For a class of n × n-matrices, we get related real solutions to the matrix equation Φt (A) = 1/n Jn by generalizing the approach of and applying the results of Zhang, Yang, and Cao [SIAM J. Matrix Anal. Appl., 21 (1999), pp. 642–645]. These solutions contain not only those obtained by Zhang, Yang, and Cao but also some which are neither diagonally nor permutation equivalent to those obtained by Zhang, Yang, and Cao. Therefore, the open problem proposed by Zhang, Yang, and Cao in the cited paper is solved

Chen, Yuming

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Mathematics Faculty Publications Mathematics 
2001 
On Real Solutions of the Equation Φt (A) = 1/n Jn 
Yuming Chen 
Wilfrid Laurier University, ychen@wlu.ca 
Follow this and additional works at: https://scholars.wlu.ca/math_faculty 
Recommended Citation 
Chen, Yuming, "On Real Solutions of the Equation Φt (A) = 1/n Jn" (2001). Mathematics Faculty 
Publications. 20. 
https://scholars.wlu.ca/math_faculty/20 
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ON REAL SOLUTIONS OF THE EQUATION Φt(A) = 1
n
Jn
∗
YUMING CHEN†
SIAM J. MATRIX ANAL. APPL. c© 2001 Society for Industrial and Applied Mathematics
Vol. 22, No. 3, pp. 965–970
Abstract. For a class of n × n-matrices, we get related real solutions to the matrix equation
Φt(A) = 1
n
Jn by generalizing the approach of and applying the results of Zhang, Yang, and Cao
[SIAM J. Matrix Anal. Appl., 21 (1999), pp. 642–645]. These solutions contain not only those
obtained by Zhang, Yang, and Cao but also some which are neither diagonally nor permutation
equivalent to those obtained by Zhang, Yang, and Cao. Therefore, the open problem proposed by
Zhang, Yang, and Cao in the cited paper is solved.
Key words. Hadamard product, diagonally equivalent, permutation equivalent
AMS subject classifications. 15A24, 93A99, 65F99
PII. S0895479800372912
1. Introduction. For a given positive integer n, let Mn(R) and GLn(R) be the
sets of all n × n real matrices and all n × n real nonsingular matrices, respectively.
Two important members of Mn(R) are the n×n identity and all-one matrix, denoted
as In and Jn, respectively.
For A = (aij) and B = (bij) in Mn(R), the Hadamard product of A and B is
defined as A ◦B = (aijbij) ∈Mn(R). Then we define
Φ : GLn(R)→Mn(R)
by
Φ(A) = A ◦A−T , A ∈ GLn(R),
where A−T means the inverse transpose, (A−1)T , of A. The mapping Φ arises in
mathematical control theory in chemical engineering design problems. The basic
question about Φ is to determine its range.
It is easy to see that every matrix in the range of Φ has row and column sums 1.
However, the converse is not true. In fact, Johnson and Shapiro [1] showed that the
equation Φ(A) = 13J3 has no real solutions. So they asked whether the equation
Φ(A) =
1
n
Jn
has a real solution. This problem was solved by Zhang, Yang, and Cao [2]. In fact,
they studied the more general problem, i.e., the existence of real solutions of the
equation
Φt(A) =
1
n
Jn(1)
for any positive integer t, where Φt is the mapping Φ applied t times.
∗Received by the editors June 5, 2000; accepted for publication (in revised form) by R. Brualdi
September 16, 2000; published electronically January 19, 2001. This research was partially supported
by the Izaak Walton Killam Memorial Postdoctoral Fellowship, University of Alberta.
http://www.siam.org/journals/simax/22-3/37291.html
†Department of Mathematics and Statistics, York University, 4700 Keele Street, Toronto, ON,
M3J 1P3 Canada. Current address: Department of Mathematical Sciences, University of Alberta,
Edmonton, AB, T6G 2G1 Canada (ychen@math.ualberta.ca).
965
966 YUMING CHEN
The purpose of this note is to obtain some new solutions to (1) by generalizing
the approach of and applying the results of [2]. Some of the solutions are neither
diagonally equivalent nor permutation equivalent to those obtained in [2]. Hence, the
open problem proposed in [2] is solved.
The organization of this note is as follows. First, by an example, we partially
answer the open problem in [2] and introduce the notion of permutation equivalence.
Then we obtain new solutions to (1) which are related to a class of n × n-matrices.
As a result, the above-mentioned open problem is solved.
2. An example. First, we recall some results about the mapping Φ.
Lemma 1 (see [1, Observations 2 and 4]). For A ∈ GLn(R),
(i) if D and E in GLn(R) are diagonal, then Φ(DAE) = Φ(A);
(ii) if P and Q in GLn(R) are permutation matrices, then Φ(PAQ) = PΦ(A)Q.
When n = 4 and t = 1, the unique solution to (1) with respect to diagonal
equivalence obtained in [2] is
A =


