Immanants of Laplacians of Graphs
Onn Chan1 and M.Y Ong2
Department of Mathematics, National University of Singapore
Singapore 117543
Abstract
Let A be a nxn complex matrix. Then it is usual to define det(A) =

sgn(  )  a ,
and per(A) =   a . When A = L(T) is a Laplacian matrix defined from a tree, we
can give a formula for det(L(T)) and per(L(T)) in terms of some numbers associated with
the matchings of the tree. Hence by studying the properties of these matching numbers,
we can deduce some properties of det(L(T)) and per(L(T)). Among the results proven is
that per(L(T)) is always even when T is any tree.
1. Introduction
Let A be a nxn matrix and f:Sn  .Then, we define the generalized matrix function
d :Mn( )  , by d (A) =  f(  )  a
. When f =  (  being a partition of n), is an
irreducible character of Sn , d ( ) is called an immanant. Let A be a nxn matrix. When f =
sgn, we get d (A) = det(A) and when f = 1, we get d (A) = per(A), where per(A) is the
permanent of A. In general, calculating d (A) for any arbitrary matrix A, can be difficult
because each   Sn contributes a term a a … a
in the calculation. However,
when A = L(T)[where L(T) is a laplacian matrix defined from some tree T], a a … a
= 0 whenever   Sn has a cycle of length  3 in its cycle decomposition, which greatly
reduces the number of terms to be considered. Furthermore, certain numbers associated
with matchings of a tree come up naturally in the computation of d (L(T)). We will
derive an expression for d (L(T)) in terms of these numbers, and investigate some
properties of these numbers.
2. Laplacian matrix of a graph
Let G = (V,E) be a graph of order n, having vertex set V = {v0,v1,…,vn }. The laplacian
matrix of G denoted L(G) = [aij] is the nxn matrix with entries
aij =
-1,
0,
deg(vi),
if i  j and {vi,vj}  E.
if i  j and {vi,vj} E
if i = j
Note that(this is used to prove(3)), det(L(G)) = 0. Because if L(G) were invertible, there
would exist a sequence of elementary matrices E1,E2,…,Ek, such that F = Ek…E2E1(L(G))
= I. But, note that if A were a matrix such that each row sum were 0, and E any
elementary matrix, then EA is still a matrix with each row sum 0. Since each row
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Senior lecturer
2
Student
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sum of L(G) is 0, then, F = Ek…E2E1(L(G)) is a matrix with each row sum 0. However, F
= I, and so we have a contradiction.
3. Computing d (L(T)) when T is a tree
Let T be a tree and L(T) = [aij]
Proposition : If   Sn has a cycle of length  3 in its cycle decomposition, then,
a a …a = 0.
Proof: Let the cycle of length  3 be (l1 l2 … lk), k  3. Suppose a a …a
 0.
Then al1l2al2l3…alkl1  0. But since l1  l2, l2  l3, … , lk  l1, then li = -1 for every 1  i  k.
This implies that we can find cycle of length greater than or equal to 3 in the tree T. But
this is impossible since by definition, T does not have any cycles.  From this, we see
that the only permutations  that can give rise to a non-zero term a a …a
are
j k
those that have cycle structure (1 ,2 ) where j+2k = n and k is an integer from 0 to n/2
.This means we can drastically cut down the number of terms in the calculation of the
permanent of L(T). We can also calculate d L(T) via numbers obtained from T itself.
This involves the notion of matchings.
Definition: For a tree of order n, define for each integer k, k  0, MT(k) =
  deg(v) ,
Where v  Ek means a vertex that is not incident with any edge in Ek.
Note that when k = 0, MT(0) is the product of the degrees of the vertices in V.
Definition : For each integer k from 0 to n/2 , let A(k)  Sn, denote the conjugacy class
where each   A(k) has cycle structure (1j,2k) where j+2k = n. Let  (k) denote the
value of a irreducible character  on the conjugacy class A(k) and let A’ =
A(0)  …  A( n/2 ). (A conjugacy class is an equivalence class of the equivalence
relation xRy    Sn such that x =  y  -1)
Proposition:
Let  be an irreducible character of Sn.
Then,
d (L(T)) =   (k) MT(k)
and when  = (n),
d (L(T)) = per(L(T)) =

