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International J.Math. Combin. Vol.1(2018), 97-108 Minimum Equitable Dominating Randic´ Energy of a Graph Rajendra P. (BharathiCollege,PGandResearchCentre,Bharathinagara,571422,India) R.Rangarajan (DOSinMathematics,UniversityofMysore,Mysuru,570006,India) [email protected],[email protected] Abstract: Let G be a graph with vertex set V(G), edge set E(G) and di is the degree of its i-th vertex vi, then the Randic´ matrix R(G) of G is the square matrix of order n, whose(i,j)-entryisequalto √d1idj ifthei-thvertexvi andj-thvertexvj ofGareadjacent, and zero otherwise. The Randic energy [3] RE(G) of the graph G is defined as the sum of the absolute values of the eigenvalues of the Randic´matrix R(G). A subset ED of V(G) is called an equitable dominating set [11], if for every vi V(G) ED there exists a vertex ∈ − vj EDsuchthatvivj E(G)and di(vi) dj(vj) 1. Inthecontrast,suchadominating ∈ ∈ | − |≤ set ED is Smarandachely if di(vi) dj(vj) 1. Recently, Adiga, et.al. introduced, the | − | ≥ minimum covering energy Ec(G) of a graph [1] and S. Burcu Bozkurt, et.al. introduced, Randic´ Matrix and Randic´ Energy of a graph [3]. Motivated by these papers, Minimum equitable dominating Randicénergy of a graph REED(G) of some graphs are worked out and boundson REED(G) are obtained. Key Words: Randic´matrix anditsenergy,minimum equitabledominatingset and minimum equitable dominating Randicénergy of a graph. AMS(2010): 05C50. §1. Introduction Let G be a graph with vertex set V(G) and edge set E(G). The adjacency matrix A(G) of the graph G is a square matrix, whose (i,j)-entry is equal to 1 if the vertices v and v are i j adjacent,otherwisezero[12]. SinceA(G)issymmetric,itseigenvaluesareallreal. Denotethem by λ ,λ ,...,λ , and as a whole, they are called the spectrum of G and denoted by Spec(G). 1 2 n The energy of graph [12] G is ε(G)= n λ . i=1| i| P The literature on energy of a graph and its bounds can refer [4,8,9,10,12]. The Randic´ matrix R(G)=(r ) of G is the square matrix of order n, where ij 1ReceivedJuly28,2017,AcceptedMarch1,2018. 98 RajendraP.andR.Rangarajan 1 , if v and v are adjacent vertices in G; (rij)= √didj i j 0, otherwise.  TheRandicenergy[3]RE(G)ofthegraphGisdefinedasthesumoftheabsolutevaluesof the eigenvaluesof the Randic´matrix R(G). Let ρ ,ρ , ,ρ be the eigenvalues ofthe Randic´ 1 2 n ··· matrix R(G). Since R(G) is symmetric matrix, these eigenvalues are real numbers and their sum is zero. Randicénergy [3] can be defined as RE(G)= n ρ i=1| i| For details of Randicénergy and its bounds, canPrefer [2, 3, 5, 6, 7]. Let G be a graph with vertex set V(G) = v ,v , ,v and ED is minimum equitable 1 2 n { ··· } dominating set of G. Minimum equitable dominating Randic´ matrix of G is n n matrix × R (G)=(r ), where ED ij 1 , if v and v are adjacent vertices in G; √didj i j (rij)= 1, if i=j and vi ∈ED; 0, otherwise. ThecharacteristicpolynomialofRED(G)isdenotedbydet(ρI RED(G))= ρI RED(G). − | − | Since R (G) is symmetric, its eigenvalues are real numbers. If the distinct eigenvalues of ED R (G) are ρ >ρ > >ρ with their multiplicities are m ,m , ,m then spectrum of ED 1 2 r 1 2 r ··· ··· R (G) is denoted by ED ρ ρ ρ 1 2 r SpecR (G)= ··· . ED m m m  1 2 r ···   The minimum equitable dominating Randicénergy of G is defined as n RE (G)= ρ . ED i | | i=1 X Example 1.1 Let W be a wheel graph, with vertex set V(G) = v ,v ,v ,v ,v , and let 5 1 2 3 4 5 { } its minimum equitable dominating set be ED = v . Then minimum equitable dominating 1 { } Randic´matrix R (W ) is ED 5 1 1 1 1 1 √12 √12 √12 √12  1 0 1 0 1  √12 3 3 RED(W5)=√112 13 0 13 0     1 0 1 0 1  √12 3 3     1 1 0 1 0  √12 3 3    0.6666 0 0.2324 1.4342 SpecR (W )= − . ED 5  1 2 1 1    MinimumEquitableDominatingRandicÉnergyofaGraph 99 The minimum equitable dominating Randicénergy of W is RE (W )=2.3332. 5 ED 5 §2. Bounds for the Minimum Equitable Dominating RandicÉnergy of a Graph Lemma 2.1 If ρ ,ρ , ,ρ are the eigenvalues of R (G). Then 1 2 n ED ··· n ρ = ED i | | i=1 X and n 1 ρ2 = ED +2 , i | | d d i j i=1 i<j X X where ED is minimum equitable dominating set. Proof (i) The sum of eigenvalues of R (G) is ED n n ρ = r = ED . i ii | | i=1 i=1 X X (ii) Consider, the sum of squares of ρ ,ρ ,...,ρ is, 1 2 n n n n n ρ2 = r r = (r )2+ r r i ij ji ii ij ji i=1 i=1j=1 i=1 i=j X XX X X6 n = (r )2+2 (r )2 ii ij i=1 i<j X X n 1 ρ2 = ED +2 . i | | d d 2 i j i=1 i<j X X Upper and lower bounds for RE (G) is similar proof to McClelland’s inequalities [10], ED are given below Theorem 2.2(UpperBound) Let G be a graph with ED is minimum equitable dominating set. Then 1 REED(G) vn ED +2 . ≤ u | | didj uu Xi<j t   Proof Let ρ ,ρ , ,ρ be the eigenvalues of R (G). By Cauchy-Schwartz inequality, 1 2 n ED ··· we have n 2 n n a b a2 b2 , (1) i i! ≤ i! i! i=1 i=1 i=1 X X X where a and b are any real numbers. 100 RajendraP.andR.Rangarajan If a =1, b = ρ in (1), we get i i i | | n 2 n n 1 ρ 12 ρ 2 , [RE (G)]2 n ED +2 i i ED | |! ≤ ! | | ! ≤ | | didj i=1 i=1 i=1 i<j X X X X   by Lemma 2.1, 1 RE (G) n ED +2 . ED v ≤u | | didj u Xi<j u t   Theorem 2.3(Lower Bound) Let G be a graph with ED is minimum equitable dominating | | set and d is degree of v . Then i i 1 REED(G) ED +2 +n(n 1)Dn2, ≥ | | d d − s i<j i j X where D = n ρ . i=1| i| Q Proof Consider n 2 n [RE (G)]2 = ρ = ρ 2+ ρ ρ . (2) ED i i i j " | |# | | | | | | i=1 i=1 i=j X X X6 By using arithmetic and geometric mean inequality, we have 1 n(n−1) 1 ρ ρ ρ ρ i j i j n(n 1) | | | | ≥  | | | | − i=j i=j X6 Y6   1 n n(n−1) ρ ρ n(n 1) ρ 2(n 1) i j i − | | | | ≥ − | | ! i=j i=1 X6 