Potential robustness measures for digraphs

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C. Kwok

Potential robustness measures for digraphs

Master thesis

Thesis advisor: Dr. F.M. Spieksma

Date master exam: 28 August 2015

Mathematisch Instituut, Universiteit Leiden

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3

1 Introduction

Whenever a connection in a network fails, it could have grave consequences. It is important that even if these malfunctions occur, the network keeps on per- forming well. This ability is called robustness. The more a network is resistant to failures, the more robust it is. In this thesis, we will consider graphs as a representation of networks. While there has been a lot of research done for undirected graphs, there is little we know about directed graphs even though directed networks play a large roll in our world. Our aim is to nd a mathematical denition and an approach to calculate the robustness of directed graphs such that it does not contradict with our sense of robustness. The most important property is that when an edge is added, the robustness should not decrease.

The same holds if an edge weight is increased. In literature, it is also desired that a direct connection is equal or more robust than an indirect connection (triangle inequality). If there is a way to calculate the robustness of a network, it is possible to compare networks and to design networks that are more robust.

First we will show how to calculate robustness for undirected graphs. This will be done by using the concept of eective resistance between two nodes, which is explained in Section 2. In Section 3, the denition of the Laplacian of a graph will be given and some properties of the Laplacian will be shown. In Section 4, we will show how the robustness is calculated by using the Laplacian pseudoinverse. In the master thesis of W. Ellens [4], it is shown that the eective resistance function has some nice properties. That is, the total eective resistance of a graph strictly decreases (the graph is more robust) when an edge is added or weights are increased. Furthermore, the eective resistance function is a metric (distance function). The proof of these properties will not be given in this thesis. It can be found in Chapter 4.3 in [4]. Another way to nd the Lapla- cian pseudoinverse will be shown in Section 5. This approach is obtained from [10]. In Section 6, an alternative expression of the eective resistance between two nodes which is obtained from [1] will be given.

We will then analyze directed graphs. In Section 7, it is shown how the method used in Section 4 fails for directed graphs. In Section 8, we analyze [10] and their denition of eective resistance for directed graphs. This is followed by Section 9 in which we propose and analyze a dierent, but similar denition. In Section 10, we will try to extend the expression found in Section 6 to directed graphs.

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2 Eective resistance 6

2 Eective resistance

A graph is denoted by G = (V; E), where V the set of N vertices and E is the set of edges or arrows of G when G is undirected respectively directed. A weighted graph has weights on the edges. The weight of edge (i; j) is denoted with wij. We assume that all weights are positive. If no weight is written next to an edge, it is assumed that the edge weight is 1.

To determine the eective resistance, the graph is viewed as an electrical circuit, where an edge (i; j) corresponds to a resistor Rij. For each pair of vertices the eective resistance between these vertices can be calculated with the well- known series and parallel manipulations. Two edges, corresponding to resistors with resistance R1 and R2, connected in series can be replaced by one edge with eective resistance R₁+ R₂. On the other hand, two resistors connected in parallel can be replaced by a resistor with resistance 1 ¹

R1+_R2¹ . Notice that from these series and parallel manipulations, the eective resistance takes both the number of paths and their length into account, intuitively measuring the presence and quality of back-up possibilities.

Now using Kirchho's rst law and Ohm's law, an expression for the eective resistance rab between vertices a and b is found. Kirchho's rst law states that at any vertex in an electrical circuit, the sum of currents owing into that vertex is equal to the sum of currents owing out of that vertex. In other words, connect a voltage between vertices a and b and let I be the net current out of a and into b. Then the current yij between vertices i and j must satisfy

X

j2N(j)

y_ij = 8>

<

>:

I if i = a;

I if i = b;

0 otherwise,

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where N(i) is the set of rst neighbors of vertex i. Now Ohm's law allows us to associate a potential v_i to any vertex i, such that for all edges (i; j) we have

y_ijR_ij = v_i v_j: (2)

Denition 2.1. The eective resistance between vertex a and b is dened as r_ab=v_a v_b

I :

Denition 2.2. The total eective resistance for undirected graphs, also known as the Kirchho index K_f, is dened as the sum of the eective resistances over all pairs of vertices:

Kf =X^N

i=1

XN j=i+1

rij:

Note that if one graph is more robust than another, the former's Kirchho index is smaller.

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3 The (weighted) Laplacian L 7

3 The (weighted) Laplacian L

For every graph G, directed or undirected, we can nd the associated Laplacian matrix.

Denition 3.1. The Laplacian L is given by L = D A;

where D is the diagonal matrix of the node out-degrees and A is the adjacency matrix corresponding to G.

Denition 3.2. For a graph with non-negative edge weights wij, the analogue of the adjacency matrix is the matrix of weights W = (w_ij). The weighted Laplacian, L^w, is given by

L^w= S W;

where S is the diagonal matrix of strengths, with S_ii=P_N

j=1w_ij.

Notice the Laplacian matrix is the weighted Laplacian matrix where the weights are either zero or one. For notational convenience we will write L for the weighted Laplacian.

We dene the resistance of an edge Rij as _w¹_ij.

Theorem 3.1. The eective resistance of an edge (i; j) is computed as follows r_ij= (e⁽ⁱ⁾_N e^(j)_N )^>Y (e⁽ⁱ⁾_N e^(j)_N ) = Y_ii+ Y_jj Y_ij Y_ji; (3) where Y 2 R^NN acts as the inverse of the Laplacian matrix on 1^?_N and as the zero map on spf1_Ng.

Proof. This proof is similar to the proof of Theorem 4.1 in [4].

We will see in Section 4 that the Laplacian pseudoinverse satises these properties mentioned in Theorem 3.1.