−1 1 1 1
1 −1 1 1
1 1 −1 1
1 1 1 −1

 .(2)
Let P be the permutation 

0 1 0 0
1 0 0 0
0 0 1 0
0 0 0 1

 .
Then, by Lemma 1,
Φ(PA) = PΦ(A) = P
(
1
4
J4
)
=
1
4
J4.
But PA is not diagonally equivalent to A. If not, let D = diag(d1, d2, d3, d4) and
E = diag(e1, e2, e3, e4) be two nonsingular diagonal matrices such that
DAE = PA.
Particularly, we have d1e1 = −1, d3e1 = 1, d1e3 = 1, and d3e3 = 1. This is impossible
since the first two equations give d1 = −d3 while the last two give d1 = d3.
With (1) and the example in mind, Lemma 1 leads us to introduce the notion of
permutation equivalence in a similar way to diagonal equivalence (see [2]).
Zhang, Yang, and Cao [2] also proved that, for n = 2 and t = 1, there is only
one nondiagonally equivalent solution to (1) and no solution to (1) when n = 2 and
t ≥ 2. This, combined with the result of Johnson and Shapiro [1], intrigues us, and
we naturally ask, for n ≥ 4, whether there exist real solutions to (1) which are neither
diagonally equivalent nor permutation equivalent to those found by Zhang, Yang, and
Cao [2]. A positive answer will be given in the next section. In this section, we give
a partial answer for the case where n = 4 and t = 1. For a = 0, let
Aa =


1 a −1 a
a 1 a −1
−1 a 1 a
a −1 a 1

 .(3)
ON REAL SOLUTIONS OF THE EQUATION Φt(A) = 1
n
Jn 967
Then |Aa| = −16a2 = 0 and
A−1a =
1
4


