A Characterization of Trace Zero Bisymmetric Nonnegative $5\Times 5

$A Characterization of Trace Zero Bisymmetric Nonnegative $5\Times 5$

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In [14], O. Spector showed that a necessary and sufficient condition for a list of 5 5 real numbers λ1 λ2 λ3 λ4 λ5 such that λ5 λ1 and λi =0 to be ≥ ≥ ≥ ≥ ≥− i=1 the eigenvalues of a 5 5 traceless symmetric nonnegative matrix is λ2 + λ5 0 5 3 × P ≤ and i=1 λi 0. In this article, we prove that the above condition is also a necessary and≥ sufficient condition of the BNIEP for traceless 5 5 matrices. We organizeP our paper as follows. In Section 2, we give sufficient× conditions of the BNIEP for 5 5 matrices. This result is Proposition 2.1. Then in Section 3, we characterize all× bisymmetric nonnegative traceless 5 5 matrices. Our main result of this characterization is Theorem 3.2. ×

2. Sufficient conditions of the BNIEP for 5 5 matrices ×

Let λ1 λ2 λ3 λ4 λ5 be real numbers. If λ1, λ2, λ3, λ4, λ5 is the ≥ ≥ ≥ ≥ 5 eigenvalues of a 5 5 nonnegative matrix, then =1 λi 0 and by the Perron- × i ≥ Frobenius Theorem, λ1 λ5 . Since we consider the BNIEP for 5 5 matrices, we ≥ | | P 5× from now on assume that λ1 λ2 λ3 λ4 λ5 λ1 and λi 0. ≥ ≥ ≥ ≥ ≥− i=1 ≥ If λ1 λ2 λ3 λ4 λ5 0 or λ1 λ2 λ3 λ4 0P> λ5 then a 5 5 ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ × nonnegative bisymmetric matrix with the eigenvalues λ1, λ2, λ3, λ4, λ5 is the matrix

λ1 + λ5 λ1 λ5 − 2 2 λ2 + λ4 λ2 λ4  −  2 2   L1 =  λ3  .  λ2 λ4 λ2 + λ4     −   2 2  λ1 λ5 λ1 + λ5   −   2 2    If λ1 0 > λ2 λ3 λ4 λ5 then a 5 5 nonnegative bisymmetric matrix ≥ ≥ ≥ ≥ × with the eigenvalues λ1, λ2, λ3, λ4, λ5 is the matrix

0 a b a λ5 − a 0 c λ4 a  5 −  L2 = b c i=1 λi c b ,  a λ4 c 0 a   − P   λ5 a b a 0  −  where   1 (λ1 + λ5)(λ2 + λ5)(λ3 + λ4) a = , 2s λ1 + λ2 λ3 + λ5 −

(λ1 + λ5)(λ2 + λ5)(λ1 + λ2 + λ4 + λ5) b = − , 2(λ1 + λ2 λ3 + λ5) s − and

(λ3 + λ4)(λ1 + λ2 + λ4 + λ5) c = − . r 2 Now we consider the case λ1 λ2 0 > λ3 λ4 λ5. If λ1 + λ3 + λ4 0 and ≥ ≥ ≥ ≥ ≥ λ2 + λ5 < 0 then the matrix L2 in the previous case is nonnegative and hence it is A CHARACTERIZATION OF TRACE ZERO BISYMMETRIC NONNEGATIVE 5 × 5 MATRICES3 our desired matrix. If λ2 + λ5 0 and λ1 + λ3 + λ4 0 then a 5 5 nonnegative ≥ ≥ × bisymmetric matrix with the eigenvalues λ1, λ2, λ3, λ4, λ5 is the matrix