Y n 2/n ρ ρ n(n 1) ρ . (3) i j i | | | | ≥ − | |! i=j i=1 X6 Y Now using (3) in (2), we get n n 2/n [RE (G)]2 ρ 2+n(n 1) ρ , ED i i ≥ | | − | |! i=1 i=1 X Y n 1 [REED(G)]2 ED +2 +n(n 1)Dn2, where D = ρi , ≥| | d d − | | i j i<j i=1 X Y 1 REED(G) ED +2 +n(n 1)Dn2. ≥ | | d d − s i<j i j X MinimumEquitableDominatingRandicÉnergyofaGraph 101 §3. Bounds for Largest Eigenvalue of R (G) and its Energy ED Proposition 3.1 Let G be a graph and ρ (G) = max ρ be the largest eigenvalue of 1 1 i n i ≤≤ {| |} R (G). Then ED 1 1 1 v ED +2 ρ1(G) ED +2 . uuun| | Xi<j didj≤ ≤s| | Xi<j didj t   Proof Consider, n 1 ρ2(G) = max ρ 2 ρ 2 = ED +2 1 1≤i≤n {| i| }≤ | i| | | didj i=1 i<j X X 1 ρ (G) ED +2 1 ≤ | | d d s i<j i j X Next, n 1 n ρ2(G) ρ2 = ED +2 1 ≥ i | | d d i j i=1 i<j X X 1 1 ρ2(G) ED +2 1 ≥ n| | d d  i j i<j X   1 1 ρ (G) ED +2 1 v ≥ un| | didj u Xi<j u t   Therefore, 1 1 1 ED +2 ρ (G) ED +2 . v 1 uun| | Xi<j didj≤ ≤s| | Xi<j didj u t   Proposition 3.2 If G is a graph and n ED +2 1 , then ≤| | i<j didj P REED(G)≤ |ED|+2nPi<j di1dj +vu(n−1)|ED|+2 di1dj − |ED|+2nPi<j di1dj !2. uu Xi<j t   Proof We know that, n n 1 1 ρ2 = ED +2 , ρ2 = ED +2 ρ2 (4) i | | d d i | | d d − 1 i j i j i=1 i<j i=2 i<j X X X X 102 RajendraP.andR.Rangarajan By Cauchy-Schwarz inequality, we have n 2 n n a b a2 b2 . i i! ≤ i! i! i=2 i=2 i=2 X X X If a =1 and b = ρ , we have i i i | | n 2 n ρ (n 1) ρ 2 i i | |! ≤ − | | i=2 i=2 X X 1 [RE (G) ρ ]2 (n 1) ED +2 ρ2 ED −| 1| ≤ − | | d d − 1 i j i<j X   1 REED(G) ≤ ρ1+vu(n−1)|ED|+2 didj −ρ21 (5) uu Xi<j t   Consider the function, 1 F(x)=x+ (n 1) ED +2 x2 . v u − | | didj −  uu Xi<j t   Then, x (n 1) F′(x)=1 − . − ED +p2 1 x2 | | i<j didj − q P Here F(x) is decreasing in 1 1 1 ED +2 , ED +2 . n| | d d  | | d d  i j i j i<j i<j X X     We know that F (x) 0, ′ ≤ x (n 1) 1 − 0. − ED +p2 1 x2 ≤ | | i<j didj − q P We have ED +2 1 x | | i<j didj . ≥s n P Since, n ED +2 1 and |ED|+2 i<j di1dj ρ , 1 ≤| | d d n ≤ i j P i<j X MinimumEquitableDominatingRandicÉnergyofaGraph 103 we have |ED|+2 i<j di1dj |ED|+2 i<j di1dj ρ ED +2 1 . s n ≤ n ≤ 1 ≤| | d d P P i j i<j X Then, equation (5) become REED(G) |ED|+2 i<j di1dj +v(n 1) ED +2 1 |ED|+2 i<j di1dj 2 . ≤ nP u − | | didj − nP !  uu Xi<j t   2 §4. Minimum Equitable Dominating RandicÉnergy of Some Graphs Theorem 4.1 If K is complete graph with n vertices, then minimum equitable dominating n Randicénergy of K is n 3n 5 RE (K )= − . ED n n 1 − Proof LetK bethecompletegraphwithvertexsetV(G)= v ,v ,...,v andminimum n 1 2 n { } equitable dominating set is ED = v , we have characteristic polynomial of R (K ) is 1 ED n { } ρ 1 1 1 1 1 1 − n−1 n−1 ··· n−1 n−1 n−1 − − − − − (cid:12) 1 ρ 1 1 1 1 (cid:12) (cid:12) n−1 n−1 ··· n−1 n−1 n−1(cid:12) (cid:12) − − − − − (cid:12) (cid:12)(cid:12) n−11 n−11 ρ ··· n−11 n−11 