1 a−1 −1 a−1
a−1 1 a−1 −1
−1 a−1 1 a−1
a−1 −1 a−1 1

 .
Therefore,
Φ(Aa) =
1
4
J4.
Now, assume that a ∈ R \ {0,−1, 1}. Obviously, Aa is not permutation equivalent
to A given in (2). Now, we claim that Aa is not diagonally equivalent to A. In fact,
if D = diag(d1, d2, d3, d4) and E = diag(e1, e2, e3, e4) are two nonsingular diagonal
matrices such that
DAE = Aa,
then we have e1 = e3, e2 = e4 = −ae1, d1 = d3, d2 = d4 = −ad1, d1e1 = −1, and
d2e2 = −1. Thus −1 = d2e2 = a2d1e1 = −a2 or a2 = 1, a contradiction.
Remark 1. To obtain the form of Aa in (3), we tried the cyclic matrix generated
by (1, a, b, c). By requiring it satisfy Φ(A) = 14J4, we get b = 1 or −1. But b = 1 is
not suitable. Taking b = −1, we have a = c. Fortunately, the matrix we get satisfies
Φ(A) = 14J4. We believe that this approach is also applicable to deal with the general
case. However, in the following section, by generalizing the approach of and using the
results of Zhang, Yang, and Cao [2], we give a new approach to the general case.
3. New solutions to Φt(A) = 1
n
Jn. In this section we always assume n ≥ 4.
Note that what is important in the presentation of Zhang, Yang, and Cao [2] are
some special properties of Jn such as J
2
n = nJn and Jn ◦ JTn = Jn. Inspired by this
observation, for k ∈ R such that k ≤ 0 or k ≥ 2n− 4 (these restrictions on k will be
clear later), we introduce a subset Mn,k(R) of Mn(R) as follows:
Mn,k(R) = {A ∈Mn(R);A◦AT = Jn, aii = 1 for i = 1, . . ., n, A2 = kIn + (n− k)A}.
Generalizing the approach of and applying some results of Zhang, Yang, and Cao [2],
we show that for each A ∈Mn,k(R) there exist corresponding solutions to (1).
Examples. Let a = (a1, . . . , an−1) ∈ Rn−1 with ai = 0 for i ∈ {1, . . . , n − 1}.
Define Aa ∈Mn(R) by
Aaij =


1 if i = j,
j−1∏
k=i
ak if i < j,
1
i−1∏
k=j
ak
if i > j.
Then it is easy to show that Aa ∈ Mn,0(R). Particularly, when ai = 1 for i ∈
{1, . . . , n− 1}, Aa = Jn. Furthermore, define A˜a ∈Mn(R) by
A˜aij =
{
Aaij if i = j,
(−1)i−j+1Aaij if i = j.
968 YUMING CHEN
Then A˜a ∈Mn,2n−4(R). Note that neither Aa nor A˜a is symmetric if there exists an
i0 ∈ {1, . . . , n− 1} such that a2i0 = 1.
The following results can be easily proved and hence the proofs are omitted.
Lemma 2. Let A ∈Mn,k(R). If a[a+b(n−k)]−b2k = 0, then aIn+bA ∈ GLn(R)
with
(aIn + bA)
−1 =
1
a[a+ b(n− k)]− b2k {[a+ b(n− k)]In − bA}
and therefore
Φ(aIn + bA) =
1
a[a+ b(n− k)]− b2k {[a
2 + ab(n− k) + b2(n− k)]In − b2Jn}.
Lemma 3. Let A ∈Mn,k(R) and λ = k. Denote
A(λ) =
1
λ− k [(λ+ n− k)In −A].(4)
Then
(i) A(λ) ∈ GLn(R) if λ = −(n−k)±
√
(n−k)2+4k
2 ;
(ii) if λ = −(n−k)±
√
(n−k)2+4k
2 , we have Φ(A(λ)) = Jn(µ), where µ = λ(λ+n−
k)− k;
(iii) A(α) and A(β) are not diagonally equivalent if α = β.
Proof. Since Jn ∈Mn,0(R), it follows from (4) that
Jn(λ) =
1
λ
[(λ+ n)In − Jn](5)
for λ = 0. Now, (i) and (ii) follow easily from Lemma 2. The proof of (iii) is similar
to that of (iii) of Lemma 2 of Zhang, Yang, and Cao [2]. This completes the proof of
the lemma.
Theorem 1. Let A ∈ Mn,k(R) and t be a positive integer. Then if n > 4, there
are 2t distinct, real values of λ such that Φt(A(λ)) = 1nJn and hence (1) has at least 2
t
nondiagonally equivalent solutions. When n = 4, there are 2t−1 distinct, real values of
λ such that Φt(A(λ)) = 14J4 and hence (1) has at least 2
t−1 nondiagonally equivalent
solutions.
Proof. We prove only the theorem for the case where n > 4. The proof is similar
for the case where n = 4. First note that the nondiagonal equivalence follows from
(iii) of Lemma 3. Second, it follows from (5) that
Jn(−n) = 1
n
Jn.(6)
Now we distinguish two cases to complete the proof.
Case 1. t = 1. Consider the equation
λ(λ+ n− k)− k = −n.(7)
Note ∆ = (n − k)2 − 4(n − k) = (n − k)(n − k − 4) > 0 since k ≤ 0 or k ≥ 2n − 4.
Thus (7) has two distinct real solutions λ1 and λ2. By (6) and (ii) of Lemma 3,
ON REAL SOLUTIONS OF THE EQUATION Φt(A) = 1
n
Jn 969
Φ(A(λ1)) = Φ(A(λ2)) = Jn(−n) = 1
n
Jn.
Case 2. t > 1. It follows from (ii) of Lemma 3, for µ = −n, that
Φ(Jn(µ)) = Jn(f(µ)),
where f(µ) = µ(µ + n). Thus, if λ = −(n−k)±
√
(n−k)2+4k
2 , using (ii) of Lemma 3
again, we have
Φt(A(λ)) = Φt−1(Φ(A(λ))) = Φt−1(Jn(µ)) = Jn(f t−1(µ)),(8)
where µ = λ(λ+n−k)−k. Lemma 3 of Zhang, Yang, and Cao [2] tells us that in the
interval (−n24 , 0] there are 2t−1 distinct, real solutions to the equation f t−1(µ) = −n,
say µ1, . . . , µ2t−1 . For i = 1, . . ., 2
t−1, consider
λ(λ+ n− k)− k = µi.(9)
Noting
∆ = (n− k)2 + 4(k + µi)
= (n2 + 4µi) + (k
2 − 2nk + 4k)
> k2 − 2nk + 4k
= k[k − (2n− 4)]
≥ 0
(from here you see why we require k ≤ 0 or k ≥ 2n − 4), we know that (9) has two
distinct, real solutions, say λi,1 and λi,2. Moreover, it is easy to see that all λ1,1, λ1,2,
. . ., λ2t−1,1, and λ2t−1,2 are distinct. Thus it follows from (6) and (8) that
Φt(A(λi,j)) = Jn(f
t−1(µi)) = Jn(−n) = 1
n
Jn, i = 1, . . . , 2
t−1, j = 1, 2,
and the proof is complete.
Remark 2. For A ∈ Mn,k(R), we can find 2t and 2t−1 mutually nondiagonally
equivalent real solutions A(λ0) to (1) for n > 4 and n = 4, respectively, where λ0
satisfies the following inverted iteration (see Theorem 1 here and Remark 1 of Zhang,
Yang, and Cao [2]):


