A Note on Frobenius Inner Product and the M-Distance Matrices of a Tree

Malaya Journal of Matematik, Vol. 8, No. 3, 1321-1327, 2020

https://doi.org/10.26637/MJM0803/0103

A note on Frobenius inner product and the m-distance matrices of a tree

P.A. Asharaf1* and Bindhu K. Thomas2

Abstract The m-distance matrix Dm of a simple connected undirected graph has an important role in computing the distance matrix D of the graph from the powers of the adjacency matrix using Hadamard product. This paper shows that for an undirected tree T with diameter d, {D0.D1,...,Dd} is an orthogonal basis for the space spanned by the binary equivalent matrices of the first d + 1 powers of the adjacency matrix of T and it gives an invertible conversion matrix for finding the m-distance matrix of T using Frobenius inner product on matrices. Keywords Adjacency matrix, Distance matrix, Binary matrix , Diameter, Hadamard product, m-distance matrix , Frobenius inner product, Frobenius norm. AMS Subject Classification 05C12, 05C40, 05C50, 05C62, 05B20, 46C05, 15A60.

1Mathematics Research Centre, Mary Matha Arts and Science College, Mananthavady-670645, Kerala, India. 2Department of Mathematics, Mary Matha Arts and Science College, Mananthavady-670645, Kerala, India. *Corresponding author: 1 [email protected]; [email protected] Article History: Received 12 April 2020; Accepted 19 August 2020 ©2020 MJM.

Contents Definition 1.2. ([3]) The diameter of a graph G is the maximum distance between any two vertices of G and it is denoted 1 Introduction...... 1321 by Diam(G). 2 Preliminaries...... 1322 Definition 1.3. (Hadamard product [4]) Consider the vector m× n 3 Main Resuls...... 1322 space R of all m × n real matrices over the real field m× n 4 Illustration...... 1325 R. For P, F ∈ R , the Hadamard product ◦ is a binary operation on m× n defined by, 5 Conclusion...... 1327 R

References...... 1327 (P ◦ F)i j := (P)i j(F)i j, ∀ i, j.

Let B = {0,1} and let Bm×n denote the set of all binary matri- 1. Introduction m×n ces in R . Consider a simple, connected, undirected graph G = (V,E) Then for P, F ∈ Bm× n, of order n with vertex set V and edge set E throughout this 1, if pi j = 1 and fi j = 1 paper unless otherwise specified. Let AG be the n × n adja- (P ◦ F)i j = th m th 0, otherwise cency matrix [1] of G. Then i j entry of AG (m power of AG), represent the number of walks of length m between the Definition 1.4. (Frobenius inner product [5]) Consider the vertices v and v of G. n× n i j vector space R over the field R. Then the Frobenius inner n× n n× n product h,iF : × → is defined by, Definition 1.1. ([2]) The distance matrix, D = (di j) of G is R R R defined as, n n n n hP,Qi = p q = (P ◦ Q) = Tr(PQT ), F ∑ ∑ i j i j ∑ ∑ i j d(vi,v j) , if i 6= j i=1 j=1 i=1 j=1 di j = 0 , if i = j n×n where P,Q ∈ R . The Frobenius norm k · k induced by this n×n 1 n×n Where, d(vi,v j) is the distance between the vertices vi and v j. inner product on R is, kPkF = (hP,PiF ) 2 , P ∈ R . Then A note on Frobenius inner product and the m-distance matrices of a tree — 1322/1327

