of squares like the results for multi-nested squares with 4 by 4 square in the centre. Proposition 3: The nested semi magic square 8 by 8 with a 6 by 6 pandiagonal square has the following space 321 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares 1 x 1 1 3 x 1 2 3 x 1 3 3 {z : z R} 1 x1 3 1 x2 3 x3 1 3 1 as a subspace of the nullspace of the square. Proof: We search for real numbers a1 ,... , a8 such that a1 C1 ... a8 C8 0 where the capital letters denote the columns of the matrix. We set and a1 a8 q sa1 . We set a2 x1 1 a x 3 2 1 a x 1 q 4 3 z , z R a x 3s 1 5 1 1 a x 2 1 6 x a 3 7 where z is an unknown real number. Now, the expression a1 C1 ... a8 C8 is the result of matrix multiplication of the square with the vector a1 a2 a 3 a4 a5 a6 a7 a 8 The result of this operation will be according to our choice a vector which has the following middle entries u11a1 u a 21 1 u a 31 1 u a 41 1 u a 51 1 u a 61 1 sa sa u a 11 1 1 1 sa sa u a 21 1 1 1 sa sa u a 31 1 1 1 sa sa u a 41 1 1 1 sa sa u a 51 1 1 1 sa sa u a 1 1 61 1 Hence, the middle entries are zero and we have to equate the first and last entry of the vector to zero. We will use this requirement to determine the values of a1 and z. Thus, we solve the following two equations: p11a1 v11a2 v12a3 v13a4 v14a5 v15a6 v16a7 r11a1 0 322 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares w a (s v )a ( s v )a ( s v )a ( s v )a 11 1 11 2 12 3 13 4 14 5 (s v )a ( s v )a t a 0 15 6 16 7 11 1 This system can be simplified to ( p11r11 )a1 v11a 2 v12 a3 v13a 4 v14 a5 v15 a6 v16 a7 0 (w t )a s(a a a a a a ) v a v a 11 11 1 2 3 4 5 6 7 11 2 12 3 v a v a v a v a 0 13 4 14 5 15 6 16 7 According to the definition of the variables we have p11 r11 4s v11 v12 v13 v14 v15 v16 w11 t11 v11 v12 v13 v14 v15 v16 2s a 2 2q a a a a a 2a 3 4 5 6 7 1 s Note that we used the sign properties of the vectors belonging to the nullspace of the 6 by 6 square. Upon using the last relations we convert the system into (4s v11 v12 v13 v14 v15 v16 )a1 v11a2 v12 a3 v13a4 v14 a5 v15a6 v16 a7 0 (v11 v12 v13 v14 v15 v16 4s)a1 v11a2 v12 a3 v13a4 v14 a5 v15a6 v16 a7 0 We recognize that one equation is redundant. We substitute the values of a 2 , a 3 , a 4 , a 5 , a 6 and a 7 . Then, we obtain from the last equation 4 (3s v11 v12 v13 v14 v15 v16 )a1 (v11 x1 v12 x 2 v13 x 3 v14 x1 v15 x 2 v16 x 3 ) z 0 3 If the coefficient of a1 is not zero, then we conclude that a1 1 z Upon substituting this value for that a1 we are done with the proof. If the coefficient of a1 is zero, then we conclude a1 can be any number, while z might be zero. In this case we can say a1 1u, 1 1, u R and we have the proof done with u instead of z. Remark: 1) We note that the sum of all entries of any vector in the subspace is zero. 2) When we consider the following numerical example for a nested semi magic square 8 by 8 8 6 2 5 3 0 1 –17 2 1 0 1 –2 1 5 0 4 2 0 3 1 2 –2 –2 0 0 1 1 4 –2 2 2 1 4 1 –3 1 2 1 1 We obtain the following reduced row echelon form 323 3 1 0 4 0 2 –1 –1 7 –2 4 0 2 1 1 –5 –17 –4 0 –3 –1 2 1 30 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 37 21 463 63 65 63 17 7 505 63 23 63 0 Hence, the nullity of the square is one. Therefore, the described subspace in proposition 3 is actually the nullspace of the square. Proposition 4: The nested semi magic square 2n by 2n with a 6 by 6 pandiagonal square has the following space n 3 n 3 n 1 n4 n 4 n 3 n 5 n2 n 1 . . . n 3 2 3 ... 