λ2 + λ5 λ2 λ5 0 0 0 − 2 2  0 0 d λ4 0  − L3 =  0 d λ1 + λ3 + λ4 d 0  ,    0 λ4 d 0 0   λ2 λ5 − λ2 + λ5     − 0 0 0   2 2    (λ3 + λ4)(λ1 + λ4) where d = − . Finally, if λ2 +λ5 0 and λ1 +λ3 +λ4 < 0 then 2 ≥ 5 r i=1 λi < λ2 +λ5 contradicting a necessary condition of the result (Lemma 1 in [8]) by R. Loewy and J. J. McDonald and hence in this case there is no nonnegative P bisymmetric matrix with the eigenvalues λ1, λ2, λ3, λ4, λ5. Next, we consider the case λ1 λ2 λ3 0 > λ4 λ5. If λ1 + λ2 + λ4 + λ5 0 ≥ ≥ ≥ ≥ ≥ and λ2 + λ5 0, then λ2 + λ4 0 and the matrix L1 is our desired matrix. If ≥ ≥ λ1 + λ2 + λ4 + λ5 0, λ2 + λ5 < 0 and λ3 + λ4 0, then the matrix ≥ ≥ λ3 + λ4 λ3 λ4 0 0 0 − 2 2  0 0 g λ5 0  − L4 =  0 g λ1 + λ2 + λ5 g 0  ,    0 λ5 g 0 0   λ3 λ4 − λ3 + λ4     − 0 0 0   2 2    (λ2 + λ5)(λ1 + λ5) where g = − , will be our desired matrix. If λ1 +λ2 +λ4 +λ5 r 2 ≥ 0, λ2 + λ5 < 0 and λ3 + λ4 < 0, then again the matrix L2 can be used. We do not have a complete solution for the case when λ1 + λ2 + λ4 + λ5 < 0. However, some partial result (Corollary 3.4) will be given at the end of Section 3.

Summing up what we have done so far, we get the following result.

Proposition 2.1. Let λ1 λ2 λ3 λ4 λ5 λ1 be real numbers such that 5 ≥ ≥ ≥ ≥ ≥ − λi 0. If λ1, λ2, λ3, λ4, λ5 satisﬁes one of the following conditions i=1 ≥ (1) λ4 0 or λ2 0, P ≥ ≤ 5 (2) λ2 0 > λ3 and λi λ2 + λ5, ≥ i=1 ≥ (3) λ3 0 > λ4 and λ1 + λ2 + λ4 + λ5 0, ≥ P ≥ then there is a 5 5 nonnegative bisymmetric matrix with the eigenvalues λ1, λ2, × λ3, λ4, λ5.

Proposition 2.1 will be employed to obtain our main result, Theorem 3.3 in Section 3. 4 SOMCHAISOMPHOTPHISUTANDKENGWIBOONTON

3. A necessary and sufficient condition of the BNIEP for 5 5 traceless matrices ×

In this section, we give a characterization of bisymmetric nonnegative traceless 5 5 matrices according to the characterization of symmetric nonnegative traceless 5 × 5 matrices given by O. Spector in [14]. To obtain our main result, we use Proposition× 2.1 in the previous section along with a useful result of Cantoni and Butler in [2] which gives a standard form of a bisymmetric matrix. This result of Cantoni and Butler (Lemma 2(ii) and Lemma 3(b) in [2]) states that a (2m + 1) A x JCJ × (2m+1) matrix Q is bisymmetric if and only if Q is of the form xT p xT J ,  C Jx JAJ where A and C are m m matrices, A = AT , JCT J = C, x is an m 1 matrix and p is a real number,× and in this case the eigenvalue of Q can be partitioned× into p √2xT the eigenvalue of A JC and the eigenvalues of . − √2x A + JC So, if p R, x is an m 1 matrix and X, Y are m m symmetric matrices, X∈ + Y ×X Y × then A = and JC = − are symmetric and hence the matrix 2 2 X + Y √2x (X Y )J 1 − Q = √2xT p √2xT J 2   J(X Y ) √2Jx J(X + Y )J −   p xT is a bisymmetric matrix with the eigenvalues obtained from the matrices x X x A JCJ √2  xT xT J  and Y . In terms of A and C, the matrix Q is of the form p . √2 √2   Jx   C JAJ    √2  Note that, Q is a nonnegative matrix if and only if p 0 and x, X + Y and X Y are nonnegative matrices. So Q is nonnegative if we≥ have the condition that p− xT , A and C are nonnegative. Therefore, to construct a (2m + 1) x A + JC × (2 m + 1) bisymmetric nonnegative matrix from m m symmetric matrices X and Y , p R and an m 1 matrix x, we need to ﬁnd× matrices A and C such that A∈ + JC A× JC p xT X = , Y = − and , A and C are nonnegative. We will 2 2 x A + JC use this method in the proof of Theorem 3.2 Before we proof our main theorem, we ﬁrst need the following auxiliary lemma.

