 Advances in Linear Algebra & Matrix Theory, 2013, 3, 22-25 http://dx.doi.org/10.4236/alamt.2013.33005 Published Online September 2013 (http://www.scirp.org/journal/alamt) Matrices That Commute with Their Conjugate and Transpose Geoffrey Goodson Towson University, Towson, USA Email: ggoodson@towson.edu Received June 25, 2013; revised July 23, 2013; accepted August 6, 2013 Copyright © 2013 Geoffrey Goodson. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. ABSTRACT It is known that if nAM is normal **AAAA, then AAAA if and only if TTAAAA. This leads to the question: do both AAAA and TTAAAA imply that A is normal? We give an example to show that this is false when , but we show that it is true when 4n2n and 3n. Keywords: Normal Matrix; Matrix Commuting with Its Conjugate and Transpose Introduction and Results Let A be an -by- normal matrix, i.e., n nA is a complex square matrix nAM, with the property that **AAAA, where *TAA is the conjugate-transpose of A. The Fuglede-Putnam Theorem tells us that if BAAB= for some n, then BM**ABBA. Suppose that AAAA, where A is the conjugate of the matrix A (so we take the complex conjugate of every entry of A). Then taking the transpos e gives TT TT*T T*T T,AAAAAAAA AAAA from the the Fuglede-Putnam Theorem. In a similar way, we see that if TTAAAA, then AAAA, so these two statements are equivalent when A is normal. The ques- tion arose in , whether the conditions TTandAAAA AAAA imply the third condition **AAAA, so that A is nor- mal. This is false when . In fact, any matrix of the 4nform where ,0ab cdabIIAI22,ababIba,,,abcd, , and not both zero, has the property that both 0cd cdAAAA2011). My solution for appeared in the spring 2012 issue, but no solution for the case has ever been given. In this paper, we give the solution for the case 2n3n3n, and for completeness, we also give the solu- tion for 2n. Specifically we prove: Theorem 1 If nAM, or , then 2n3nAAAA and TTAAAA imply that A is normal. Proof. We need the following preliminary result, which is a direct consequence of Theorem 2.3.6 in  (using the fact that for nAM, ABiC where and are real then BCAAAA if and only if BC CB), and stated explicitly in [1,2]. Theorem 2 Let nAM, , with 3nAAAA. Then there exists a real orthogonal matrix nQM such that is of the form: TQAQ123000000 kAAA,A  where each iA, 1ik (for some ) is a 1-by-1 matrix or a 2-by-2 matrix. k and TTAAAA, but A is not normal. In this paper, we prove that if nAM where or 2n3n, then these conditions do imply that A is normal. This result was first pro- posed as a problem by the current author in the Interna- tional Linear Algebra Society journal IMAGE (fall Example 1. Note that if TAQQ , real ortho- gonal, QAAAA and TTAAAA if and only if  and TT. Also note that if TAA and AAAA. then A is normal since *AA in this case. with TTAAAALemma 1 If 2AM , then A is Copyright © 2013 SciRes. ALAMT G. GOODSON 23ei Proof. Suppose that, with ther symmetric or of the form b. a,,ababa,,,,bAabcdcd TTAAAA, then 2b a222ac bdac bdcd2222acab cdab dcbd ac bd. Hence and 22bc bc,ab cd Case 1. so thatA is symmetrir c. Case 2. c, then ab bdab bd o0bab bd. If, then bA is symmetric. If 0b, a Propon 1 If ad nd ab Aba.ositi 2AM with TTAAAA and AAAA, then A is al. normProof. e From th Lemma 1, we have two cases. Ifab, ,ab, then AbaA is normal. On the other hand, if A is symmetricwith AAAA, then since *AA in this case, we must have **AAAA, so A is al. Eample 2. We now look at the case of normx3AM. We