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: [email protected]
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
** TT It is known that if A M n is normal AAAA , then AAAA if and only if AAAA . This leads to the question: do both AAAA and AATT AA imply that A is normal? We give an example to show that this is false when n 4 , but we show that it is true when n 2 and n 3 .
Keywords: Normal Matrix; Matrix Commuting with Its Conjugate and Transpose
Introduction and Results 2011). My solution for n 2 appeared in the spring 2012 issue, but no solution for the case n 3 has ever Let A be an n -by- n normal matrix, i.e., A is a been given. In this paper, we give the solution for the complex square matrix A M , with the property that n case n 3 , and for completeness, we also give the solu- AA* AA*, where A* AT is the conjugate-transpose tion for n 2 . Specifically we prove: of A . The Fuglede-Putnam Theorem tells us that if ** Theorem 1 If A M n , n 2 or n 3, then AB = BA for some B M, then A BBA . Suppose TT n AAAA and AA AA imply that A is normal. that AA AA, where A is the conjugate of the matrix Proof. We need the following preliminary result, A (so we take the complex conjugate of every entry of which is a direct consequence of Theorem 2.3.6 in [3] A ). Then taking the transpose gives (using the fact that for A M n , A BiC where B ATTAAAAAAAAAAA TT *T T* T T, and C are real then AAAA if and only if BC CB ), and stated explicitly in [1,2]. from the the Fuglede-Putnam Theorem. In a similar way, Theorem 2 Let A M , n 3 , with AAAA . we see that if AAAATT , then AAAA , so these two n Then there exists a real orthogonal matrix QM n statements are equivalent when A is normal. The ques- T such that QAQ is of the form: tion arose in [2], whether the conditions A AAAAand AAAATT 1 0 A 2 imply the third condition AAAA** , so that A is nor- 00A3 , mal. This is false when n 4 . In fact, any matrix of the 00 0Ak IIab cd ab form A , where Iab , 0 Iab ba where each Ai , 1 ik (for some k ) is a 1-by-1 abcd,,, , cd220 , c and d not both zero, matrix or a 2-by-2 matrix. T has the property that both AA AA and ATTAAA , Example 1. Note that if A QQ, Q real ortho- TT but A is not normal. In this paper, we prove that if gonal, AAAA and AA AA if and only if TT T A M where n 2 or n 3 , then these conditions and . Also note that if A A n * do imply that A is normal. This result was first pro- and AAAA . then A is normal since A A in this posed as a problem by the current author in the Interna- case. TT tional Linear Algebra Society journal IMAGE (fall Lemma 1 If A M 2 with AAAA , then A is
Copyright © 2013 SciRes. ALAMT G. GOODSON 23
ab or either symmetric or of the form ,,ab . ba abx222 xy x ab xy a22 b y 2 y Proof. Suppose that Aabcd,,,, , with 2 cd xy TT AAAA , then ab220 axby a2b2 a c bd a22 c ab cd 0.ab22 bxay 2222 22 2 ac bd c d ab dc b d ax by ay bx x y Hence bc22 and ab cd ac bd . It follows that xy 0 , and A is normal. Case 1. bc , so that A is symmetric. axy Case 2. bc, then ab bd ab bd or Case 2: If 0 bz then ab bd . If b 0 , then A is symmetric. If b 0 , 00c ab ad and A . axya00 a 00 axy ba 0000,bzxb xb bz Propositi on 1 If A M with AAAATT and 2 00cyzc yzc 00 c AAAA , then A is norm al. Proof. From the Lemma 1, we have two cases. If or 222 ab axybxyzcy A , ab, , then A is normal. On the 22 ba bx yz b z cz 2 other hand, if A is symmetric with AAAA , then cy cz c since A* A in this case, we must have AA** AA, aaxay2 so A is norm al. ax x22 b xy bz . Ex ample 2. We now look at the case of A M 3 . We start with a lemma: 222 ay xy bz c y z Lemma 2 Suppose A M with AA AA , 22 22 22 3 Hence xy 0 , x z , yz0 , and AAAATT and A QQT for some real o rtho gonal x iy , x z , y iz and also ay cy , so y 0 (giving diagonal and A normal) or ac . Sup- matrix Q M3 where is of one of the two pose y 0 so that ac . forms: see Equation (1). Case 2(a). If xz 0 , then then A is normal. ax bx yz y b a and x ib a z so that abx aibaba Proof. Case 1: ba y. Now we require 0.biab However, this matrix 00 00 a TT , so that also has the property that , which gives in abxa b00a b abxEquation (2). It follows from equating the entries in the ba y b a00 b a ba y, (1, 2) position 22 22 2 00 xy xy 00a b22 ab a b ab , or ab0 ,
abx axy ba y,or 0 bz , ab ,,,,,,cxyz , (1) 00 00c
a222 i a b2 ab 2 ab 23 ab a222 b a i a22 b2 ab 2 ab 23 ab a2 b 2 2222 22 02biabab 0 biaba2b (2) 00 aa2 0 0 2
Copyright © 2013 SciRes. ALAMT 24 G. GOODSON
so ab , and hence is diagonal and A is normal. ay y d e ye ay y d e ye Case 2(b). This is where xz 0 , and since ax bx xy we have yab, so that and simplifying gives yd a ya d, so if ad a iab ab yad we have 0,biab and this is treated in a yad Equating entries in the (2, 3) position gives: 00 a similar way. y ea dyyeyea dyye, TT Proposition 2 If A M 3 with AA AA and AAAA , then A is normal. and this reduces to: yd a yd a, so if ad , Proof. We show that every case reduces to the case of yad , contradicting the above. We conclude that the Lemma 2. From Theorem 1, every matrix with yad T A QQ (Q real orthogonal) can be chosen t o be ad and is of the form one of the following three forms: aiaeiy axy abx ia e a y , (I) 0 bz, (II) cd y, or (III) 00e 00c 00e and we can apply Lemma 2. The other possibility is that axy y 0 x , so that bc and is either of the form 0 bc. ab0 0 de bd0 , a symmetric matrix (when cb ), or of We have dealt with Case (I) in Lemma 2, so consider 00e Case (II): TT gives in Equation (3). abiy It follows that bxc222, cyb222, and the form ba y (when cb , since in this xy220 , so that x iy . 00e Case 1. xi0y, then since bx dy ey , biy dy ey , so bide. case ad ). Case 2. xiy 0 , then bx dy ey gives Also xeaxcy gives xeaxcix , so that ciea= , so has the form bied , and ax cy xe gives ciae , so aieiyd that has the form iea d y aiediy 00e ia e d y. Now we use the fa ct that . This gives in E qua- 00e tion (4). On equat ing entries in the (1, 3) position we have: We proceed exact ly as in Ca se 1 to re duce to the
abxacbxyxeac222 d 2 2 abcdaxcy 222 22 ac bd xy c d y ey ab cd b d bx dy (3) 2222 xe ye e ax cy xb yd x y e
adeeaiadeidd2 eiayiydeiye iaea idea ead e d2 yea dyye 2 00e a2 d e e a ia d e id d e iay iy d e iye ia e a id e a e a d e d2 y e a dy ye (4) 00e 2
Copyright © 2013 SciRes. ALAMT G. GOODSON 25
situation of Lemma 2. doi:10.3103/S0278641910020019 axy [2] G. R. Goodson and R. A. Horn, “Canonical Forms for Normal Matrices That Commute with Their Complex In Case (III), where 0 bc, we proceed ex- Conjugate,” Linear Algebra and Its Applications, Vol. 0 de 430, No. 4, 2009, pp. 1025-1038. actly as in Case (II) to deduce the result. doi:10.1016/j.laa.2008.09.039 [3] R. A. Horn and C. R. Johnson, “Matrix Analysis,” Cam- bridge University Press, New York, 1985. REFERENCES doi:10.1017/CBO9780511810817
[1] Kh. Ikramov, “On the Matrix Equation XX XX ,” Mos- cow University Computational Mathematics and Cyber- netics. Vol. 34, No. 2, 2010, pp. 51-55.
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