λt = −n,
λk =
−n±
√
n2+4λk+1
2 , k = 1, . . . , t− 1,
λ0 =
−(n−k)±
√
(n−k)2+4(λ1+k)
2 .
Remark 3. For a = (a1, . . . , an−1) ∈ Rn−1 with ai = 0 for i ∈ {1, . . . , n− 1}, let
Aa and A˜a be defined as in the examples above. Then we can easily show that the
solutions associated with Aa and A˜a are diagonally equivalent to those associated with
A1 and A˜1, respectively, where 1 = (1, . . . , 1) ∈ Rn−1. But the solutions associated
with A1 and A˜1 are neither diagonally equivalent nor permutation equivalent. Note
that the solutions associated with A1 are just those obtained by Zhang, Yang, and
Cao [2] and hence the open problem, whether there are real solutions to (1) which
are not diagonally equivalent to those found in [2] when n ≥ 4, proposed by them is
970 YUMING CHEN
solved. Moreover, when n = 4, t = 1 and a ∈ R \ {0,−1, 1}, the solutions given in
section 2 are not associated with any A ∈Mn,k(R).
REFERENCES
[1] C. R. Johnson and H. M. Shapiro, Mathematical aspects of the relative gain array (A ◦A−T ),
SIAM J. Algebraic Discrete Methods, 7 (1986), pp. 627–644.
[2] X. Zhang, Z. Yang, and C. Cao, Real solutions of the equation Φt(A) = 1
n
Jn, SIAM J. Matrix
Anal. Appl., 21 (1999), pp. 642–645.