n×n the metric dF induced by this norm on R is dF (P,Q) = Remark 2.3. (i)D m ∈ Bn×n, ∀ m = 0,1,2,... n×n kP − QkF ,P,Q ∈ R . (ii)D 0 = In 1 1 T 2 Remark 1.5. (i) kPkF = (hP,PiF ) 2 = Tr(PP ) , P ∈ (iii)D + D + ... + D = n×n 0 1 d R 1 1 ... 1 (ii) Let X,Y ∈ 1×n. Then hX,Yi = n x y and R F ∑i=1 i i 1 1 ... 1   1 ......  n ! 2 Jn =   = The matrix of ones 2 ......  d f (X,Y) = kX −YkF = ∑(xi − yi) ......  i=1   1 1 ... 1 which is the Euclidean distance between the vectors X and Y of n. Diam(G) = Max{m : D 6= 0} R (iv) m m (iii) For P ∈ Bn×n, 0, if i 6= j ! 1 (v)D i ◦ D j = , for 0 ≤ i, j ≤ d. n n 2 Di, if i = j d (P,0) = kP − 0k = kPk = p2 n×n F F F ∑ ∑ i j i.e., {D0,D1,...,Dd} is an orthogonal subset R i=1 j=1 with respect to the binary operation ◦. 2 Then (dF (P,0)) represents the number of 1’s in P. Theorem 2.4. ([6]) Let AG be the adjacency matrix of G = For P,Q ∈ Bn×n, (V,E) with diameter d. Then β2 = {D0,D1,...,Dd} is a Lin- n×n 1 early independent subset of the vector space R . n n ! 2 2 dF (P,Q) = kP − QkF = ∑ ∑(pi j − qi j) Theorem 2.5. ([6]) Let AG be the adjacency matrix of G = i=1 j=1 (m) (V,E) with diameter d. Then for 1 ≤ m ≤ d, Dm = AG − m−1 (m) (s) 2 δ ∑ A ◦ A , where Dm is the m-distance matrix of Then (dF (P,Q)) represent the number of non identical entries s=0 G G in P and Q and the number of identical entries in P and Q is G. 2 2 n − (dF (P,Q)) 3. Main Resuls

2. Preliminaries Theorem 3.1. Let AG be the adjacency matrix of G = (V,E). Definition 2.1. ([6]) The function δ : R → B = {0,1} is de- Let d = diam(G). Then for 0 ≤ K, m ≤ d, fined by (m) Dm, if K = m Dk ◦ A = 0, a ≤ 0 G 0, if m < K. δ(a) := ,a ∈ 1, otherwise R Proof. Case(i): K = m m×n Also, δ : R → Bm×n is defined by δ(D) = (δ(D))i j := (m) (m) m×n m Dm ◦ AG = (Dm)i j(AG )i j (δ(di j)), ∀ i, j and D ∈ R . Then δ(AG) ∈ Bn×n, ∀ m ∈ i j {0,1,2,...} and it is the equivalent binary matrix representa- (m) 1, if (D ) = 1 and (A ) = 1 Am A(m) = (Am) = m i j G i j tion of G. Let G δ G . Then 0, otherwise  (m) 1, if there exist a walk of length m 1, if d(vi,v j) = m and (A )i j = 1 (m)  = G (AG )i j = from vi to v j in G 0, otherwise.  0, otherwise (m) But d(vi,v j) = m ⇒ (AG )i j = 1 Also, for C,F ∈ Bm×n, δ(C ◦ F) = δ(C) ◦ δ(F). (m) (m) Dm ◦ AG = (Dm)i j(AG )i j Definition 2.2. (m-distance matrix [6]) Consider the graph ∴ i j G = (V,E) with n vertices. Then the m- distance matrix D m 1, if d(v ,v ) = m of G is an n × n symmetric binary matrix defined by = i j = (D ) 0, otherwise m i j 1, if d(vi,v j) = m (m) (Dm)i j := 0, otherwise That is, (Dm ◦ AG )i j = Dm. Case(ii): K > m (m) (m) where, d(vi,v j) is the distance between the vertices vi and v j. (Dk ◦ AG )i j = (Dk)i j(AG )i j

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If (Dk)i j = 1, then d(vi,v j) = K. Lemma 3.4. Let G = (V,E) be a connected undirected graph That is, if there exist a shortest path of length K between of order n(n > 1). Let di j denote the distance between vi and vi and v j. But then this path cannot be traversed by any of v j. Then there always exist at least one walk between vi and (m) v j of length di j + 2r, ∀ r = 0,1,2,.... That is, the walk between vi and v j of length< K. Thus (AG )i j = 0.

