It is possible to bound the number of connected components of the positive and negative supports of an eigenvector of a graph.
Proof:
For a subset
of the vertex set of
, let
be the diagonal matrix with ones on the positions indexed by elements
of
and zeros elsewhere.
Let
run through the connected components of
and
. Put
.
Then the space
has dimension
.
For simplicity, replace
by the adjacency matrix of
,
and then replace
by
.
Define a real symmetric matrix
by
.
Then
has zero row sums and nonpositive off-diagonal entries,
so
is positive semidefinite and by Lemma 1.13.2
the number of zero eigenvalues equals
.
It follows that for
we have
.
This means that
intersects the space
spanned by the eigenvectors with negative eigenvalue in 0,
and
follows.
The vectors
with
correspond to eigenvectors
with eigenvalue 0 of
, and altogether there are
independent such. This proves the other inequality.
Examples
(i) Let
be connected and bipartite, and let
be the smallest
eigenvalue of
. The corresponding eigenvector
has different
signs on the two sides of the bipartition, so
and
are the two sides of the bipartition,
is the total number of vertices
and
. We have equality in both inequalities.
(ii) Let
be the star
. The spectrum is
,
,
. Let
be an eigenvector
with eigenvalue
. We can choose
to be nonzero off the
apex of the star. Then
and
,
, again equality in both inequalities.
(iii) Let
be the Petersen graph. It has spectrum
,
,
. Let
be an eigenvector with eigenvalue
that vanishes on 4 points, so that
induces
with spectrum
,
. We find
and
,
, again equality in both inequalities.
(iv) Let
be the path of length
(with
vertices).
The eigenvalues are
for
. The eigenvector
corresponding to
has
sign changes, so that
.
If
then
has no zero entries, so that
.
Now we have equality in both inequalities.
If
, then
has
zero entries, so that
.
Also, the eigenvalue
is the
-th of each component
of
, so that
and
, again with equality
in both inequalities.
Remark
By interlacing, the numbers
and
for
are upper bounded by the corresponding numbers for
,
but it is not true that
.
For example, in case (ii) above we have
but the multiplicity of
in the spectrum of
is
.