MT(k) ---(1)
We will only show (1). The general proof is similar.
Proof:
Note that per (L(T)) =
 a
=
=
 a
+ 0
=
 (  a
 a
+
 a
(follows from the earlier proposition in 3)
)
---(2)
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We first fix k, for some k from 1 to n/2 . We shall prove that   a = MT(k),
which upon substitution into (2) above, will give the result.
Define N(k) to be {   A(k) |  a
 0 }. Let H(k) be the set of all k-matchings of T.
Define f:N(k)  H(k) to be the function such that for each  , f(  ) = {[vi,v
 (i )  i}
Then f is bijective. And for each   N(k),  a
=  deg(v) .
Then
 a
=
]|
  deg(v) =MT(k), since as  runs through N(k), f(  ) runs
through H(k). Clearly
 a
=
 a
and we are done. 
Because (as will be shown later) MT(k) is even for all k when T is a tree of odd order,
then from the above proposition, d (L(T)) is even for all irreducible characters  of Sn.
This is however not necessarily true when for L(T) when T is a tree of even order, for
there may exist integers h such that MT(h) is odd.
4. Properties of MT(k)
Some properties are:-a) MT(n/2) = 0 or 1 when T is a tree of order n, n even,
b) MT(0)  MT(1) for all trees of order n > 1,
c)

MT(k) =

MT(k),
---(3)
d) MT(k) is even for all k, when the order of T is odd, and
e)

MT(k) is even for all trees T.
We will only show that MT(k) is even for all k, when T is a tree of odd order.
But to do this we will need a result on trees. Let n  1 . Let Y = (Vy,Ey) be a tree of order
n. In general, given any vertex q in Ey, we can form two trees Xq and Zq where
Xq has vertex set Vy  {u1,u2} and edge set Ey  { {u1,u2} ,{u1,q}}
Zq has vertex set Vy  {u1,u2} and edge set Ey  { {u1,q} ,{u2,q}}
Lemma: Given any tree T of order n+2 (n  1) , we can find a tree Y of order n, such that
T = Xq ,
or
T = Zq ,
for some vertex q of Y.
The proof consists of taking a maximal path in T, and then deleting two vertices
appropriately to form the required Y.
Definition: Let T be a tree of order n and vertex set V. Let k be an integer.
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HT,k is the set of all k-matchings in T.
LT,k = {Ek  HT,k | whenever u is an even vertex , u  Ek}
When T is a tree of odd order n, MT(1) is even. This can be shown, but will not be done
so here. We will need this later in the main inductive proof. Also, it is clear that MT(0) is
even, since there is always a vertex of even order in a tree of odd order. Then, since
MT(0) is the product of the degrees of all vertices, the product is even.
Theorem: When T is a tree of odd order, MT(k) is even for all k.
Proof: We just show that LT,k contains an even number of matchings, whence the result
is quite clear. We do this by induction.
Induction hypothesis:Let m  3, with m odd. Then for all trees of odd order n, 3  n  m,
and for every k with 1  k  n/2 , LT,k has even number of matchings.
Then, note that when m = 3, LT,1 has two 1-matchings. Also, since LT,1 always contains
an even number of 1-matchings, we are now trying to prove that LT,k contains an even
number of matchings when k  2.
Let T be a tree of order m+2, m  3 and m odd. Assume the inductive hypothesis is true.
Then from the above lemma, T = Xq or Zq for some vertex q in a tree Y of order m.
We will only show the case when T = Xq. Since the case T = Zq is similar.
We now suppose the degree of q in T is odd. For k  2, define two sets Q0 and Q1 as
follows.
Q0 = { Ek  LT,k | Ek = Ek-1  {u1,u2} for some Ek-1  LY,k-1},and
Q1 = { Ek  LT,k | Ek = Ek-1  {q,u1} or Ek = Ek-1  {u1,u2} },
where Ek-1  HY,k-1 is such that whenever u is even, u  Ek-1.
Then, when k  2, LT,k = Q0  Q1 and both Q0 and Q1 have an even number of matchings.
Suppose now that the degree of q in T is even. Then define for k  2,
Q = { Ek  LT,k | Ek = Ek-1  {q,u1} if Ek-1 leaves q uncovered
or
Ek = Ek-1  {u1,u2} if Ek-1 covers q }
Again Q = LT,k (k  2)and Q contains an even number of vertices by construction.

References
1 Hsiu –Khuern Tang, Generalized Matrix Functions, Honours Thesis, National Univ. of
Singapore 1995
2 Onn Chan and T.K Lam, Hook immanantal inequalities for Laplacians of trees,
Linear Algebra Appl. 261:23-47(1997)
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