(∵ K > m) (di j+2r) (m) AG = 1,∀r = 0,1,2,.... ∴ (DK)i j(AG )i j = 0 i j (m) If (DK)i j = 0, then also (DK)i j(AG )i j = 0. Proof. Since G is a connected undirected graph, there exist (m) Therefore, (DK ◦ AG )i j = 0, ∀ i, j, ∀ K > m. at least one path in between vi and v j. Let Pi j : v0(= vi) − (m) v1 − v2 − ... − vd −1 − vd (= v j) denote the shortest path ie., DK ◦ AG = 0 if K > m. Hence i j i j between vi and v j of length di j. If we traverse back and forth once along the last edge in P , then we get a walk v (= (m) Dm, if K = m i j 0 DK ◦ A = , for K ≤ d. G 0, if m < K vi) − v1 − v2 − ... − vdi j−1 − vdi j (= v j) − vdi j−1 − vdi j (= v j) of length di j + 2 between vi and v j. Also If we traverse back and forth twice along the last edge in Pi j, then we get a walk

v0(= vi)−v1 −v2 −...−vdi j−1 −vdi j −vdi j−1 −vdi j −vdi j−1 − Theorem 3.2. Let AG be the adjacency matrix of G = (V,E). vdi j (= v j) of length di j + 4 between vi and v j. If we proceed (0) (1) (d) like this, then we always get a walk of length di j + 2r, ∀ Let d = diam(G). Then β1 = {AG ,AG ,...AG } is a linearly n×n r = 0,1,2,... between v and v in G. This walk will reﬂect independent subset of the vector space R . i j th (di j+2r) as 1 in the i j entry of AG . Proof. Suppose (di j+2r) ∴ AG = 1,∀ r = 0,1,2,... (0) (1) (d) i j c0 · AG + c1 · AG + ... + cd · AG = 0 (3.1) for some c0,c1,...,cd ∈ R. By taking Hadamard product ◦ by Theorem 3.5. Let T = (V,E) be an undirected tree of order Dd n(n > 1). Let di j denote the distance between vi and v j. Then, (3.1) ⇒ D ◦(c · A(0) + c · A(1) + ... + c · A(d)) = 0 d 0 G 1 G d G (m) 1, if m = di j + 2r,∀r = 0,1,2,... AT = (0) (1) (d) i j 0, otherwise c0 · (Dd ◦ AG ) + c1 · (Dd ◦ AG ) + ... + cd · (Dd ◦ AG ) = 0 (d) Proof. Let Ti j denote the unique path in T from vi to v j of ⇒ 0 + 0 + ...0 + cd · (Dd ◦ AG ) = 0(∵ Theorem 3.1) distance di j. Consider a walk Wi j from vi to v j such that length ⇒ cd · Dd = 0 ⇒ cd = 0, as Dd 6= 0 of Wi j, l(Wi j) > di j. Now delete the edges of Ti j from Wi j. Let (0) (1) (d−1) 1 1 (3.1) ⇒c0 · AG + c1 · AG + ... + cd−1 · AG = 0 Wi j be the remaining part of the walk Wi j. Then Wi j should (3.2) 1 be disconnected. Otherwise Wi j and some of the edges of Ti j together would form a cycle. But T has no cycle. By taking Hadamard product ◦ by Dd−1 on both side of equa- 1 Let Wi j,1, Wi j,2, ..., Wi j,s be the component walks of Wi j. tion (3.2). Then Wi j,h ∩ Ti j, (1 ≤ h ≤ s) should be a single vertex vwi j,h, W T (0) (1) (d−1) otherwise some of the edges of i j,h and i j together form a (3.2) ⇒ Dd−1 ◦ (c0 · A + c1 · A + ... + cd−1 · A ) = 0 G G G cycle. Since each vertex vwi j,h is a part of the walk Wi j, Wi j,h (0) (1) should be either the single vertex vw ,h or a closed walk from c0 · (Dd−1 ◦ A ) + c1 · (Dd−1 ◦ A ) + ... i j G G v to v . (d−1) wi j,h wi j,h + cd−1 · (Dd−1 ◦ AG ) = 0 But length of a closed walk in a tree is always even. So 1 s (d−1) l(Wi j) = ∑ l(Wi j,h) must be even. ⇒0 + 0 + ... + 0 + cd−1 · (Dd−1 ◦ A ) = 0 h=1 G 1 s Let l(Wi j) = ∑h=1 l(Wi j,h) = 2r, for some r ∈ {0,1,2,...} ⇒cd−1Dd−1 = 0 ⇒ cd−1 = 0, as Dd−1 6= 0 ⇒ l(W ) = l(T ) + l(W 1) = d + s l(W ) = d + 2r, (0) (1) (d−1) i j i j i j i j ∑h=1 i j,h i j (3.2) ⇒c0 · AG + c1 · AG + ... + cd−1 · AG = 0 for some r ∈ {0,1,2,...} ⇒ length of every walk from vi to v j must be of the form Continue the above process further a ﬁnite number of times, di j + 2r, for r ∈ {0,1,2,...}. we get c0 = 0, c1 = 0,...,cd = 0. By Lemma 3.4, there always exist a walk between vi and v j (0) (1) (d) ⇒ β1 = {AG ,AG ,...,AG } is a linearly independent subset of length di j + 2r, for all r = 0,1,2,.... So n×n of R . (m) 1, if m = di j + 2r,∀r = 0,1,2,... AT = (0) (1) (d) i j 0, otherwise Remark 3.3. β1 = {AG ,AG ,...,AG } is a basis for span (0) (1) (d) {AG ,AG ,...,AG }.