n−11(cid:12)(cid:12) |ρI −RED(Kn)|=(cid:12)(cid:12)(cid:12) −... −... ... ... −... −... −... (cid:12)(cid:12)(cid:12) . (cid:12) (cid:12) (cid:12) 1 1 1 ρ 1 1 (cid:12) (cid:12) n−1 n−1 n−1 ··· n−1 n−1(cid:12) (cid:12) − − − − − (cid:12) (cid:12) 1 1 1 1 ρ 1 (cid:12) (cid:12) n−1 n−1 n−1 ··· n−1 n−1(cid:12) (cid:12) − − − − − (cid:12) (cid:12)(cid:12)(cid:12) n−−11 n−−11 n−−11 ··· n−−11 n−−11 ρ (cid:12)(cid:12)(cid:12)n×n (cid:12) (cid:12) (cid:12) (cid:12) R = R R , k = 3,4, ,n 1,n. Then, we get ρ+ 1 common from R to R k′ k − 2 ··· − n 1 3 n − and we have (cid:16) (cid:17) ρ 1 1 1 1 1 1 − n−1 n−1 ··· n−1 n−1 n−1 − − − − − (cid:12) 1 ρ 1 1 1 1 (cid:12) (cid:12) n−1 n−1 ··· n−1 n−1 n−1(cid:12) (cid:12) − − − − − (cid:12) (cid:12) 0 1 1 0 0 0 (cid:12) n 2(cid:12) − ··· (cid:12) |ρI−RED(Kn)|= ρ+ n 1 1 − (cid:12)(cid:12)(cid:12) ... ... ... ... ... ... ... (cid:12)(cid:12)(cid:12) . (cid:18) − (cid:19) (cid:12) (cid:12) (cid:12) 0 1 0 1 0 0 (cid:12) (cid:12) − ··· (cid:12) (cid:12) (cid:12) (cid:12) 0 1 0 0 1 0 (cid:12) (cid:12) − ··· (cid:12) (cid:12) (cid:12) (cid:12) 0 1 0 0 0 1 (cid:12) (cid:12) − ··· (cid:12)n n (cid:12) (cid:12) × (cid:12) (cid:12) (cid:12) (cid:12) 104 RajendraP.andR.Rangarajan C =C +C + +C , we get the characteristic polynomial 2′ 2 3 ··· n n 2 1 − 2n 3 n 3 ρI R (K ) = ρ+ ρ2 − ρ+ − , ED n | − | n 1 − n 1 n 1 (cid:18) − (cid:19) (cid:20) (cid:18) − (cid:19) (cid:18) − (cid:19)(cid:21) 1 (2n 3) √4n 3 (2n 3)+√4n 3 SpecRED(Kn) = n−−1 −2(n−−1) − −2(n−1) − . n 2 1 1  −   The minimum equitable dominating Randicénergy of K is n 3n 5 RE (K )= − . ED n n 1 2 − Theorem 4.2 If S ,(n 4) is star graph with n vertices, then minimum equitable dominating n ≥ Randicénergy of S is n RE (S )=n. ED n Proof Let S ,(n 4) be the star graph with vertex set V(G) = v ,v , ,v and n 1 2 n ≥ { ··· } minimumequitabledominatingsetisED = v ,v , ,v ,wehavecharacteristicpolynomial 1 2 n { ··· } of R (S ) is ED n ρ 1 1 1 1 1 1 − √−n 1 √−n 1 ··· √−n 1 √−n 1 √−n 1 − − − − − (cid:12) 1 ρ 1 0 0 0 0 (cid:12) (cid:12)(cid:12)√−n−1 − ··· (cid:12)(cid:12) (cid:12) 1 0 ρ 1 0 0 0 (cid:12) ρI RED(Sn) =(cid:12)(cid:12)(cid:12)(cid:12)√−n...−1 ... −... ·..·.· ... ... ... (cid:12)(cid:12)(cid:12)(cid:12) . | − | (cid:12) (cid:12) (cid:12)(cid:12)(cid:12)√−n11 0 0 ··· ρ−1 0 0 (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)√−n−−11 0 0 ··· 0 ρ−1 0 (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)√−n−11 0 0 ··· 0 0 ρ−1(cid:12)(cid:12)(cid:12)(cid:12)n×n (cid:12) (cid:12) R =R R , k =3,4,(cid:12) ,n. Then taking (ρ 1) common from R to R(cid:12) , we get k′ k− 2 ··· − 3 n ρ 1 1 1 1 1 1 − √−n 1 √−n 1 ··· √−n 1 √−n 1 √−n 1 − − − − − (cid:12) 1 ρ 1 0 0 0 0 (cid:12) (cid:12)√−n 1 − ··· (cid:12) (cid:12) − (cid:12) (cid:12) 0 1 1 0 0 0 (cid:12) (cid:12) − ··· (cid:12) |ρI −RED(Sn)|=(ρ−1)n−2(cid:12)(cid:12)(cid:12) ... ... ... ... ... ... ... (cid:12)(cid:12)(cid:12) . (cid:12) (cid:12) (cid:12) 0 1 0 1 0 0 (cid:12) (cid:12) − ··· (cid:12) (cid:12) (cid:12) (cid:12) 0 1 0 0 1 0 (cid:12) (cid:12) − ··· (cid:12) (cid:12) (cid:12) (cid:12) 0 1 0 0 0 1 (cid:12) (cid:12) − ··· (cid:12)n n (cid:12) (cid:12) × (cid:12) (cid:12) (cid:12) (cid:12) C =C +C + +C . Then, the characteristic polynomial 2′ 2 3 ··· n ρI R (S ) =ρ(ρ 1)n 2(ρ 2), ED n − | − | − − MinimumEquitableDominatingRandicÉnergyofaGraph 105 0 1 2 SpecRED(Sn)= . 1 n 2 1  −    The minimum equitable dominating Randicénergy of S is RE (S )=n. n ED n 2 Theorem 4.3 If K , where m< n and m n 2 is complete bipartite graph with m+n m,n | − |≥ vertices, then minimum equitable dominating Randicénergy of K is RE (K )=m+n. m,n ED m,n Proof Let K , where m < n and m n 2 be the complete bipartite graph with m,n | − | ≥ vertex set V(G) = v ,v ,...,v ,u ,u , ,u and minimum equitable dominating set is 1 2 m 1 2 n { ··· } ED =V(G), we have characteristic polynomial of R (K ) is ED m,n ρ 1 0 ... 0 0 1 1 1 1 − √−mn √−mn ··· √−mn √−mn (cid:12) 0 ρ 1 ... 0 0 1 1 1 1 (cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ... −... ... ... ... √−m...n √−m...n ·..·.· √−m...n √−m...n(cid:12)(cid:12)(cid:12)(cid:12) (cid:12) (cid:12) (cid:12)(cid:12)(cid:12) 0 0 ··· ρ−1 0 √−m1n √−m1n ··· √−m1n √−m1n(cid:12)(cid:12)(cid:12) ρI R (K ) =(cid:12)(cid:12)(cid:12) 0 0 ··· 0 ρ−1 √−m1n √−m1n ... √−m1n √−m1n(cid:12)(cid:12)(cid:12). | − ED m,n | (cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n ρ−1 0 ··· 0 0 (cid:12)(cid:12)(cid:12) (cid:12) 1 1 1 1 0 ρ 1 0 0 (cid:12) (cid:12)(cid:12)√−mn √−mn ··· √−mn √−mn − ··· (cid:12)(cid:12) (cid:12)(cid:12) ... ... ... ... ... ... ... ... ... ... (cid:12)(cid:12) (cid:12) (cid:12) (cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n 0 0 ··· ρ−1 0 (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n 0 0 ··· 0 ρ−1(cid:12)(cid:12)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) R =R R , k =(cid:12) 1,2,3, ,m 1andR =R R , d=m+2,m+3, ,m+(cid:12) n. k′ k− m ··· − d′ d− m+1 ··· Then, taking (ρ 1) common from R to R and R to R , we get 1 m 1 m+2 m+n − − 1 0 0 1 0 0 0 0 ··· − ··· (cid:12) 0 1 0 1 0 0 0 0 (cid:12) (cid:12) (cid:12) (cid:12)(cid:12)(cid:12) ... ... ·..·.· ... −... ... ... ·..·.· ... ... (cid:12)(cid:12)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 0 0 1 1 0 0 0 0 (cid:12) (cid:12) ··· − ··· (cid:12) (ρ 1)m+n−2(cid:12)(cid:12)(cid:12) 0 0 ··· 0 ρ−1 √−m1n √−m1n ··· √−m1n √−m1n(cid:12)(cid:12)(cid:12). − (cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n ρ−1 0 ··· 0 0 (cid:12)(cid:12)(cid:12) (cid:12) 0 0 0 0 1 1 0 0 (cid:12) (cid:12) (cid:12) (cid:12)(cid:12)(cid:12) ... ... ·..·.· ... ... −... ... ·..·.