On Real Solutions of the Equation Φ\u3csup\u3e\u3cem\u3et\u3c/em\u3e\u3c/sup\u3e (\u3cem\u3eA\u3c/em\u3e) = 1/\u3cem\u3en\u3c/em\u3e J\u3csub\u3e\u3cem\u3en\u3c/em\u3e\u3c/sub\u3e

Wilfrid Laurier University

Wilfrid Laurier University Scholars Commons @ Laurier Mathematics Faculty Publications Mathematics 2001 On Real Solutions of the Equation Φt (A) = 1/n Jn Yuming Chen Wilfrid Laurier University, ychen@wlu.ca Follow this and additional works at: https://scholars.wlu.ca/math_faculty Recommended Citation Chen, Yuming, "On Real Solutions of the Equation Φt (A) = 1/n Jn" (2001). Mathematics Faculty Publications. 20. https://scholars.wlu.ca/math_faculty/20 This Article is brought to you for free and open access by the Mathematics at Scholars Commons @ Laurier. It has been accepted for inclusion in Mathematics Faculty Publications by an authorized administrator of Scholars Commons @ Laurier. For more information, please contact scholarscommons@wlu.ca. ON REAL SOLUTIONS OF THE EQUATION Φt(A) = 1nJn∗YUMING CHEN†SIAM J. MATRIX ANAL. APPL. c© 2001 Society for Industrial and Applied MathematicsVol. 22, No. 3, pp. 965–970Abstract. For a class of n × n-matrices, we get related real solutions to the matrix equationΦt(A) = 1nJn by generalizing the approach of and applying the results of Zhang, Yang, and Cao[SIAM J. Matrix Anal. Appl., 21 (1999), pp. 642–645]. These solutions contain not only thoseobtained by Zhang, Yang, and Cao but also some which are neither diagonally nor permutationequivalent to those obtained by Zhang, Yang, and Cao. Therefore, the open problem proposed byZhang, Yang, and Cao in the cited paper is solved.Key words. Hadamard product, diagonally equivalent, permutation equivalentAMS subject classiﬁcations. 15A24, 93A99, 65F99PII. S08954798003729121. Introduction. For a given positive integer n, let Mn(R) and GLn(R) be thesets of all n × n real matrices and all n × n real nonsingular matrices, respectively.Two important members of Mn(R) are the n×n identity and all-one matrix, denotedas In and Jn, respectively.For A = (aij) and B = (bij) in Mn(R), the Hadamard product of A and B isdeﬁned as A ◦B = (aijbij) ∈Mn(R). Then we deﬁneΦ : GLn(R)→Mn(R)byΦ(A) = A ◦A−T , A ∈ GLn(R),where A−T means the inverse transpose, (A−1)T , of A. The mapping Φ arises inmathematical control theory in chemical engineering design problems. The basicquestion about Φ is to determine its range.It is easy to see that every matrix in the range of Φ has row and column sums 1.However, the converse is not true. In fact, Johnson and Shapiro [1] showed that theequation Φ(A) = 13J3 has no real solutions. So they asked whether the equationΦ(A) =1nJnhas a real solution. This problem was solved by Zhang, Yang, and Cao [2]. In fact,they studied the more general problem, i.e., the existence of real solutions of theequationΦt(A) =1nJn(1)for any positive integer t, where Φt is the mapping Φ applied t times.