3.1 Properties of the Laplacian

In this section some properties of the Laplacian are shown.

Lemma 3.1. The (weighted) Laplacian corresponding to an undirected graph is symmetric, while the (weighted) Laplacian corresponding to a directed graph can be non-symmetric.

Proof. For undirected graphs, we have

(i; j) 2 E , (j; i) 2 E

Thus, Lij = Lji for all i; j 2 V . On the otherhand, if G is a directed graph, it is possible that (i; j) 2 E, while (j; i) =2 E i.e. Lij = wij and Lji= 0.

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Whether or not the Laplacian is symmetric, the sum of each row is always equal to zero, thus L1_N = 0 i.e. 0 is always an eigenvalue of L and 1_N its corresponding eigenvector.

The following lemma shows that the eigenvalues of the Laplacian L of an undirected graph are all positive.

Lemma 3.2. Let G be an undirected graph. The (weighted) Laplacian L of G is a positive semi-denite matrix.

Proof. Taken from [4]. First convert G in a directed graph, by choosing an arbitrary direction for each edge in G. Now dene an arc-vertex incidence matrix M, i.e. for an arc a and a vertex v:

M_av = 8>

<

>:

pw_ij if a = (i; j) and v = i;

pwij if a = (i; j) and v = j;

0 otherwise.

Then the Laplacian L satises L = M^>M and we have for all z 2 R^N : z^>Lz = z^>M^>Mz

= (Mz)^>(Mz) 0;

where the last inequality holds, since it is the inner product of the vector Mz with itself.

Lemma 3.3. Let G be a directed graph. Then all eigenvalues of the Laplacian L of G have non-negative real part.

Proof. It follows from Gerschgorin circles [8] that all eigenvalues of L have non- negative real part.

As mentioned before, zero is always an eigenvalue of L. The algebraic multiplicity of zero is given in the next theorem. However, we introduce a denition rst.

Denition 3.3. A component C of a directed graph G is called a strongly connected component if for all vertices i; j of C, there is a directed path from i to j and a directed path from j to i. A directed path from i 6= j to j is a sequence of arcs which connects a sequence of vertices starting from i and ending in j such that all arcs have the same direction. Assume that there is always a directed path from i to i for every i 2 G. A strongly connected closed component C is a strongly connected component such that for all vertices j not belonging to C we have w_ij = 0 for all vertices i in C.

Theorem 3.2. Let G be an undirected graph. The algebraic multiplicity of the eigenvalue zero of the (weighted) Laplacian L of G is equal to the number of connected components of G. If G were a directed graph, then it is equal to the number of strongly connected closed components of G.

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Proof. The following proof is from [4].

Let G be an undirected graph and L its (weighted) Laplacian. For every connected component C, let y^C2 Rⁿbe the vector with y_i^C= 1 if i is a vertex of C and y^C_i = 0 otherwise. Notice that y^C is an eigenvector of L with eigenvalue 0, since Ly^C = 0. Furthermore, the set of these eigenvectors is linearly independent. Thus, we will prove that all eigenvectors with eigenvalue zero are linear combinations of these eigenvectors.

Now suppose x is an eigenvector of L with eigenvalue zero, i.e. Lx = 0, then we have for all i 2 V

xi

Xn j=1

wij = Xn j=1

xjwij: (4)

Now let C be an arbitrary component and i such that xi = maxj2Cxj, then using (4) we nd

x_iXⁿ

j=1

w_ij =Xⁿ

j=1

x_jw_ij Xⁿ

j=1

x_iw_ij = x_iXⁿ

j=1

w_ij:

Thus x_i = x_j for all neighbours j of i, since in that case we have w_ij > 0.

Similarly, we see that x_j = x_k for every neighbor k of j. By repeating this argument, we see that all eigenvectors x have x_i= x_j when i and j are part of the same component. Therefore, all eigenvectors with eigenvalue zero are linear combinations of the vectors y^C.

If G were a directed graph, we let C be a strongly connected closed component and then the proof still holds.

For example, notice in Example 8.1 there is only one strongly connected closed component, that is, f3g.

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4 The Laplacian pseudoinverse L⁺ 10

4 The Laplacian pseudoinverse L

⁺

As mentioned before there are multiple ways to calculate the eective resistance for undirected graphs. In [4] it is shown how one can calculate the eective resistance for undirected graphs using the Laplacian. Leaving the proofs aside, we will repeat some of the results.

Since the Laplacian has an eigenvalue zero, it is not invertible. However, if we restrict the linear transformation to the subspace orthogonal to the null space spf1Ng, the matrix can be inverted. Let L⁺be the matrix that corresponds to this inverse on 1^?_N and to the zero map on spf1Ng. In other words:

Denition 4.1. The Laplacian pseudoinverse L⁺ is dened as the matrix satisfying

L⁺1_N = 0 and for every w?1N:

L⁺w = v , Lv = w and v?1_N:

This denition is a specic case of the Moore-Penrose pseudoinverse for general mn matrices provided that the graph is undirected. In Section 5.2, it is shown that it is unique.

One way to construct the Laplacian pseudoinverse is as follows. Since the Lapla- cian is symmetric, it has an orthonormal set of eigenvectors v₁; : : : ; v_N. Let U be the matrix that has these eigenvectors as its columns, consequently U ¹= U^>, and let D be the matrix with the corresponding eigenvalues on its diagonal and zero elsewhere. Notice that _i= 0 for some i = 1; : : : ; N, because 0 is an eigenvalue of L. Without loss of generality, assume 1 = 0. Furthermore i 0 for all i, since L is a positive semi-denite matrix, see Lemma 3.2. Thus, D is of the form:

D = 0 BB B@

0 0 : : : 0 0 2 : : : 0 ... ... ... ...