start with a lemma: Lemma 2 Suppose 3AM with AA AA, TTAAAA and TAQQ  where for somrthoe real ogonal matrix Q3M is of one of the two forms: see Equ(1). ation then A is normal. Proof. Case 1: 00abxbay . Now we require , so that ,TT 000000 00abxa babxbayb ab abayxy xya b  2222222222222200.abx xyxxya byyxyab axbyab bxayax byay bxxyor  0xyA is normal. It follows that , and 000ax ybzcCase 2: If 00 000000,00 00axyaa axybzxb xbbzcyzcyzcc then       222222222222.axybxyzcybx yzbzczcycz caax ayaxxbxy bzayxybzcyzor    220xy22, Hence xz, , and 220yzxzxiy, , yizay cy0y and also , so  (giving acA normal) or  diagonal and . Sup- pose ac0y.  so that 0xzCase 2(a). If axbxyzy ba, then xib az so that and 0.00aibababiaba   However, this matrix also has the property that   , which gives in Equation (2). It follows from equating the entries in the (1, 2) position 22 2222ab ab ab ab , or 20ab , ,or0, ,00 00abx axybaybz abc  ,,,,,, (1)cxyz 222 2222222 2222 2302 2(2)00 0aia babababa bia babababa bbiabab biababaa          22 22 22222300aCopyright © 2013 SciRes. ALAMT G. GOODSON 24 so , and hence is diagonal and ab A is normal. Case 2(b). This is where 0xz, and since we have, so that and this is treated in a si Propositiax bx xy yab0,0a iababbiab   0amilar way. on 2 If 3AM h Twit TAAAA and AAAA, then A is normal. Proof. We show that every cathe Lemma 2. From Theorem se reduces to the case of 1, every matrix  with TAQQ (Q real orthogonal) can be choso be three forms: (I) , or (III) have dealt within Lemma 2, s Case (II): gives in Equation (3). It follows that 22, and , so that en tone of the following0z, (II) 00c 00e0bc. ax yb0axydeabxcdyWe Case (I) o consider TT  22bxc, 22cyb220xyxiy0, then . since Case 1. xiyeybx dyey, biy dy, so bide. Also xea ives xcy gxeaxcix , so th=cieas the form at ha, so  00aieiyi ye ea dde use the faNow wct t h at is gives in E  . Thqua-tion (4). On ing 3) position we have: equat entries in the (1,ayydeyeayy deye and simplifying gives yd ayad , so if ad we have yadyad Equating entries in the (2, 3) position gives: ,yeady yeyeadyye  and this reduces to: yd ayd a, so if ad, yadyad, contradicting the above. We conclu de thatad and  is of the form aiaeiya eaye,00i  apply Lemma 2. The other possibility is that and we can0yx, sothat bc  and  is either of the form 0000abbdecb, a symmetric matrix (when ), or of the form 00a yeabiyb  (whe cb, since in this ncase ad). Case 2. 0xiy, then gives bx dyeybied gives ciae, and axcyxe, so that  has the form aiediye.ia edy00  We procely as in Caduceed exactse 1 to re  t 2o the 222 22222222(3)abxacb xyxeac abcdaxcyac bdxycdyeyab cdbdbx dyxeax cyxbydxye    22ye ed    22222200(4)00adeea iadeiddiaeaideaeadedyeady yeeadee aiadeiddeiay iydeiyeiae aideae adedye adyyee   eiayiydeiye  Copyright © 2013 SciRes. ALAMT G. GOODSON 25 situation of Lemma 2. In Case (III), where , we proceed ex- actly as in Case (II) to deduce the result. REFERENCES  Kh. Ikramov, “On the Matrix Equation 00axybcdeXX XX,” Mos- cow University Computational Mathematics and Cyber- 2, 2010, pp. 51-55. doi:10.3103/S0278641910020019  G. R. Goodson and R. A. Horn, “Canonical Forms for Normal Matrices That Commute with Their Complex Conjugate,” Linear Algebra and Its Applications, Vol. 430, No. 4, 2009, pp. 1025-1038. doi:10.1016/j.laa.2008.09.039  R. A. Horn and C. R. Johnson, “Matrix Analysis,” Cam- bridge University Press, New York, 1985. doi:10.1017/CBO9780511810817netics. Vol. 34, No. Copyright © 2013 SciRes. ALAMT