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d (m) Theorem 3.6. Let AT be the adjacency matrix of a tree T = ⇒ ∑K=0 AT ◦ DK is a linear combination of D0,D1,...,Dd. (V,E) of order n(n > 1). Let d = diam(T). Then for 0 ≤ K ≤ d (m) d i.e., ∑K=0 AT ◦DK = ∑K=0 cm,KDK, for cm,0,cm,1,...,cm,d ∈ d, B = {0,1}. Now (K+2r) (m) (m) (m) AT = AT ◦ Jn = AT ◦ (D0 + D1 + ··· + Dd ),(∵ Remark2.3(iii)) (i) DK ◦ AT = DK. d d (m) (K+2r+1) = ∑ AT ◦ DK = ∑ cm,K DK (ii) DK ◦ AT = 0, ∀ r = 0,1,2,.... K=0 K=0 (m) Proof. ⇒ AT is a linear combination of D0,D1,...,Dd. For ﬁnding the scalars cm,k, consider (m) (m) DK ◦ AT = (DK)i j(AT )i j (m) i j hAT ,D jiF = hcm,0 · D0 + cm,1 · D1 + ··· + cm,d · Dd,D jiF (m) = cm,0 · hD0,D jiF + cm,1 · hD1,D jiF + ··· = 1, if (DK)i j = 1 and (AT )i j = 1 0, otherwise + cm,d · hDd,D jiF

= 0 + 0 + ··· + cm, jhD j,D jiF + 0 + ··· + 0 (DK)i j = 1 ⇒ di j = K. (K+2r) = cm, jhD j,D jiF (∵ Remark2.3(v)) But then (AT )i j = 1, ∀ r = 0,1,2,.... (∵ Theorem 3.5) (K+2r) (m) ie., (DK)i j = 1 ⇒ (AT )i j = 1, ∀ r = 0,1,2,.... hAT ,D jiF ⇒ cm, j = , j = 0,1,...,d hD j,D jiF (K+2r) DK ◦ AT = (m) ∴ i j hAT ,D jiF the scalars, cm, j = ∈ B = {0,1},∀0 ≤ m, j ≤ d, 1, if K = di j hD j,D jiF = (DK)i j,∀i, j and r = 0,1,2,.... 0, otherwise So,