0 0 : : : N

1 CC CA:

Then the Laplacian satises L = UDU ¹= UDU^>. The pseudoinverse is then as follows

L⁺= UD⁺U ¹= UD⁺U^>; with D⁺:

D⁺= 0 BB B@

0 0 : : : 0 0 ¹₂ : : : 0 ... ... ... ...

0 0 : : : ¹_N 1 CC CA:

Furthermore, L⁺ is similar to D⁺, thus the eigenvalues of L⁺ are 0;¹₂; : : : ;¹_n. Also, L⁺ is symmetric:

(L⁺)^>= (UD⁺U^>)^>= U(D⁺)^>U^>

= UD⁺U^>= L⁺

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Lemma 4.1. The Laplacian pseudoinverse L⁺ satises L⁺=

Z ₁

0

e ^Lt 1 N1_N1^>_N

dt:

Proof. Let L = UDU ¹ with U and D as above. Note the rst column of U is

p1

N; : : : ;^p¹_N_>

and we have an orthonormal set of eigenvectors. Observe:

e ^Lt=X¹

k=0

( 1)^kL^kt^k k! =X¹

k=0

( 1)^k(UDU ¹)^kt^k k!

=X¹

k=0

( 1)^kUD^kU ¹t^k

k! = UX¹

k=0

( 1)^kD^kt^k

k! U ¹

= U 0 BB B@

1 0 : : : 0

0 e ²^t : : : 0 ... ... ... ...

0 0 : : : e ^N^t 1 CC CAU ¹

= 0 B@

N1 : : : _N¹ ... ... ...

N1 : : : _N¹ 1

CA + e ²^tv₂^>v₂+ : : : + e ^N^tv_N^>v_N

Thus we nd Z ₁

0

e ^Lt 1 N1N1^>_N

dt =

Z ₁

0 e ²^tv₂^>v2+ : : : + e ^N^tv^>_NvN dt

= 1

2v₂^>v₂+ : : : + 1

Nv_N^>v_N

= UD⁺U ¹= L⁺:

Since the Laplacian pseudoinverse is given by L⁺ = UD⁺U^>, it acts as the inverse of the Laplacian matrix on 1^?_N and as the zero map on spf1Ng. This leads to the following corollary.

Corollary 4.1. The eective resistance of an edge (i; j) is computed as follows r_ij = (e⁽ⁱ⁾_N e^(j)_N )^>L⁺(e_N⁽ⁱ⁾ e^(j)_N ) = L⁺_ii+ L⁺_jj 2L⁺_ij:

Now the total eective resistance is given by the following theorem.

Theorem 4.1. The Kirchho index Kf satises

Kf = N XN

i=2

1

^L_i ; (5)

with ^L_i's the eigenvalues of L.

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Proof.

K_f =X^N

i=1

XN j=i+1

r_ij =1 2

XN i=1

XN j=1

r_ij

=1 2

XN i=1

XN j=1

(L)⁺_ii+ (L)⁺_jj 2(L)⁺_ij

= NX^N

i=1

(L⁺_ii) 1^>_NL⁺1N

= N Tr(L⁺) = NX^N

i=2

1

^L_i :

4.1 L

⁺

: Example undirected graph

Using the Laplacian pseudoinverse, the total eective resistance can be calculated. Observe a rather easy example.

Example 4.1. Let G be an undirected graph with weights 1:

2

1 3 with the Laplacian matrix L =

0

@ 1 1 0

1 2 1

0 1 1

1 A :

The eigenvalues of L are ₁ = 0; ₂ = 1; ₃ = 3. Since L is symmetric, it has an orthonormal set of eigenvectors, explicitly

v1=

1 3

p3;1 3

p3

; v2=

1 2

p2; 0;1 2

p2

; v3=

1 6

p6; 1 3

p6;1 6

p6

:

Let U be the matrix that has these eigenvectors as its columns:

U = 0

@

13

p3 ¹₂p

2 ¹₆p

1 6

3

p3 0 ¹₃p

1 6

3

p3 ¹₂p

2 ¹₆p 6

1

A , U ¹= U^>= 0

@

13

p3 ¹₃p 3 ¹₃p

1 3

2

p2 0 ¹₂p

1 2

6

p6 ¹₃p 6 ¹₆p

6 1 A : Then let D be the matrix with the eigenvalues on the diagonal and zero elsewhere:

D = 0

@ 0 0 0 0 1 0 0 0 3

1 A :

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Note L = UDU ¹= UDU^>. Then D⁺ is given by

D⁺= 0

@ 0 0 0 0 1 0 0 0 ¹₃

1 A ; and so the Laplacian pseudoinverse L⁺ is

L⁺= UD⁺U ¹= UD⁺U^>= 1 9

0

@ 5 1 4

1 2 1

4 1 5

1 A :

We have L⁺1 = 0 and for every w?1 :

L⁺w = v such that Lv = w and v?1:

For example take w = 0

@ 1 01

1

A, then v = ¹₉ 0

@ 6 33

1

A and Lv = w.

The total eective resistance of this graph is Kf = 3 1 +¹₃

= 4.

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5 Dierent approach for the undirected graph 14

5 Dierent approach for the undirected graph

In this section, a dierent way to construct the Laplacian pseudoinverse is shown. This matrix, which we will refer to as X in this section, is thus identical to the Laplacian pseudoinverse L⁺for undirected graphs. However, in Section 8 the matrix X will be redened such that it is unique and symmetric for directed graphs, while acting as the Laplacian pseudoinverse for undirected graphs. Ef- fective resistance is then dened using the redened X. Consequently, the square root of the eective resistance is a metric on the nodes of any connected directed graph. The following results are from [10].