4 5 n 1 1 x 1 2 ... n 3 4 n 1 1 3 n 3 1 2 x ... 2 3 4 n 1 x 1 2 ... n 3 3 3 4 n 1 {z : z R} x 1 2 ... n 3 1 3 4 n 1 n3 1 2 x2 3 4 ... n 1 n 3 1 2 . x ... 3 3 4 n 1 n 3 2 3 ... 1 4 5 n 1 . . . n 4 n 3 n 5 n 2 n 1 n3 n 4 n 1 n3 as a subspace of the nullspace of the square for all n 4. Proof: We will use mathematical induction in our proof. We start with Basis step: when n 4 we obtain the semi magic square 8 by 8, which has according to proposition 3 a subspace as claimed. We continue now with 324 IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares n k Induction step: we suppose that the given form of the nullspace is true for the nullspace of the multi-nested square 2k by 2 k , i.e. there exists a subspace of , which has the following structure k 3 k 3 k 1 k 4 k 4 k 3 k 5 k 2 k 1 . . . 2 3 ... k 3 1 4 5 k 1 k 3 1 2 y 1 x ... y 1 3 4 k 1 2 . x 1 2 ... k 3 2 3 4 k 1 . . k 3 1 2 x3 3 4 ... k 1 yk z 1 2 ... k 3 yk 1 x1 3 4 k 1 k 3 1 2 . ... . x2 3 4 k 1 . k 3 1 2 y . x ... 3 3 4 k 1 2k k 3 2 3 ... 1 4 5 k 1 . . . k 4 k 3 k 5 k 2 k 1 k 3 k 1 k 4 k 3 We construct now a subspace of the nullspace of the multi-nested square ,zR 2k 2 by 2 k 2 . We search for a1 ,... , a2 k 2 such that the following relation holds a1 C1 ... a2k 2 C2k 2 0 where the capital letters denote the columns of the square. We choose a2 a 3 a 4 . . . a 2k y 1 y 1 2 . 1 1 . . . q . z yk ks . yk 1 1 . 1 1 . 1 . y 2k 325 a1 a2 k 2 , q sa1 . We choose further IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares Since the statement of the proposition holds for nested 2k by 2 k n k square inside the multi-nested square each entry is equal to we conclude that the multiplication of the multi- 2k 2 by 2 k 2 by this vector yields a vector, where q . Hence, the multiplication of the multi-nested square 2k 2 by 2 k 2 by a1 a 2 a3 . ) . . a 2k 2 yields p m ( m 1) a1 sa sa1 1 sa1 p ( m 1)( m 1) a1 sa1 . . . . . . . . . p a sa 1( m 1) 1 sa1 1 u1( m 1) a1 sa sa1 1 u 2 ( m 1) a1 sa1 sa1 u 3( m 1) a1 sa1 sa1 u 4 ( m 1) a1 sa1 sa1 u 5( m 1) a1 sa1 sa1 sa 1 u 6 ( m 1) a1 sa1 sa1 w a sa1 1( m 1) 1 . . . . . . . . . sa 1 sa w( m 1)( m 1) a1 1 sa w 1 m ( m 1) a1 sa1 p m ( m 1) a1 p ( m 1)( m 1) a1 p1( m 1) a1 u1( m 1) a1 u 2 ( m 1) a1 u 3( m 1) a1 u 4 ( m 1) a1 u 5( m 1) a1 u 6 ( m 1) a1 w1( m 1) a1 w( m 1)( m 1) a1 wm ( m 1) a1 All the entries are zero except the first and last entry. We will use this requirement to determine z and obtain therefore the following equations p( m1)( m1) a1 p( m1) m a2 ... p( m1)1ak 2 v( m1)1ak 1 v( m1) 2 ak v( m1)3 ak 1 v( m1) 4 ak 2 v( m1)5 ak 3 v( m1)6 ak 4 r( m1)1ak 5 ... r( m1)( m1) a1 0 w( m1)( m1) a1 ( s p( m1) m )a 2 ... ( s p( m1)1 )a k 2 ( s v( m1)1 )a k 1 ( s v( m1) 2 )a k ( s v( m1)3 )a k 1 ( s