Lemma 3.1. Let 1 λ2 λ3 > 0 > λ4 λ5 1 be real numbers such that 5 ≥ ≥5 3 ≥ ≥ − 1+ λi =0 and 1+ λ 0. If 1+ λ2 + λ4 + λ5 < 0, then i=2 i=2 i ≥ λ3λ5 P λ2λ4 λ3λ5 λ3 Pλ5 0 and λ2λ4 λ3λ5 λ3 λ5 . − − − ≥ − − − ≥ λ2 + λ4

Proof. Suppose that 1 + λ2 + λ4 + λ5 < 0. Since 1 λ2 λ3 > 0 > λ4 λ5 1 ≥ ≥ ≥ ≥− and 1+λ2+λ4+λ5 < 0, we have λ2+λ5 < 0. Since 1 > λ2 λ3 and λ5 > λ4 1, ≥ − ≥− A CHARACTERIZATION OF TRACE ZERO BISYMMETRIC NONNEGATIVE 5 × 5 MATRICES5 by the rearrangement inequality, we have λ5 + λ2λ4 λ3 λ2 + λ3λ4 λ5. So, − − ≥− − λ5 + λ2λ4 λ3 λ3λ5 λ2 + λ3λ4 λ5 λ3λ5 = λ3(λ4 λ5) (λ2 + λ5) 0. − − − ≥− − − − − ≥ Therefore, we have λ2λ4 λ3λ5 λ3 λ5 0. − − − ≥ Since λ5 = (1 + λ2 + λ3 + λ4), − 5 3 3 3 3 3 0 1+ λ =1+ λ2 + λ3 + λ4 (1 + λ2 + λ3 + λ4) . ≤ i − i=2 X Simplifying the right-hand side of the previous inequality, we obtain 2 2 0 λ2λ5 λ2λ4 + λ2λ3λ5 + λ2λ3 + λ2λ5 λ2λ4 + λ3λ4λ5 + λ3λ4 + λ4λ5. Therefore,≤ − − 2 2 λ2λ5 λ2λ4 λ2λ3λ5 λ2λ3 λ2λ5 + λ2λ4 λ3λ4λ5 λ3λ4 λ4λ5 ≥ − − − − − − = (λ2λ4 λ3λ5 λ3 λ5)(λ2 + λ4). − −λ3λ5− Since λ2 + λ4 < 0, we have λ2λ4 λ3λ5 λ3 λ5. λ2 + λ4 ≤ − − − Now, we are ready to prove the following result.

Theorem 3.2. Let λ1 λ2 λ3 > 0 > λ4 λ5 λ1 be real numbers such that 5 ≥ ≥ ≥ ≥− i=1 λi =0 and λ1 + λ2 + λ4 + λ5 < 0. Then λ1, λ2, λ3, λ4, λ5 is the eigenvalue of 5 a 5 5 nonnegative bisymmetric matrix if and only if λ3 0. P × i=1 i ≥ Proof. Normalizing by the Perron root, without loss ofP generality, we assume that λ1 = 1. If there is a 5 5 nonnegative bisymmetric matrix Q with the eigenvalues × 3 1, λ2, λ3, λ4, λ5, then Q is a nonnegative diagonalizable matrix and hence 1 + 5 5 λ3 = Tr(Q3) 0. So 1+ λ3 0 is a necessary condition. i ≥ i ≥ i=2 i=2 X 5 X 3 Assume that 1+ λ 0. Then, by Lemma 3.1, we have λ2λ4 λ3λ5 λ3 λ5 i ≥ − − − ≥ i=2 X λ3λ5 0 and λ2λ4 λ3λ5 λ3 λ5 . Let A and C be nonnegative matrices such − − − ≥ λ2 + λ4 that λ2 + λ4 √ λ2λ4 (λ2 + λ4) √ λ2λ4 A JC = − and A + JC = − − . − √ λ2λ4 0 √ λ2λ4 0 − − 0 √ λ2λ4 0 0 Then A = − and C = . Note that A JC has √ λ2λ4 0 (λ2 + λ4) 0 − − − λ2 and λ4 as its eigenvalues. Now, let 0 a b Ua,b = a (λ2 + λ4) √ λ2λ4 ,  − −  b √ λ2λ4 0 − where a,b are nonnegative real numbers. We want to show that there are real numbers a and b such that the matrix Ua,b has the eigenvalues 1, λ3, λ5. Suppose that 1, λ3, λ5 are the eigenvalues of Ua,b. Then by comparing the coeﬃ- cients of (x 1)(x λ3)(x λ5) and the characteristic polynomial of Ua,b we have the system of− equations− − 2 2 2 a + b = λ2λ4 λ3λ5 λ3 λ5 and (λ2 + λ4)b 2ab λ2λ4 = λ3λ5. − − − − − − − p 6 SOMCHAISOMPHOTPHISUTANDKENGWIBOONTON