Let = IN 1

N1N1^>_N, denote the orthogonal projection matrix onto the subspace of R^N perpendicular to 1N. This subspace will be denoted with 1^?_N. is a symmetric matrix. Since the entries of the rows of L add up to 0 (i.e. the rows are perpendicular to 1), we have L1_N = 0, L = L and L^> = L^> for any graph. Note, if the graph is balanced, i.e. the out-degree and in-degree are equal, which is true for undirected graphs, L = L^>= L^>= L also hold.

Let Q 2 R^{(N 1)N} be a matrix whose rows form an orthonormal basis for 1^?_N. Thus, we require h

p1

N1N Q^>i

to be an orthogonal matrix. It follows,

Q1N = 0; QQ^>= IN 1 and Q^>Q = : (6) The last result holds since h

p1

N1N Q^>i

is an orthogonal matrix, thus

p1

N1_N Q^>

p1

N1_N Q^>

_>

= I_N

and by adding the ^p¹_N1N column to Q, we essentially add _N¹1N1^>_N to Q^>Q.

X will be constructed such that X is equal on 1^?_N to the inverse of L and 0 otherwise. Thus, it satises the denition of the Laplacian pseudoinverse in Section 4. This can be achieved by taking X as the unique generalized inverse of L.

5.1 Reduced Laplacian L

We construct X by using the reduced Laplacian. This matrix acts as the Lapla- cian L on 1^?_N.

Let b1; : : : ; bN 1be an orthonormal basis for 1^?_N and let Q 2 R^{(N 1)N} be the matrix formed with these basis vectors as rows. Then for any v 2 R^N, we can write v =P_{N 1}

i=1 cibi+ cN1N, with c1; : : : ; cN 2 R. By taking the innerproduct with any bi on both sides, we nd bi v = ci. Thus, v := Qv is a coordinate vector (with respect to the chosen basis) of the orthogonal projection of v onto 1^?_N. For any M 2 R^NN, let M 2 R^{N 1N 1}be the linear transformation with respect to b₁; : : : ; b_{N 1} with M := QMQ^>. Then M acts as M on 1^?_N, i.e.

Mv = (QMQ^>)Qv = QMv = QMv = Mv for any v 2 1^?_N:

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Thus, on 1^?_N, the Laplacian matrix L is equivalent to

L = QLQ^>; (7)

which we refer to as the reduced Laplacian. We see that L is symmetric if and only if the graph is balanced, since in that case we have L = L^>:

L^> = (QLQ^>)^>= QL^>Q^>= QLQ^>= L:

Theorem 5.1. Let L be the Laplacian matrix, then L given in (7) has the same eigenvalues, except for a single zero eigenvalue, as L. Also, if v is an eigenvector with eigenvalue 6= 0 of L, then v on 1^?_N is an eigenvector with eigenvalue of L.

Proof. This proof is partially taken from [9].

Dene the matrix

V =

₁

pN1^>_N Q

; (8)

with Q satisfying (6). Then V V^>= I_N and V^>V = I_N hold, which means V^>

is the inverse of V , i.e. V^> = V ¹: Now we see that L is similar to V LV^> for the invertible matrix V , thus L and V LV^> have the same eigenvalues. Using the fact L1_N = 0, it follows V LV^> =

0 0 0 L

with L given in (7) and we see that this is a block matrix. Thus, the eigenvalues of V LV^>, which are the same as the eigenvalues of L, are the solutions of det(IN 1 L) = 0, i.e.

zero and the eigenvalues of L. We conclude that L has the same eigenvalues as L except for a single zero eigenvalue.

Let v be an eigenvector of L with eigenvalue . Then, we have Lv= QLQ^>Qv= QLv

= QLv= Qv= v: Thus, v is an eigenvector with eigenvalue of L.

Hence, it follows from Theorems 3.2 and 5.1 that for a connected graph, L is invertible since L only has one 0 eigenvalue.

5.2 Generalized inverse

Denition 5.1. For A 2 R^mn, the generalized inverse of A is dened as a matrix A⁺2 R^nm satisfying the so-called Moore-Penrose criteria:

1. AA⁺A = A 2. A⁺AA⁺= A⁺ 3. (AA⁺)^>= AA⁺ 4. (A⁺A)^>= A⁺A

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Theorem 5.2. The generalized inverse of A 2 R^mn is unique.

Proof. Let B; C 2 R^nm both be a generalized inverse of A 2 R^mn, thus they satisfy the Moore-Penrose criteria. Then observe that

AB = (AB)^>= B^>A^> = B^>(ACA)^> = B^>A^>C^>A^>

= (AB)^>(AC)^>= ABAC = AC:

Analogously we nd BA = CA. It follows,

B = BAB = BAC = CAC = C;

thus the generalized inverse of A 2 R^mn is unique.

As mentioned before, X will be constructed as the unique generalized inverse of L. Thus, it must satisfy the Moore-Penrose criteria. This can also be formulated dierently.

Lemma 5.1. If X satises the following:

XL = LX = and X = X = X; (9)

then X satises the Moore-Penrose criteria.

Proof. Let X be a solution to (9). Since L is symmetric, we have:

1. LXL = L = L 2. XLX = X = X 3. (LX)^>= ^>= = LX 4. (XL)^>= ^>= = XL:

Thus it satises the Moore-Penrose criteria.

5.3 Dening X

Now using the reduced Laplacian L we give an explicit construction for X:

X = Q^>L ¹Q: (10)

Observe:

X1_N = Q^>L ¹Q1_N = 0, since Q1_N = 0;

Xv = QXv = QQ^>L ¹Qv = L ¹Qv = L ¹v for any v 2 R^N:

Thus, X acts as the inverse of L, which is equivalent to the inverse of L on 1N

and X is 0 elsewhere.