∗Received by the editors June 5, 2000; accepted for publication (in revised form) by R. BrualdiSeptember 16, 2000; published electronically January 19, 2001. This research was partially supportedby the Izaak Walton Killam Memorial Postdoctoral Fellowship, University of Alberta.http://www.siam.org/journals/simax/22-3/37291.html†Department of Mathematics and Statistics, York University, 4700 Keele Street, Toronto, ON,M3J 1P3 Canada. Current address: Department of Mathematical Sciences, University of Alberta,Edmonton, AB, T6G 2G1 Canada (ychen@math.ualberta.ca).965966 YUMING CHENThe purpose of this note is to obtain some new solutions to (1) by generalizingthe approach of and applying the results of [2]. Some of the solutions are neitherdiagonally equivalent nor permutation equivalent to those obtained in [2]. Hence, theopen problem proposed in [2] is solved.The organization of this note is as follows. First, by an example, we partiallyanswer the open problem in [2] and introduce the notion of permutation equivalence.Then we obtain new solutions to (1) which are related to a class of n × n-matrices.As a result, the above-mentioned open problem is solved.2. An example. First, we recall some results about the mapping Φ.Lemma 1 (see [1, Observations 2 and 4]). For A ∈ GLn(R),(i) if D and E in GLn(R) are diagonal, then Φ(DAE) = Φ(A);(ii) if P and Q in GLn(R) are permutation matrices, then Φ(PAQ) = PΦ(A)Q.When n = 4 and t = 1, the unique solution to (1) with respect to diagonalequivalence obtained in [2] isA =−1 1 1 11 −1 1 11 1 −1 11 1 1 −1 .(2)Let P be the permutation 0 1 0 01 0 0 00 0 1 00 0 0 1 .Then, by Lemma 1,Φ(PA) = PΦ(A) = P(14J4)=14J4.But PA is not diagonally equivalent to A. If not, let D = diag(d1, d2, d3, d4) andE = diag(e1, e2, e3, e4) be two nonsingular diagonal matrices such thatDAE = PA.Particularly, we have d1e1 = −1, d3e1 = 1, d1e3 = 1, and d3e3 = 1. This is impossiblesince the ﬁrst two equations give d1 = −d3 while the last two give d1 = d3.With (1) and the example in mind, Lemma 1 leads us to introduce the notion ofpermutation equivalence in a similar way to diagonal equivalence (see [2]).Zhang, Yang, and Cao [2] also proved that, for n = 2 and t = 1, there is onlyone nondiagonally equivalent solution to (1) and no solution to (1) when n = 2 andt ≥ 2. This, combined with the result of Johnson and Shapiro [1], intrigues us, andwe naturally ask, for n ≥ 4, whether there exist real solutions to (1) which are neitherdiagonally equivalent nor permutation equivalent to those found by Zhang, Yang, andCao [2]. A positive answer will be given in the next section. In this section, we givea partial answer for the case where n = 4 and t = 1. For a = 0, letAa =1 a −1 aa 1 a −1−1 a 1 aa −1 a 1 .