We observe that the ﬁrst equation represent a circle with radius

r = λ2λ4 λ3λ5 λ3 λ5 − − − p λ3λ5 and the second equation gives the hyperbola with b-intercept at b = ± λ2 + λ4 r λ3λ5 in ab-coordinates. By Lemma 3.1, r2 . So, the hyperbola intersects b- ≥ λ2 + λ4 2 axis inside the circle. Using the quadratic formula to the equation (λ2 + λ4)b − − 2ab√ λ2λ4 = λ3λ5 we have − − 2 2a√ λ2λ4 4a λ2λ4 + 4(λ2 + λ4)λ3λ5 b = − ± − . 2(λ2 + λ4) p− We see that the function f defined by 2 2a√ λ2λ4 + 4a λ2λ4 + 4(λ2 + λ4)λ3λ5 f(a)= − − 2(λ2 + λ4) p− for a 0, is increasing and f(a) 0 for a 0. Therefore the hyperbola and the circle≥ are intersect at some point≥ in the first≥ quadrant. So, there are nonnegative real numbers a0 and b0 satisfying our system and hence the matrix Ua0,b0 has the x A JCJ √2  xT xT J  eigenvalue 1, λ3, λ5. Finally, we define the matrix Q = 0 where  √2 √2   Jx   C JAJ  √2  T   x = a0 b0 . Then Q is our desired matrix.   Now, we state and prove our main result for this section.

Theorem 3.3. Let λ1 λ2 λ3 λ4 λ5 λ1 be real numbers such that 5 ≥ ≥ ≥ ≥ ≥ − i=1 λi =0. Then λ1, λ2, λ3, λ4, λ5 is the eigenvalue of a 5 5 nonnegative bisymmetric matrix if and only if × P 5 3 λ2 + λ5 0 and λ 0. ≤ i ≥ i=1 X 5 3 Proof. We have seen that i=1 λi 0 is a necessary condition. In [8], R. Loewy and J. J. McDonald proved that for≥ any 5 5 nonnegative symmetric matrix A P × with the eigenvalues α1 α2 α3 α4 α5, Tr(A) α2 + α5. So, for any ≥ ≥ ≥ ≥ ≥ traceless bisymmetric matrix Q with the eigenvalues α1 α2 α3 α4 α5, we ≥ ≥ ≥ ≥ have α2 + α5 Tr(Q) = 0. So, we prove the direct part of our assertion. ≤ 5 3 Conversely, assume that λ2 + λ5 < 0 and λ 0. i=1 i ≥ Case 1: If λ5 0, then all λi’s are 0 and the zero 5 5 matrix is our desired matrix. ≥ P × Case 2: If λ4 0 > λ5 or λ2 0 > λ3 or λ1 0 > λ2, then we can apply Propositon 2.1. ≥ ≥ ≥ Case 3: Suppose that λ3 0 > λ4. If λ1 + λ2 + λ4 + λ5 0, then we can use ≥ ≥ Proposition 2.1. If λ1 + λ2 + λ4 + λ5 < 0, then λ3 = 0 and hence Theorem 3.2 can be applied. 6 A CHARACTERIZATION OF TRACE ZERO BISYMMETRIC NONNEGATIVE 5 × 5 MATRICES7

Finally, we give a partial result concerning the case that we do not have a complete answer, that is the case of ﬁnding a 5 5 bisymmetric nonnegative matrix × having the eigenvalues λ1, λ2, λ3, λ4, λ5 satisfying the condition λ1 λ2 λ3 5 ≥ ≥ ≥ 0 > λ4 λ5 λ1, λi > 0 and λ1 + λ2 + λ4 + λ5 < 0. Note that, in this ≥ ≥ − i=1 condition, we have λ1 = 0 and λ3 = 0. Therefore, without loss of generality, we P6 6 may assume that λ1 = 1 in this condition.