Corollary 5.1. Let X = [xi;j], then the eective resistance is given by

rij= (e⁽ⁱ⁾_N e^(j)_N )^>X(e⁽ⁱ⁾_N e^(j)_N ) = xi;i+ xj;j xi;j xj;i: (11)

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Since L is symmetric for undirected graphs, so is L ¹ and thus X is also symmetric:

X^> = (Q^>L ¹Q)^>= Q^>(L ¹)^>Q = Q^>L ¹Q = X Consequently, we can write

r_ij = x_i;i+ x_j;j 2x_i;j:

Notice that we always have rij= rji, even if X is not symmetric.

Lemma 5.2. X constructed as in (10) satises (9) when the graph is undirected and is thus the Moore-Penrose generalized inverse of L. Hence, it is unique.

Proof. Using (6) and L = L = L, we nd

XL = XL = Q^>L ¹QLQ^>Q = Q^>L ¹LQ = Q^>Q = and analogously we nd LX = . Furthermore,

X = Q^>L ¹Q = Q^>L ¹QQ^>Q = Q^>L ¹Q = X and again in the same way X = X also holds.

Note L is not unique, since it depends on the choice of Q. However X is independent of the choice of Q as long it satises (6).

Theorem 5.3. X is independent of the choice of Q.

Proof. Let Q and Q⁰ both satisfy (6). Dene P := Q⁰Q^>. P is orthogonal:

P^>P = (Q⁰Q^>)^>Q⁰Q^>= QQ^0>Q⁰Q^>= QQ^>= IN 1

and analogously we have P P^>= IN 1, thus P^> is the inverse of P . Write Q⁰= Q⁰ = Q⁰Q^>Q = P Q:

Hence,

X⁰ := Q^0>(Q⁰LQ^0>) ¹Q⁰= Q^>P^>(P QLQ^>P^>) ¹P Q

= Q^>(QLQ^>) ¹Q = X:

Thus X is independent of the choice of Q.

5.4 X: Total eective resistance

Theorem 5.4. The Kirchho index satises

K_f = NX^N

i=2

^X_i

with ^X_i 's the eigenvalues of X.

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Proof. Take the matrix V as in (8). Then X is similar to V XV^> for V , thus X and V XV^> have the same eigenvalues. Since X = Q^>L ¹Q, we can write V XV^> as the following: V XV^> =

0 0

0 L ¹

: Thus, X has a single 0 eigenvalue and its remaining eigenvalues are the same as L ¹. However, the eigenvalues of L ¹ are one over the eigenvalues of L and those are the same as the eigenvalues of L, excluding the zero eigenvalue. This implies that X has a single 0 eigenvalue and its remaining eigenvalues are one over the eigenvalues of L. Then using (5) we have

Kf = N XN

i=2

1

^L_i = N XN i=2

^X_i

with ^L_i and ^X_i the i'th eigenvalue of L respectively X.

5.5 X: Example undirected graph

We will now show the calculation of X corresponding to Example 4.1. Take

Q = ^p¹² ^p¹² 0

p1

6 p1

6 p2

6

! , Q^>=

0 B@

p1

2 p1

1 6 p2 p1

0 ^p²6₆

1 CA :

Then we nd

L = QLQ^>= _p₃⁵² ^p³¹²

12 9

6

!

and L ¹= 1 6

3 p

p 3

3 5

:

X is given by X = ¹₉ 0

@ 5 1 4

1 2 1

4 1 5

1

A, which is the same as L⁺ and we also have Kf = 3 1 + ¹₃

= 4.

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6 Computing eective resistance using determinants 19

6 Computing eective resistance using determinants

As we have seen in Section 4, the eective resistance between node i and j is given by

r_ij = L⁺_ii+ L⁺_jj L⁺_ij L⁺_ji:

We will show another expression for rij that is quite simple. This result holds for any weighted Laplacian of an undirected graph.

First dene L(i) to be the submatrix obtained by deleting the i-th row and i-th column of the Laplacian matrix L of a graph G. The submatrix L(i; j) is obtained by deleting the i-th and j-th rows and i-th and j-th columns of L. The following theorem and proof are from [1].

Theorem 6.1. Let G = (V; E) be an undirected connected graph with n 3 vertices, and 1 i 6= j n. Let L(i) and L(i; j) be the above dened submatrices of the Laplacian matrix of G. Then the eective resistance between node i and j is given by

rij= det L(i; j)

det L(i) : (12)

Proof. Associate to each node vi of the graph G a variable xi and dene the auxiliary function

f(x1; : : : ; xn) = X

k<l;l2N(k)

(xk xl)² (13)

where N(k) denotes the set of rst neighbors of v_k. The following relation is a result known in the theory of electrical networks [2]. For i 6= j,

rij = max

1

f(x₁; : : : ; x_n)

xⁱ= 1; xj= 0; 0 xk 1; k = 1; : : : ; n

: (14) Then using (13) and (14), we derive for k 6= i; j :

@f

@xk = 2 X

l2N(k)

(x_k x_l) = 0: (15)

Since the vertices of G are numbered arbitrarily, without loss of generality, we may consider the special case in which i = n 1 and j = n. Thus xn 1= 1 and xn = 0 may hold. Let A = (aij) be the adjacency matrix of G and denote the submatrix L(n 1; n) with Ln 2, then we can write the Laplacian matrix as

L =

L_{n 2} B B^> D

; where we have

B = 0 B@

a1;n 1 a1;n

... ...