(K+2r) (m) ⇒ DK ◦ AT = DK, ∀ r = 0,1,2,.... AT =cm,0 · D0 + cm,1 · D1 + ··· + cm,d · Dd By Theorem 3.5 there does not exist a walk of length di j + (m) (m) hAT ,D0iF hAT ,D1iF 2r + 1 between vi and v j, ∀ r = 0,1,2,.... = · D0 + · D1 + ··· (di j+2r+1) hD0,D0iF hD1,D1iF ⇒ AT = 0, ∀ i, j and r = 0,1,2,.... i j (m) hAT ,DdiF ⇒ D ◦ A(K+2r+1) = 0 ∀ r = 0,1,2,.... + · Dd K T hDd,DdiF d (m) (m) hA ,D jiF Remark 3.7. Let AT be the adjacency matrix of a tree T = T AT = ∑ D j (V,E) of order n(n > 1). Let d = diam(T). Then for 0 ≤ K ≤ j=0 hD j,D jiF (m) d. For m < K, AT ◦ DK = 0, (by Theorem 3.1). ⇒ β2 = {D0,D1,...,Dd} is an orthogonal basis for span For m > K, (0) (1) (d) {AT ,AT ,...,AT }. (i)D ◦A(K+2r) = D . K T K Remark 3.9. (i) We have

(K+2r+1) (m) (ii)D K ◦A = 0. ∀r = 0,1,2,..., (by Theorem 3.6) d T (m) hAT ,D jiF AT = ∑ D j(∵ Theorem3.8) j=0 hD j,D jiF (m) DK, m = K + 2r,∀r = 0,1,2,... hA(m),D i d−1 hA(m),D i ∴ AT ◦ DK = (m) T d F T j F 0, otherwise AT = Dd + ∑ D j hDd,DdiF j=0 hD j,D jiF (m) (m) Theorem 3.8. Let AT be the adjacency matrix of T = (V,E) hA ,D i d−1 hA ,D i T d F D = A(m) − T j F D with d = diam(G). Then β2 = {D0,D1,...,Dd} is an orthog- hD ,D i d T ∑ hD ,D i j (0) (1) (d) d d F j=0 j j F onal basis for span {AT ,AT ,...AT }. (0) (1) (d) ⇒ The orthogonal basis for span {AT ,AT ,...,AT } Proof. By Theorem 2.4, β2 = {D0,D1,...,Dd} is linearly obtained from the basis β1 by Gram-Schmidt process is independent. By Remark 2.3(v) β2 is orthogonal also. So (m) nothing but β2 itself. it is enough to prove that AT is a linear combination of D ,D ,...,D , for 0 ≤ m ≤ d. By Remark 3.7, (m) d 0 1 d (ii)A T = ∑ j=0 cm, jD j, by Theorem 3.8, where (m) DK, m = K + 2r,∀r = 0,1,2,... (m) A ◦ DK = hAT ,D jiF ∴ T 0, otherwise cm, j = ,m, j = 0,1,...,d. hD j,D jiF

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This can be written in the matrix form as given below,

 (0)    AT c00 c01 ··· c0d D0  (1) c c ··· c D AT   10 11 1d  1       ·   ······  ·    =    (3.3)  ·   ·····  ·     ······  ·   ·     (d) cd0 cd1 ··· cdd Dd AT Let  (0)     AT c00 c01 ··· c0d D0  (1) c c ··· c D Figure 1 AT   10 11 1d   1        ·   ······   ·  Y =  ,C =  ,X =    ·   ·····   ·     ······   ·   ·      (d) cd0 cd1 ··· cdd Dd AT Then (3.3) ⇒

