18.085 PROBLEM SET 5 SOLUTIONS
2.4.5 If there is an arrow from i to j in the graph, then some row of A is
[... − 1...1...], with -1 in position i and 1 in position j ; this row multiplied by u gives
−ui + uj . Hence Au = 0 implies ui = uj . If i and j are not connected by an edge,
there is still some path i − i1 − ... − ik − j in the graph. Then the above argument
gives ui = ui1 , ui1 = ui2 ,..., ui
= uj , hence ui = uj . Therefore the only solution to
k
C
 C 


Au = 0 is given by u =  .  .
 .. 
C
Pn
2
T
aki = ±1 if there is an edge
2.4.6 We have (A A)ii =
k=1 aki = deg i, since P
between k and i, and 0 otherwise. Hence tr(AT A) = ni=1 deg i = 2m, because in
summing all edges connected to all nodes each edge is counted twice (once for each
endpoint).
2.4.7 We have

−1
A= 0
0
0
−1
0
0
0
−1

c1
1

0
1  ⇒ K = AT CA = 
 0
1
−c1

0
c2
0
−c2

−c1

−c2
,

−c3
c1 + c2 + c3
0
0
c3
−c3
which is singular. Grounding
node 4 has the eect of removing the 4th row and

c1
column of K : Kred =  0
0
0
c2
0
0
0  , det Kred = c1 c2 c3 .
c3

−1 1
0
0 0
 0 −1 1
0 0 
.
2.4.9 For the path graph with edges 1 → 2 → 3 → 4 → 5, A = 
 0
0 −1 1 0 
0
0
0 −1 1


1 −1 0
0
0
 −1 2 −1 0
0 


. Grounding node 5 gives K = (AT A)red =
0
−1
2
−1
0
Hence AT A = 


 0
0 −1 2 −1 
0
0
0 −1 1




1 −1 0
0
4 3 2 1
 −1 2 −1 0 
 3 3 2 1 
−1




 0 −1 2 −1 , which we recognize as T4 . Hence K =  2 2 2 1  , det K =
0
0 −1 2
1 1 1 1
(2i−1)π
1. The eigenvalues of K are λi = 2(1 − cos 9 ). They can also be approximated
by using eig(K) in Matlab: λ1 = .12, λ2 = 1, λ3 = 2.34, λ4 = 3.53.
2.4.13 The numbers of spanning trees are 3 and 8 respectively.
T
2.4.17 a. In A A, an entry ij is non-zero if and only if it's on diagonal or there is
an edge between i and j . Hence the number of 0's in AT A is 81−9−2(nr.edges) = 48.

1
18.085 PROBLEM SET 5 SOLUTIONS
b.
(AT A)ii = deg i, so D = diag
2
3
2
3
4
3
2
2
2 . (The nodes
3
of the 3x3 grid are labelled row by row from left to right: 1,2,3; 4,5,6; 7,8,9)
c. d55 = deg 5 = 4 (4 edges connect to 5). This is also why there are four −1's
in row 5 - their column indices correspond to the nodes connected to 5.
2.4.18 Grounding node 9, we have

Kred = (AT A)red

Solving Kred u = fred





=










=





1
0
0
0
0
0
0
0
2
−1
0
−1
0
0
0
0
−1
3
−1
0
−1
0
0
0
0
−1
2
0
0
−1
0
0
−1
0
0
3
−1
0
−1
0
0
−1
0
−1
4
−1
0
−1
0
0
−1
0
−1
3
0
0






 in Matlab we nd u = K −1 fred
red





−1
(u is just the rst column of Kred
)

0
0 

0 

0 
.
−1 

0 

−1 
3


1.5
 1 


 .75 


 1 
.

=

 .75 
 .5 


 .75 
.5
0
0
0
−1
0
0
2
−1