a_{n 2;n 1} a_{n 2;n} 1

CA and D = P_N

i=1an 1;i an 1;n

an;n 1 P_N

i=1an;i

! :

(20)

Then from (15) the following can be derived for all k 6= n 1; n:

X

l2N(k)

(x_k x_l) = X

l2N(k)nfn 1;ng

(x_k x_l) + (x_k x_{n 1})1^(k)_{n 1}+ (x_k x_n)1^(k)_n

= 0; (16)

with

1^(k)_{n 1}=

(1 if vn 1 is adjacent to vk, 0 otherwise:

and 1^(k)n similar. Now since xn 1= 1 and xn= 0, we have X

l2N(k)nfn;n 1g

(xk xl) + xk1_{n 1}^(k) + xk1^(k)_n = 1 1^(k)_{n 1}

Thus L_{n 2}x = b; (17)

where x = (x₁; : : : ; x_{n 2})^> and b = (b₁; : : : ; b_{n 2})^>, with b_k= 1 if the vertices vn 1 and vk are adjacent, otherwise bk = 0, for all k = 1; : : : ; n 2.

Since xn 1= 1; xn= 0 holds and L is symmetric, the following relations hold:

x^>B(xn 1; xn)^>= (xn 1; xn)B^>x =

n 2X

i=1

an 1;ixi= X

k2N(n 1)

xk

and (x_{n 1}; x_n)D(x_{n 1}; x_n)^>= d_{n 1};

with d_{n 1}denoting the degree of the vertex v_{n 1}. Now rewrite (13) as follows f(x1; : : : ; xn) = (x^>; xn 1; xn)L(x; xn 1; xn)

= x^>Ln 2x + (xn 1; xn)B^>x + x^>B(xn 1; xn)^>

+ (x_{n 1}; x_n)D(x_{n 1}; x_n)^>

= x^>b 2 X

k2N(n 1)

xk+ dn 1

= dn 1

X

k2N(n 1)

xk;

(18)

where the last equation holds because x^>b =

n 2X

i=2

an 1;ixi= X

k2N(n 1)

xk:

Since G is an undirected, connected graph, its Laplacian matrix L is positive semi-denite. Thus, the submatrix L_{n 2} is a positive denite matrix, which means its inverse (L_{n 2}) ¹ exists. Denote the (i; j)-th entry of (L_{n 2}) ¹ with t_ij, then x = (L_{n 2}) ¹b implies

xk= X

l2N(n 1)

tkl: (19)

We will use the following lemma.

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Lemma 6.1. The following equation holds:

det L(i; j) = det L(j; i):

Proof. We have

L(i; j) = (L(j; i))^>and detL(i; j) = det(L(i; j))^>: Thus, it follows

detL(i; j) = detL(j; i):

Let L_{n 2}(k; l) be the submatrix obtained by removing from L_{n 2}the k-th row and the l-th row. Then using Cramer's rule and Lemma 6.1 we obtain

t_kl= ( 1)^k+ldetLn 2(l; k)

detL_{n 2} = ( 1)^k+ldetLn 2(k; l)

detL_{n 2} : (20) Next we can rewrite (18) using (19) and (20) into

f(x1; : : : ; xn) = dn 1

X

k2N(n 1)

X

l2N(n 1)

( 1)^k+ldetLn 2(k; l)

detL_{n 2} : (21) Now look at detL(n) by expanding it with respect to its last column, which is the (n 1)-th column of L:

detL(n) = d_{n 1}detL_{n 2} X

k2N(n 1)

( 1)^{k+n 1}detL(n; k; n 1); (22)

where L(n; k; n 1) is the submatrix obtained by removing the k-th row and the (n 1)-th column from L(n). Furthermore, we expand detL(n; k; n 1) with respect to its last row,

detL(n; k; n 1) = X

l2N(n 1)

( 1)^{l+n 1}detL_{n 2}(k; l):

Substituting this in (22) yields detL(n) = d_{n 1}detL_{n 2} X

k2 (n 1)

( 1)^{k+n 1} X

l2N(n 1)

( 1)^{l+n 1}detL_{n 2}(k; l)

= dn 1detLn 2

X

k2N(n 1)

X

l2N(n 1)

( 1)^k+ldetLn 2(k; l); (23)

which results in the following expression when substituted back in (21) f(x1; : : : ; xn) = detL(n)

detLn 2: Finally, substituting this in (14), gives us (12).

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According to Kirchho's theorem, the number of spanning trees is equal to any cofactor of the Laplacian matrix of G. Thus

det L(i) = t(G);

where t(G) is the number of spanning trees of G. Now, according to [7], det L(i; j) is equal to the number of trees of G⁰, where G⁰ is the graph where vertex vi and vertex vj is contracted to one vertex v. Thus,

detL(i; j) = t(G⁰) and we can write

rij= t(G⁰) t(G):

This can be interpreted as follows. The eective resistance between vertex i and j is equal to the number of trees of G containing the edge (i; j) divided by the total number of spanning trees of G. This can be seen as how much the edge (i; j) contributes to the number of spanning trees of G.