Y = CX (3.4) Here the maximum distance d = 3, A(0) = I,A(1) = δ(A ) = where, X, Y are (d + 1) × 1 column matrices whose elements T T T A are n × n binary matrices and C is a (d + 1) × (d + 1) real T matrix. But by Theorem 3.1, cii = 1 and ci j = 0, when i < j,(i, j = 0,1,...,d). Which implies that C is a unit lower triangular matrix. −1 ∴ C is invertible and C is a lower triangular unit matrix with |C| = 1. So, by multiplying C−1 on both sides od equation (3.4). Then (3.4) ⇒ 1 1 1 0 1 0 X = C−1Y (3.5) 1 1 1 0 1 0   2 1 1 1 0 1 0 A =  , Then (3.3) ⇒ T 0 0 0 4 0 1      −1  (0) 1 1 1 0 2 0 D0 c00 c01 ··· c0d AT 0 0 0 1 0 1 (1) D1 c10 c11 ··· c1d  A       T  1 1 1 0 1 0  ·   ······   ·    =     1 1 1 0 1 0  ·   ·····   ·          (2) (2) 1 1 1 0 1 0  ·   ······   ·  A = δ(A ) =     T T 0 0 0 1 0 1 Dd cd0 cd1 ··· cdd A(d)   T 1 1 1 0 1 0 0 0 0 1 0 1 ⇒ C−1 and C are the conversion matrices for getting the   orthogonal basis β2 from β1 and vice versa. 0 0 0 0 0 0 0 0 0 0 0 0   (2) (1) 0 0 0 0 0 0 4. Illustration A ∗ A =   T T 0 0 0 0 0 0   Consider the following tree T = (V,E). Then the adja- 0 0 0 0 0 0 cency matrix AT of T is 0 0 0 0 0 0 1 2 3 4 5 6   1 0 0 0 1 0 0 0 0 0 4 0 1 0 0 0 4 0 1 2 0 0 0 1 0 0   3 0 0 0 4 0 1 AT = 3 0 0 0 1 0 0 A =   T 4 4 4 0 5 0 4 1 1 1 0 1 0   5 0 0 0 1 0 1 0 0 0 5 0 2 6 0 0 0 0 1 0 1 1 1 0 2 0