Corollary 3.4. Let 1 λ2 λ3 > 0 > λ4 λ5 1 be real numbers such that 1+ 5 ≥ ≥ ≥ ≥− λ3λ5 i=2 λi > 0 and 1+λ2 +λ4 +λ5 < 0. If λ2λ4 λ3λ5 λ3 λ5 − , then − − − ≥ 1+ λ3 + λ5 there is a 5 5 nonnegative bisymmetric matrix with the eigenvalues 1, λ2, λ3, λ4, λ5. P × Proof. Again, by setting λ2 + λ4 √ λ2λ4 1+ λ3 + λ5 √ λ2λ4 A JC = − , A + JC = − − √ λ2λ4 0 √ λ2λ4 0 − − 0 a b and Ua,b = a 1+ λ3 + λ5 √ λ2λ4 and then comparing the coeﬃcient of char-  −  b √ λ2λ4 0 − acteristic polynomials of Ua,b and the polynomials (x 1)(x λ3)(x λ5), we have the system − − − 2 2 2 a + b = λ2λ4 λ3λ5 λ3 λ5 and (1 + λ3 + λ5)b 2ab λ2λ4 = λ3λ5. − − − − − − Using the idea in the proof of Theorem 3.2, the inequality conditionp λ2λ4 λ3λ5 λ3λ5 − − λ3 λ5 − give us a nonnegative solution (a0,b0) of the above system. − ≥ 1+ λ3 + λ5 Then the matrix

1+ λ2 + λ3 + λ4 + λ5 a0 1+ λ3 + λ5 λ2 λ4 √ λ2λ4 0 − − 2 − √2 2  b0  √ λ2λ4 0 0 0  − √2   a0 b0 b0 a0    Q =  0   √2 √2 √2 √2   b0   √ λ2λ4   0 0 0   √2 −   1+ λ3 + λ5 λ2 λ4 a0 1+ λ2 + λ3 + λ4 + λ5   − − 0 √ λ2λ4   2 √ 2   2 −  is our desired matrix. 

Example 3.5. The set of real numbers 1, 0.3, 0.2, 0.7, 0.8 satisﬁes the condition of Theorem 3.2. Using the construction{ in the proof− of− Theorem} 3.2, we let A and C be two matrices such that 0.4 √0.21 0.4 √0.21 A JC = − and A + JC = − √0.21 0 √0.21 0 0 a b and let Ua,b = a 0.4 √0.21 be such that Ua,b has the eigenvalues 1, 0.2,   b √0.21 0 0.8. That is, we want to ﬁnd nonnegative real numbers a and b satisfying the − 8 SOMCHAISOMPHOTPHISUTANDKENGWIBOONTON system 2 2 2 a + b =0.55 and 0.4b 2√0.21ab =0.16. − 51 3√273 One nonnegative solution (a0,b0) of this system is given by a0 = − r 200 59+3√273 and b0 = . Therefore the matrix r 200 a0 0 √0.21 0 0.4 √2  b0  x √0.21 0 0 0 A JCJ √ 2  2  xT xT  a0 b0 b0 a0  Q =  J =  0  , 0  √ √ √ √  2 x 2  2 2 2 2   C J JAJ  b0     0 0 0 √0.21  2   √2     a0   0.4 0 √0.21 0     √2  T   where x = a0 b0 , is a nonnegative bisymmetric matrix with the eigenvalues 1, 0.3, 0.2, 0.7, 0.8. − −

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Department of Mathematics and Computer Science, Faculty of Science, Chulalongkorn University, Phyathai Road, Patumwan, Bangkok 10330, Thailand E-mail address: [email protected] and [email protected]