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
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 n
A
M is normal

**
A
AAA, then
A
AAA

if and only if TT
A
AAA. This leads to the
question: do both
A
AAA and TT
A
AAA 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 n
A
is a
complex square matrix

n

A
M, with the property that
**
A
AAA, where *T
A
A is the conjugate-transpose
of
A
. The Fuglede-Putnam Theorem tells us that if
BA
A
B= for some n, then BM**
A
BBA. Suppose
that
A
A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,
A
AAAAAAA AAAA
from the the Fuglede-Putnam Theorem. In a similar way,
we see that if TT
A
AAA, then
A
AAA, so these two
statements are equivalent when
A
is normal. The ques-
tion arose in [2], whether the conditions
TT
and
A
AAA AAAA
imply the third condition **
A
AAA

, so that
A
is nor-
mal.
This is false when . In fact, any matrix of the
4n
form where ,
0
ab cd
ab
II
AI



22
,
ab
ab
Iba




,,,abcd, , and not both zero,
has the property that both
0cd cd
A
AAA
2011). 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 n
A
M

, or , then 2n3n
A
AAA

and TT
A
AAA

imply that
A
is normal.
Proof. We need the following preliminary result,
which is a direct consequence of Theorem 2.3.6 in [3]
(using the fact that for n
A
M

,
A
BiC where
and are real then B
C
A
AAA

if and only if
BC CB

), and stated explicitly in [1,2].
Theorem 2 Let n
A
M

, , with 3n
A
AAA

.
Then there exists a real orthogonal matrix


n
QM
such that is of the form:
T
QAQ
1
2
3
0
00
000 k
A
A
A,
A
























 


where each i
A
, 1ik

 (for some ) is a 1-by-1
matrix or a 2-by-2 matrix. k
 and TT
A
AAA

,
but
A
is not normal. In this paper, we prove that if
n
A
M 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 T
A
QQ , real ortho-
gonal, Q
A
AAA

and TT
A
AAA if and only if

 and TT

. Also note that if T
A
A

and
A
AAA

. then
A
is normal since *
A
A

in this
case.
with TT
A
AAALemma 1 If 2
A
M

, then
A
is
C
opyright © 2013 SciRes. ALAMT

G. GOODSON 23
ei
Proof. Suppose that, with
ther symmetric or of the form b.
a

,,
ab
a
ba




,,,,
b
Aabcd
cd





TT
A
AAA, then
2
b a
2
22
ac bd
ac bdcd






22
22
acab cd
ab dcbd





ac bd.



Hence and
22
bc
bc,ab cd Case 1. so that
A
is symmetrir c.
Case 2. c, then ab bdab bd o
0
b
ab bd. If, then b
A
is symmetric. If 0b

,
a
Propon 1 If
ad nd ab


Aba




.
ositi 2
A
M with TT
A
AAA and
A
AAA, then
A
is al. norm
Proof. e From th Lemma 1, we have two cases. If
ab


, ,ab, then
Aba

A
is normal. On the
other hand, if
A
is symmetricwith
A
AAA, then
since *
A
A in this case, we must have **
A
AAA

,
so
A
is al.
Eample 2. We now look at the case of
norm
x3
A
M

. We
start with a lemma:
Lemma 2 Suppose 3
A
M with AA AA,
TT
A
AAA and T
A
QQ

where
for somrthoe real ogonal
matrix Q

3
M

is of one of the two
forms: see Equ(1). ation
then
A
is normal.
Proof. Case 1:
00
abx
bay



 


 
. Now we require
, so that
,
TT
 
00
00
00 00
abxa babx
bayb ab abay
xy xy
a b



 









222
222
2
22
22
222
0
0.
abx xyx
xya byy
xy
ab axby
ab bxay
ax byay bxxy






or

























0xy
A
is normal. It follows that , and

0
00
ax y
bz
c




 Case 2: If






00 00
0000,
00 00
axyaa axy
bzxb xbbz
cyzcyzcc
then

 

 


 

 

 
222
22
2
2
22
222
.
axybxyzcy
bx yzbzcz
cycz c
aax ay
axxbxy bz
ayxybzcyz
or


 













 







22
0xy22 , Hence

x
z

, , and
22
0yz
x
z
x
iy

,

,
y
izay cy0y
and also , so


(giving ac
A
normal) or

diagonal and

. Sup-
pose ac
0y

.

so that


0xzCase 2(a). If


axbxyzy ba, then 

x
ib az so that and




0.
00
aibaba
biab
a
 




 






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 22
22ab ab ab ab , or 20ab
,
,or0, ,
00 00
abx axy
baybz ab
c



 



,,,,,, (1)cxyz 






222 2222
222 2
2
22 23
02 2(2)
00 0
aia babababa bia babababa b
biabab biabab
aa
 
 
 
 
 
 
 
 
 
22 2
2 2
2
2223
0
0
a



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 Propositi
ax bx xy yab


0,
0
a iabab
biab
  

 



0a
milar way.
on 2 If 3
A
M h T wit T
A
AAA and
A
AAA, then
A
is normal.
Proof. We show that every ca
the Lemma 2. From Theorem se reduces to the case of 1, every matrix

with
T
A
QQ (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 t
one of the following
0z

, (II) 
00c


 00e



0bc


.
ax
 
y
b
0
axy
de



abx
cdy


We Case (I) o consider
TT
 
22
bxc, 22 cyb
22
0xy
x
iy
0, then . since Case 1. xiy
ey
bx dyey,
biy dy, so

bide.
Also
x
ea ives xcy g
x
eaxcix , so th

=cieas the form at

ha, so 



00
aieiy
i y
e





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 yad
yad



Equating entries in the (2, 3) position gives:


,
y
eady yeyeadyye 
and this reduces to:

 
yd ayd a



, so if ad

,
yad
yad

, contradicting the above. We conclu

de that
ad

and

is of the form

ai
aeiy
a eay
e


,
00
i




 






apply Lemma 2. The other possibility is that and we can
0
y
x

, sothat bc

 and  is either of the form
0
0
00
ab
bd
e











cb, a symmetric matrix (when ), or of

the form
00
a y
e
abiy
b




 






(whe cb, since in this n
case ad

).
Case 2. 0xiy

, then gives bx dyey


bied

 gives


ciae, and axcyxe

, so
that

has the form

ai ediy
e


.ia edy
00




 






We procely as in Caduceed exactse 1 to re

t
2





o the
222 2
22222
22
(3)
abxacb xyxeac abcdaxcy
ac bdxycdyeyab cdbdbx dy
xeax cyxbydxye

  

 




2
2
ye e


d



 




 
 
 
2
2
2
2
2
2
00
(4)
00
adeea iadeidd
iaeaideaeadedyeady ye
e
adee aiadeiddeiay iydeiye
iae aideae adedye adyye
e

 


 





 







eiayiydeiye 

G. GOODSON 25
situation of Lemma 2.
In Case (III), where , we proceed ex-
actly as in Case (II) to deduce the result.
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0
0
axy
bc
de





XX XX,” Mos-
cow University Computational Mathematics and Cyber-
2, 2010, pp. 51-55.
doi:10.3103/S0278641910020019
[2] G. R. Goodson and R. A. Horn, “Canonical Forms for
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Conjugate,” Linear Algebra and Its Applications, Vol.
430, No. 4, 2009, pp. 1025-1038.
doi:10.1016/j.laa.2008.09.039
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