v( m1) 4 )a k 2 ( s v( m1)5 )a k 3 ( s v( m1)6 )a k 4 ( s r( m1)1 )a k 5 ... t ( m1)( m1) a1 0 This linear system can be simplified to ( p( m1)( m1) r( m1)( m1) )a1 p ( m1) m a 2 ... p ( m1)1 a k 2 v( m1)1 a k 1 v( m1) 2 a k v( m1)3 a k 1 v( m1) 4 a k 2 v( m1)5 a k 3 v( m1) 6 a k 4 r( m1)1 a k 5 ... r( m1) m a 2 k 1 0 326 a1 . We IJRRAS 8 (3) ● September 2011 ElShokry & Al-Ashhab ● The Nullspace of Nested Magic Squares ( w( m 1)( m 1) t ( m 1)( m 1) )a1 s(a 2 a 3 ... a 2 k 1 ) p ( m 1) m a 2 ... p ( m 1)1 a k 2 v ( m 1)1 a k 1 v ( m 1) 2 a k v ( m 1)3 a k 1 v ( m 1) 4 a k 2 v ( m 1)5 a k 3 v ( m 1) 6 a k 4 r( m 1)1 a k 5 ... r( m 1) m a 2 k 1 0 According to the definition of our variables we have p( m1)( m1) r( m1)( m1) (k 1) s p( m1) m p( m1)( m1) ... p( m1)1 v( m1)1 v( m1) 2 v( m1)3 v( m1) 4 v( m1)5 v( m1) 6 r( m1)1 ... r( m1) m w( m1)( m1) t ( m1)( m1) p( m1) m ... p( m1)1 v( m1)1 v( m1) 2 v( m1)3 v( m1) 4 v( m1)5 v( m1) 6 r( m1)1 ... r( m1) m (k 1) s a 2 a3 ... a 2 k 1 2q 2a1 s According to the definition of r( m1)( m1) and t ( m1)( m1) we obtain the following linear system ((k 1) s p( m1) m ... p( m1)1 v( m1)1 v( m1) 2 v( m1) 3 v( m1) 4 v( m1) 5 v( m1) 6 r( m1)1 ... r( m1) m )a1 p( m1) m a 2 ... p( m1)1 a k 2 v( m1)1a k 1 v( m1) 2 a k v( m1) 3 a k 1 v( m1) 4 a k 2 v( m1) 5 a k 3 v( m1) 6 a k 4 r( m1)1a k 5 ... r( m1) m a 2 k 1 0 ( p ( m 1) m ... p ( m 1)1 v( m1)1 v( m 1) 2 v( m1)3 v( m 1) 4 v( m1)5 v( m 1) 6 r( m 1)1 ... r( m 1) m (k 1) s)a1 p ( m 1) m a 2 ... p ( m 1)1 a k 2 v( m1)1 a k 1 v( m1) 2 a k v( m 1)3 a k 1 v( m 1) 4 a k 2 v( m1)5 a k 3 v( m1) 6 a k 4 r( m1)1 a k 5 ... r( m1) m a 2 k 1 0 We recognize that one equation is redundant. We substitute the values of a 2 , …, a 2 k 1 and obtain the following equation k 1 (ks p( m 1) m ... p ( m 1)1 v( m 1)1 v( m 1) 2 v( m 1)3 v( m 1) 4 v( m 1) 5 v( m1) 6 k r( m 1)1 ... r( m 1) m )a1 ( p ( m 1) m y1 ... p ( m 1)1 y k 3 v( m 1)1 y k 2 v( m 1) 2 y k 1 v( m 1) 3 y k v( m 1) 4 y k 1 v( m 1)5 y k 2 v( m1) 6 y k 3 r( m 1)1 y k 4 ... r( m1) m y 2 k ) z 0 We can say using a similar argument to the one in the proof of proposition 3 a1 k 1 z In analogy to that proof we are done. ACKNOWLEDGMENT This paper is extracted from a Msc. Thesis by Ayoub ElShokry (supervised by Saleem Al-Ashhab) at Al-Albayt University in 2007 REFERENCES [1]. Al-Ashhab, S., Computer Solutions of Magic 44 Squares and Semi – Magic 44 Squares Problem , Dirasat, Natural and Engineering Sciences, Vol. , 25 , No , 3 , 1998 , pp . (445-450). [2]. Al-Ashhab, S., Theory of Magic Squares / Programming and Mathematics, Royal Scientific Society, Jordan, 2000 (in Arabic). [3]. Anton, H., Elementary Linear Algebra, 7th edition, Jon Wiley, New York, 1994. [4]. Rosser, B. & Walker, R., J., On the Transformation Group for Diabolic Magic Squares of Order Four, Amer. Math. Soc., Bult XLIV, 1938, pp. 416-420. [5]. Trenkler, D. & Trenkler, G., Magic Squares, Melancholy and The Moore-Penrose Inverse, The Bulletin of The International Linear Algebra Society, No.27 , 2001, pp (3-10). 327
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