Remark: Since any cofactor of the Laplacian is equal to the number of spanning trees, we have detL(i) = detL(j) for every i; j 2 V . Using Lemma 6.1, we can write

r_ij= det L(i; j)

det L(i) = det (L(j; i))^>

det L(j) =det L(j; i) det L(j) = r_ji:

6.1 Example undirected graph using determinants

Recall Example 4.1:

2

1 3 with the Laplacian matrix L =

0

@ 1 1 0

1 2 1

0 1 1

1 A :

Using the corresponding Laplacian pseudoinverse of this graph 1

9 0

@ 5 1 4

1 2 1

4 1 5

1 A ;

yields

r12= L⁺₁₁+ L⁺₂₂ 2L⁺₁₂= 5 9 +2

9 +2 9 = 1;

r13= L⁺₁₁+ L⁺₃₃ 2L⁺₁₃= 5 9 +5

9 +8 9 = 2;

r23= L⁺₂₂+ L⁺₃₃ 2L⁺₂₃= 2 9 +5

9 +2 9 = 1:

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These are the same as using (12):

r12= detL(1; 2) detL(1) = 1;

r₁₃= detL(1; 3) detL(1) = 2;

r23= detL(2; 3) detL(2) = 1:

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7 L⁺: Directed graph 24

7 L

⁺

: Directed graph

It is not always possible to compute L⁺ for directed graphs as in Section 4 for undirected graphs. The matrix U such that L = UDU ¹ can not always be constructed.

Example 7.1. Consider 2

1 3 with L =

0

@ 1 1 0

0 1 1

0 0 0

1 A :

The eigenvalues of L are ₁= 0 with multiplicity two and ₂ = 1 with multiplicity one. The eigenvectors are

v₁=

1 3

p3;1 3

p3

and v₂= (1; 0; 0) :

This means that L is not diagonalizable; we can't construct U such that L = UDU ¹. Furthermore, even if we were to restrict the graphs, in [6] it has been argued that the robustness if dened through L⁺could decline when an edge is added, which contradicts with our concept of robustness. Thus, this approach is not suitable for directed graphs.

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8 X: Directed graph 25

8 X: Directed graph

Here as well, for directed graphs we can't always compute X in the same manner as in Section 5. From Theorem 5.1 it follows that in the case of a directed graph, the algebraic multiplicity of the eigenvalue zero is equal to the number of strongly connected closed components, which is not always equal to one.

Therefore, the inverse of L does not necessarily need to exist.

Therefore in [10], the authors propose a dierent way to construct X when the graph is directed. This approach will lead to a matrix X that is unique and symmetric. Furthermore, it is equal to the Laplacian pseudoinverse when calculated for undirected graphs and in Section 8.2 it is shown that the square root of the eective resistance is a metric on the nodes of any connected directed graph. The symmetry of X will be used to prove this. If we would take X as the Moore-Penrose inverse neither the eective resistance nor the square root of it is a metric.

In section Conclusion and discussion, we will discuss the eective resistance function satisfying symmetry.

The approach used in [10] considers the following Lyapunov equation:

A + A + B = 0; (24)

with A; B square matrices, A the conjugate transpose of A and unknown.

Theorem 8.1. If A has only eigenvalues with a negative real part, then

= Z ₁

0 e^A^tBe^Atdt (25)

is the unique solution to (24).

Proof. This proof is taken from [3].

Let be given as in (25). Then we have A + A =

Z ₁

0

d dt

he^A^tBe^Ati

dt = e^A^tBe^At¹

0 = B;

which means that (25) is a solution for (24). To prove the uniqueness of (25), we dene a linear map : Cⁿⁿ ! Cⁿⁿ, with () = A + A. Now for all B 2 Cⁿⁿ, the equation () = B has a solution. Thus, the dimension of the image of is n². Note that the dimension of the domain of is also n². Therefore, ker( ) = 0 holds. We conclude () = B has a unique solution for each B.

Lemma 8.1. Furthermore, if B is positive denite, then is also positive denite.

Proof. Let z 2 Cⁿ and z the conjugate transpose of z, then observe zz = z

Z ₁

0 e^A^tBe^Atdtz = Z ₁

0 ze^A^tBe^Atzdt (26)

= Z ₁

0 (e^Atz)B e^Atz

dt > 0: (27)

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The last inequality follows since B is positive denite.

Now by taking A = L^> and B = IN 1, (24) can be rewritten to

L + L^> = I_{N 1}: (28)

It follows from Lemma 8.1 that is positive denite. We will see that is also symmetric.

Lemma 8.2. satisfying (28) is symmetric.

Proof. Let satisfy (28). Then from Theorem 8.1 it follows that

= Z ₁

0 e ^Lte ^L^>^tdt:

Observe the following:

^>= Z ₁

0

e ^Lte ^L^>^t_>

dt = Z ₁

0 e ^Lte ^L^>^tdt = :

Thus, is symmetric.

Construct X as

X = 2Q^>Q; (29)

with Q satisfying (6) and the unique solution to the Lyapunov equation (28).

Note that since is symmetric, X is also symmetric for any graph. Now if L is symmetric, we have

=1

2L ¹: (30)

Corollary 8.1. If L is symmetric, X in (29) is equal to the Laplacian pseudoinverse L⁺.

Proof. Straightforward.

Corollary 8.2. satises

= Z ₁

0 Qe ^Lte ^L^>^tQ^>dt:

Proof. Let satises (25) with A = L^> and B = I_{N 1}. Then we can write

(27)

the following:

= Z ₁

0 e ^Lte ^L^>^tdt

= Z ₁

0

X1 k=0

( 1)^k(QLQ^>)^kt^k k!

X1 k=0

( 1)^k(QL^>Q^>)^kt^k

k! dt

= Z ₁

0 QX¹

k=0

( 1)^kL^kt^k

k! Q^>QX¹

k=0

( 1)^k(L^>)^kt^k k! Q^>dt

= Z ₁

0 QX¹

k=0

( 1)^kL^kt^k k! X¹

k=0

( 1)^k(L^>)^kt^k k! Q^>dt

= Z ₁

0 Qe ^Lte ^L^>^tQ^> 1 NQ

X1 k=0

( 1)^kL^kt^k

k! 1N1^>_Ne ^L^>^tQ^>dt

= Z ₁

0 Qe ^Lte ^L^>^tQ^> 1

NQ1_N1^>_Ne ^L^>^tQ^>dt

= Z ₁

0 Qe ^Lte ^L^>^tQ^>dt:

The following lemma shows us that what kind of eigenvalues X has.