· ... ... (cid:12)(cid:12)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 0 0 0 0 1 0 1 0 (cid:12) (cid:12) ··· − ··· (cid:12) (cid:12) (cid:12) (cid:12) 0 0 0 0 1 0 0 1 (cid:12) (cid:12) ··· − ··· (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 106 RajendraP.andR.Rangarajan C =C +C + +C , m′ +1 m+1 m+2 ··· m+n 1 0 0 1 0 0 0 0 ··· − ··· (cid:12) 0 1 0 1 0 0 0 0 (cid:12) (cid:12) (cid:12) (cid:12)(cid:12)(cid:12) ... ... ·..·.· ... −... ... ... ·..·.· ... ... (cid:12)(cid:12)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 0 0 1 1 0 0 0 0 (cid:12) (cid:12) ··· − ··· (cid:12) (ρ 1)m+n−2(cid:12)(cid:12)(cid:12) 0 0 ··· 0 ρ−1 √−mnn √−m1n ··· √−m1n √−m1n(cid:12)(cid:12)(cid:12). − (cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n ρ−1 0 ··· 0 0 (cid:12)(cid:12)(cid:12) (cid:12) 0 0 0 0 0 1 0 0 (cid:12) (cid:12) (cid:12) (cid:12)(cid:12)(cid:12) ... ... ·..·.· ... ... ... ... ·..·.· ... ... (cid:12)(cid:12)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 0 0 0 0 0 0 1 0 (cid:12) (cid:12) ··· ··· (cid:12) (cid:12) (cid:12) (cid:12) 0 0 0 0 0 0 0 1 (cid:12) (cid:12) ··· ··· (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) The characterist(cid:12)ic polynomial ρI RED(Km,n) =ρ(ρ 1)m+n−2(ρ 2), (cid:12) | − | − − 0 1 2 SpecR (K )= . ED m,n 1 m+n 2 1 −   The minimum equitable dominating Randicénergy of K is RE (K )=m+n. m,n ED m,n 2 Theorem4.4 IfK ,(n 3)iscocktailpartygraphwith2nvertices, thenminimumequitable n 2 × ≥ dominating Randicénergy of K is 4n 6. n×2 n−−1 Proof LetK ,(n 3)bethecocktailpartygraphwithvertexsetV(G)= v ,v , ,v , n 2 1 2 n × ≥ { ··· u ,u , ,u andminimumequitabledominatingsetisED = v ,u ,wehavecharacteristic 1 2 n 1 1 ··· } { } polynomial of R (K ) is ED n 2 × λ 1 0 −1 −1 −1 −1 −1 −1 − 2n−2 2n−2 ··· 2n−2 2n−2 ··· 2n−2 2n−2 (cid:12) 0 λ 1 −1 −1 −1 −1 −1 −1 (cid:12) (cid:12) − 2n−2 2n−2 ··· 2n−2 2n−2 ··· 2n−2 2n−2(cid:12) (cid:12) (cid:12) (cid:12) −1 −1 λ −1 −1 0 −1 −1 (cid:12) (cid:12)2n−2 2n−2 2n−2 ··· 2n−2 ··· 2n−2 2n−2(cid:12) (cid:12) (cid:12) (cid:12) −1 −1 −1 λ −1 −1 −1 −1 (cid:12) (cid:12)2n−2 2n−2 2n−2 ··· 2n−2 2n−2 ··· 2n−2 2n−2(cid:12) (cid:12)(cid:12)(cid:12) ... ... ... ... ... ... ... ... ... ... (cid:12)(cid:12)(cid:12) |ρI−RED(Kn×2)|=(cid:12)(cid:12)(cid:12)(cid:12)2n−−12 2n−−12 2n−−12 2n−−12 ··· λ 2n−−12 ··· 2n−−12 0 (cid:12)(cid:12)(cid:12)(cid:12). (cid:12)(cid:12)(cid:12)2n−−12 2n−−12 0 2n−−12 ··· 2n−−12 λ ··· 2n−−12 2n−−12(cid:12)(cid:12)(cid:12) (cid:12)(cid:12) ... ... ... ... ... ... ... ... ... ... (cid:12)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) −1 −1 −1 −1 −1 −1 λ −1 (cid:12) (cid:12)2n−2 2n−2 2n−2 2n−2 ··· 2n−2 2n−2 ··· 2n−2(cid:12) (cid:12) (cid:12) (cid:12) −1 −1 −1 −1 0 −1 −1 λ (cid:12) (cid:12)2n−2 2n−2 2n−2 2n−2 ··· 2n−2 ··· 2n−2 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)