(3)ON REAL SOLUTIONS OF THE EQUATION Φt(A) = 1nJn 967Then |Aa| = −16a2 = 0 andA−1a =141 a−1 −1 a−1a−1 1 a−1 −1−1 a−1 1 a−1a−1 −1 a−1 1 .Therefore,Φ(Aa) =14J4.Now, assume that a ∈ R \ {0,−1, 1}. Obviously, Aa is not permutation equivalentto A given in (2). Now, we claim that Aa is not diagonally equivalent to A. In fact,if D = diag(d1, d2, d3, d4) and E = diag(e1, e2, e3, e4) are two nonsingular diagonalmatrices such thatDAE = Aa,then we have e1 = e3, e2 = e4 = −ae1, d1 = d3, d2 = d4 = −ad1, d1e1 = −1, andd2e2 = −1. Thus −1 = d2e2 = a2d1e1 = −a2 or a2 = 1, a contradiction.Remark 1. To obtain the form of Aa in (3), we tried the cyclic matrix generatedby (1, a, b, c). By requiring it satisfy Φ(A) = 14J4, we get b = 1 or −1. But b = 1 isnot suitable. Taking b = −1, we have a = c. Fortunately, the matrix we get satisﬁesΦ(A) = 14J4. We believe that this approach is also applicable to deal with the generalcase. However, in the following section, by generalizing the approach of and using theresults of Zhang, Yang, and Cao [2], we give a new approach to the general case.3. New solutions to Φt(A) = 1nJn. In this section we always assume n ≥ 4.Note that what is important in the presentation of Zhang, Yang, and Cao [2] aresome special properties of Jn such as J2n = nJn and Jn ◦ JTn = Jn. Inspired by thisobservation, for k ∈ R such that k ≤ 0 or k ≥ 2n− 4 (these restrictions on k will beclear later), we introduce a subset Mn,k(R) of Mn(R) as follows:Mn,k(R) = {A ∈Mn(R);A◦AT = Jn, aii = 1 for i = 1, . . ., n, A2 = kIn + (n− k)A}.Generalizing the approach of and applying some results of Zhang, Yang, and Cao [2],we show that for each A ∈Mn,k(R) there exist corresponding solutions to (1).Examples. Let a = (a1, . . . , an−1) ∈ Rn−1 with ai = 0 for i ∈ {1, . . . , n − 1}.Deﬁne Aa ∈Mn(R) byAaij =1 if i = j,j−1∏k=iak if i < j,1i−1∏k=jakif i > j.Then it is easy to show that Aa ∈ Mn,0(R). Particularly, when ai = 1 for i ∈{1, . . . , n− 1}, Aa = Jn. Furthermore, deﬁne A˜a ∈Mn(R) byA˜aij ={Aaij if i = j,(−1)i−j+1Aaij if i = j.968 YUMING CHENThen A˜a ∈Mn,2n−4(R). Note that neither Aa nor A˜a is symmetric if there exists ani0 ∈ {1, . . . , n− 1} such that a2i0 = 1.The following results can be easily proved and hence the proofs are omitted.Lemma 2. Let A ∈Mn,k(R). If a[a+b(n−k)]−b2k = 0, then aIn+bA ∈ GLn(R)with(aIn + bA)−1 =1a[a+ b(n− k)]− b2k {[a+ b(n− k)]In − bA}and thereforeΦ(aIn + bA) =1a[a+ b(n− k)]− b2k {[a2 + ab(n− k) + b2(n− k)]In − b2Jn}.Lemma 3. Let A ∈Mn,k(R) and λ = k. DenoteA(λ) =1λ− k [(λ+ n− k)In −A].(4)Then(i) A(λ) ∈ GLn(R) if λ = −(n−k)±√(n−k)2+4k2 ;(ii) if λ = −(n−k)±√(n−k)2+4k2 , we have Φ(A(λ)) = Jn(µ), where µ = λ(λ+n−k)− k;(iii) A(α) and A(β) are not diagonally equivalent if α = β.Proof. Since Jn ∈Mn,0(R), it follows from (4) thatJn(λ) =1λ[(λ+ n)In − Jn](5)for λ = 0. Now, (i) and (ii) follow easily from Lemma 2. The proof of (iii) is similarto that of (iii) of Lemma 2 of Zhang, Yang, and Cao [2]. This completes the proof ofthe lemma.Theorem 1. Let A ∈ Mn,k(R) and t be a positive integer. Then if n > 4, thereare 2t distinct, real values of λ such that Φt(A(λ)) = 1nJn and hence (1) has at least 2tnondiagonally equivalent solutions. When n = 4, there are 2t−1 distinct, real values ofλ such that Φt(A(λ)) = 14J4 and hence (1) has at least 2t−1 nondiagonally equivalentsolutions.Proof. We prove only the theorem for the case where n > 4. The proof is similarfor the case where n = 4. First note that the nondiagonal equivalence follows from(iii) of Lemma 3. Second, it follows from (5) thatJn(−n) = 1nJn.