Lemma 8.3. X satisfying (29) has a single 0 eigenvalue and its remaining eigenvalues are twice those of .

Proof. Take V = h

p1

N1N Q^> i

with Q as in Section 5. Then V is an orthogonal matrix. X is similar to V^>XV =

0 0^>

0 2

. Hence, X has a single 0 eigenvalue and its remaining eigenvalues are twice those of .

Now in [10] they dene the eective resistance between nodes in the graph as the following.

Denition 8.1. Let G be a connected graph with N nodes and Laplacian matrix L. Then the eective resistance between nodes i and j in G is dened as

rij = (e⁽ⁱ⁾_N e^(j)_N )^>X(e_N⁽ⁱ⁾ e^(j)_N ) = xi;i+ xj;j 2xi;j; (31) where

X = 2Q^>Q; L + L^>= IN 1; L = QLQ^>; (32) and Q satisfying (6).

Remark: This expression for the eective resistance comes from (3). However X dened as in (32) does not satisfy all the properties mentioned in Theorem 3.1. X acts as the zero map on spf1_Ng, but it does not act as the inverse of the Laplacian matrix on 1^?_N; XL = does not hold.

By taking (31) as the denition for the eective resistance between two nodes, the denition of the Kirchho index for undirected graphs can be generalized

(28)

by summing over all distinct eective resistances of pairs of vertices of a graph, directed or undirected.

Denition 8.2. The total eective resistance, K_f, of a graph is the sum over all distinct eective resistances of pairs of vertices:

Kf= XN i=1

XN j=i+1

rij (33)

Theorem 8.2. The Kirchho index satises

Kf = N XN i=1

^X_i ;

with ^X_i 's the eigenvalues of the matrix X satisfying (32).

Proof. Similarly to the proof of Theorem 4.1, the Kirchho index can be written as

K_f =X^N

i=1

XN j=i+1

r_ij= NX^N

i=1

(x_i;i) 1^>_NX1_N

= N XN

i=1

(xi;i) 1^>_N(2Q^>Q)1N = N XN i=1

xi;i

= N Tr(X) = NX^N

i=1

^X_i ;

with ^X_i 's the eigenvalues of X.

Then from Lemma 8.3 it follows

Kf = 2N

N 1X

i=1

_i;

with _i's the eigenvalues of .

8.1 Eective resistance for the directed graph is well-dened

To conrm that the concept of eective resistance for directed graphs given in Denition 8.1 is indeed well-dened, observe the following two lemmas from [10].

Lemma 8.4. The value of the eective resistance between two nodes in a connected directed graph is independent of the choice of Q.

Proof. Let Q an Q⁰be two matrices satisfying (6) and let the eective resistance between nodes i and j be rij, r⁰_ij, computed using Q and Q⁰ respectively. Fur- thermore, dene W = Q⁰Q^>. Then we have W Q = Q⁰Q^>Q = Q⁰ = Q⁰ and

(29)

W W^>= Q⁰Q^>(Q⁰Q^>)^> = Q⁰Q^>QQ^0>= Q⁰Q^0>= Q⁰Q^0>= I_{N 1}. W^>W is computed analogously. Now using (7) and the above we can write

L⁰= Q⁰LQ^0>= W QL(W Q)^>= W QLQ^>W^> = W LW^>:

Substituting this expression in (28) for L⁰, we see = W^>⁰W . And thus we nd X = 2Q^>Q = 2Q^>W^>⁰W Q = 2Q^0>⁰Q⁰= X⁰, hence rj;k= r_j;k⁰ . From this lemma it follows that the eective resistance is independent of the choice of Q as long as it satises (6). The eective resistance is also independent of the labelling of the nodes, which will be proven in the following lemma.

Lemma 8.5. The value of the eective resistance between two nodes in a connected directed graph is independent of the labelling of the nodes.

Proof. Let L and L⁰ be two Laplacian matrices associated with the same graph, but with dierent labellings of the nodes. Then L⁰ can be found from L by permuting its rows and columns. That is, there exists a matrix P such that L⁰ = P LP^>. Note that since P is a permutation matrix, P has exactly one 1 in every row and column, and every other entry is 0. Consequently, we have P ¹= P^>; P 1N = 1N and 1^>_NP = 1^>_N. Furthermore, we have

P^>= (QP Q^>)^>= QP^>Q^> = QP ¹Q^>= P ¹; P Q = QP Q^>Q = QP = QP (I_N 1

N1_N1^>_N)

= QP QP 1

N1_N1^>_N = QP

and analogously Q^>P = P Q^>. Now using (7) and the properties above, it follows

L⁰= QL⁰Q^>= QP LP^>Q^>= P QLQ^>P^> = P LP^>:

Then substituting this expression in (28) for L⁰ yields = P^>⁰P . Hence, X = 2Q^>Q = 2Q^>P^>⁰P Q = 2(P Q)^>⁰QP

= 2P^>Q^>⁰QP = P^>X⁰P:

Thus, if P permutes node i to node k and node j to node l, r⁰_kl= rij holds.

8.2 The square root of the eective resistance is a metric

The following theorem is taken from [10].

Theorem 8.3. The square root of the eective resistance is a metric on the nodes of any connected directed graph. That is, we have for all nodes i; j; k

rij 0;

rij = 0 , i = j;

r_ij= r_ji; and prik+ prkj prij:

Potential robustness measures for digraphs

C. Kwok