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0 0 0 1 0 1 (3) (3) (3) (1) (3) (2) D3 =AT − δ(AT ∗ I + AT ∗ AT + AT ∗ AT ) 0 0 0 1 0 1     0 0 0 0 0 1 (3) (3) 0 0 0 1 0 1 A = δ(A ) = 0 0 0 0 0 1 T T 1 1 1 0 1 0     0 0 0 0 0 1 0 0 0 1 0 1 =    0 0 0 0 0 0 1 1 1 0 1 0   0 0 0 0 0 0     0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 1 0 0   (3) (1) 0 0 0 1 0 0 Let A ∗ A =  , T T 1 1 1 0 1 0 0 0 0 1 0 1  (0)     AT D0 0 0 0 0 1 0  (1) AT  D1   Y = (2) ,X =   0 0 0 0 0 0   D2 AT    0 0 0 0 0 0 (3) D3   AT (3) (2) 0 0 0 0 0 0 A ∗ A =   G T 0 0 0 0 0 0   Then Y = CX, where 0 0 0 0 0 0 0 0 0 0 0 0 (m) hAT ,D jiF cm, j = ,m, j = 0,1,2,3 hD j,D jiF (1) AT ∗ I =0 (0) D0 =AT = I (0) hA ,D i 6 (1) (1) (1) (1) T 0 F D =A − (A ∗ I) = A − 0 = A c00 = = = 1,c01 = c02 = c03 = 0 1 T δ T T T hD0,D0iF 6 (1) hAT ,D0iF 0 (2) (2) (1) c10 = = = 0, AT ∗ I + AT ∗ AT hD0,D0iF 6     (1) 1 0 0 0 0 0 0 0 0 0 0 0 hAT ,D1iF 10 c11 = = = 1,c12 = c13 = 0 0 1 0 0 0 0 0 0 0 0 0 0 hD1,D1iF 10     0 0 1 0 0 0 0 0 0 0 0 0 A(2) D A(2) D =   +   h T , 0iF 6 h T , 1iF 0 0 0 0 1 0 0 0 0 0 0 0 0 c20 = = = 1,c21 = = = 0,     hD0,D0iF 6 hD1,D1iF 10 0 0 0 0 1 0 0 0 0 0 0 0 (2) hAT ,D2iF 14 0 0 0 0 0 1 0 0 0 0 0 0 c22 = = = 1,c23 = 0, hD ,D iF 14 1 0 0 0 0 0 2 2 (3) (3) 0 1 0 0 0 0 hAT ,D0iF 0 hAT ,D1iF 10   c30 = = = 0,c31 = = = 1, 0 0 1 0 0 0 hD0,D0iF 6 hD1,D1iF 10 =   0 0 0 1 0 0 (3) (3)   hAT ,D2iF 0 hAT ,D3iF 6   c32 = = = 0,c33 = = = 1. 0 0 0 0 1 0 hD2,D2iF 14 hD3,D3iF 6 0 0 0 0 0 1 So, D =A(2) − (A(2) ∗ I + A(2) ∗ A(1)) 2 T δ T T T 1 0 0 0  1 0 0 0   0 1 1 0 1 0 0 1 0 0 −1  0 1 0 0 C =  ,C =   1 0 1 0 1 0 1 0 1 0 −1 0 1 0       1 1 0 0 1 0 0 1 0 1 0 −1 0 1 =  0 0 0 0 0 1   1 1 1 0 0 0 Then Y = CX, also X = C−1Y 0 0 0 1 0 0 Y = CX ⇒ X = C−1Y ⇒ 0 0 0 1 0 0 (0) (0) AT = D0 D0 = AT 0 0 0 1 0 0   (1) (1) (3) (3) (3) 2 0 0 0 1 0 0 AT = D1 D1 = AT (A ∗ I + A ∗ A + A ∗ A ) =   T T T 1 1 1 0 1 0 (2) (0) (2)   A = D0 + D2 D2 = (−1)A + A 0 0 0 1 0 1 T T T (3) (1) (3) 0 0 0 0 1 0 AT = D1 + D3 D3 = (−1)AT + AT

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5. Conclusion

Generally, β2 = {D0,D1,...,Dd} is not an orthogonal (0) (1) (d) basis for span {AG ,AG ,...,AG }, for a simple connected undirected graph G with diameter d. But we proved that this is true for an undirected tree T with diameter d and also derived an invertible conversion matrix for computing one basis β1 from the other basis β2 and vice versa. Further study may be done on exploring some other connected undirected graphs having this property other than trees.

References [1] L. Graham and L. Lovasz, Distance matrix polynomials of trees, Advances in Mathematics, 29(1978), 60–88. [2] Bapat, Graphs and Matrices, Universitext, Springer, 2010. [3] Narsingh Deo, Graph Theory with applications to Engi- neering and Computer Science, Courier Dover Publica- tions, 2016. [4] F. Harary, Graph Theory, Addison-Wesley, 1994. [5] Xiao-Qing Jin, Seak-WengVong, An Introduction to Ap- plied Matrix Analysis, P.112, Higher Education Press Ltd, 2016. [6] P. A. Asharaf and Bindhu K. Thomas , Distance matrix from adjacency matrix using Hadamard product, Malaya Journal of Matematik, 8(3)(2020), 877–881.

????????? ISSN(P):2319 − 3786 Malaya Journal of Matematik ISSN(O):2321 − 5666 ?????????

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