(6)Now we distinguish two cases to complete the proof.Case 1. t = 1. Consider the equationλ(λ+ n− k)− k = −n.(7)Note ∆ = (n − k)2 − 4(n − k) = (n − k)(n − k − 4) > 0 since k ≤ 0 or k ≥ 2n − 4.Thus (7) has two distinct real solutions λ1 and λ2. By (6) and (ii) of Lemma 3,ON REAL SOLUTIONS OF THE EQUATION Φt(A) = 1nJn 969Φ(A(λ1)) = Φ(A(λ2)) = Jn(−n) = 1nJn.Case 2. t > 1. It follows from (ii) of Lemma 3, for µ = −n, thatΦ(Jn(µ)) = Jn(f(µ)),where f(µ) = µ(µ + n). Thus, if λ = −(n−k)±√(n−k)2+4k2 , using (ii) of Lemma 3again, we haveΦt(A(λ)) = Φt−1(Φ(A(λ))) = Φt−1(Jn(µ)) = Jn(f t−1(µ)),(8)where µ = λ(λ+n−k)−k. Lemma 3 of Zhang, Yang, and Cao [2] tells us that in theinterval (−n24 , 0] there are 2t−1 distinct, real solutions to the equation f t−1(µ) = −n,say µ1, . . . , µ2t−1 . For i = 1, . . ., 2t−1, considerλ(λ+ n− k)− k = µi.(9)Noting∆ = (n− k)2 + 4(k + µi)= (n2 + 4µi) + (k2 − 2nk + 4k)> k2 − 2nk + 4k= k[k − (2n− 4)]≥ 0(from here you see why we require k ≤ 0 or k ≥ 2n − 4), we know that (9) has twodistinct, real solutions, say λi,1 and λi,2. Moreover, it is easy to see that all λ1,1, λ1,2,. . ., λ2t−1,1, and λ2t−1,2 are distinct. Thus it follows from (6) and (8) thatΦt(A(λi,j)) = Jn(ft−1(µi)) = Jn(−n) = 1nJn, i = 1, . . . , 2t−1, j = 1, 2,and the proof is complete.Remark 2. For A ∈ Mn,k(R), we can ﬁnd 2t and 2t−1 mutually nondiagonallyequivalent real solutions A(λ0) to (1) for n > 4 and n = 4, respectively, where λ0satisﬁes the following inverted iteration (see Theorem 1 here and Remark 1 of Zhang,Yang, and Cao [2]):λt = −n,λk =−n±√n2+4λk+12 , k = 1, . . . , t− 1,λ0 =−(n−k)±√(n−k)2+4(λ1+k)2 .Remark 3. For a = (a1, . . . , an−1) ∈ Rn−1 with ai = 0 for i ∈ {1, . . . , n− 1}, letAa and A˜a be deﬁned as in the examples above. Then we can easily show that thesolutions associated with Aa and A˜a are diagonally equivalent to those associated withA1 and A˜1, respectively, where 1 = (1, . . . , 1) ∈ Rn−1. But the solutions associatedwith A1 and A˜1 are neither diagonally equivalent nor permutation equivalent. Notethat the solutions associated with A1 are just those obtained by Zhang, Yang, andCao [2] and hence the open problem, whether there are real solutions to (1) whichare not diagonally equivalent to those found in [2] when n ≥ 4, proposed by them is970 YUMING CHENsolved. Moreover, when n = 4, t = 1 and a ∈ R \ {0,−1, 1}, the solutions given insection 2 are not associated with any A ∈Mn,k(R).REFERENCES[1] C. R. Johnson and H. M. Shapiro, Mathematical aspects of the relative gain array (A ◦A−T ),SIAM J. Algebraic Discrete Methods, 7 (1986), pp. 627–644.[2] X. Zhang, Z. Yang, and C. Cao, Real solutions of the equation Φt(A) = 1nJn, SIAM J. MatrixAnal. Appl., 21 (1999), pp. 642–645.

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On Real Solutions of the Equation Φ\u3csup\u3e\u3cem\u3et\u3c/em\u3e\u3c/sup\u3e (\u3cem\u3eA\u3c/em\u3e) = 1/\u3cem\u3en\u3c/em\u3e J\u3csub\u3e\u3cem\u3en\u3c/em\u3e\u3c/sub\u3e

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