LINEAR
On
ALGEBRA
AND
ITS
a Characterization
WERNER
APPLICATIONS
of Tridiagonal
C. RHEINBOLDT
AND
Computer Science Center
8, 87-90
Matrices by M. Fiedler*
ROGER A. SHEPHERD Department of Mathematics
University of Maryland
University of Maryland
College Park. Maryland
College Park, Maryland
Communicated
87
(1974)
by Hans Schneider
ABSTRACT In this note two new proofs are given of the following characterization M. Fiedler:
Let C,, n > 2, be the class of all symmetric,
with the property that rank (A + D) > n for any A E C, there exists a permutation and irreducible.
Recently,
theorem of
real matrices A of order n
1 for any diagonal real matrix D. Then matrix P such that PAPT
is tridiagonal
M. Fiedler [l] obtained the following characterization
of tridiagonal
of a class
matrices.
THEOREM.
Let C,, n > 2, be the class of all symmetric, real matrices A
of order n z&h the property
that rank(A
+ D) 3 n -
1 for any diagonal
real matrix D. Then for any A E C, there exists a permutation matrix P such that PAPT
is tridiagonal and irreducible.
The purpose of this note is to give two new proofs of this result which appear to be more direct than the proof by Fiedler. A preliminary must
and self-evident
be irreducible.
symmetric submatrix
Gaussian in C,.
The
remark
observation
elimination
More precisely,
applied
is that
central
any matrix
to both
to a matrix
A in C,
proofs
in Cnfl
is that
leads to a
for any index k, 1 < k < n + 1, we can
choose d, such that ulclc+ d, # 0, so this diagonal element
can be used as
*This research was supported in part by Grant GJ-106’7 from the National Science Foundation to the Computer Science Center of the University of Maryland. 0 American
Elsevier Publishing
Company,
Inc., 1974
88
WERNER
C. RHEINBOLDT
AND ROGER
A. SHEPHERD
the pivot for an elimination step. Since simultaneous permutation of the rows and columns does not lead out of Cnfl, it is no restriction to assume that k = 1 and that A E C,+l is partitioned as follows:
(1) all let
Here the irreducibility of A ensures that a # 0. Now with dr # a=-
1
(2)
all + dl ’ where I is then x n identity matrix. Then for any D = diag(d,, ds, we obtain L(A + D)LT = with
b =
all + dl o
diag(d,,. . ., d,+l).
.?4Lll)
B = A - aaaT,
(3)
From
n < rank(A + D) = rank L(A + D)LT = rank(B +
b) +
1,
it now follows that B E C,. This opens up the possibility of proving the theorem by induction on n. For n = 2 the statement is trivial, and it is useful to note that it also holds for n = 3. In fact, in that case the irreducibility requires that aij = aji = 0 for at most one index pair i # i. For, if aij # 0, i, i = 1, 2, 3, i # j, then for
4 =
-all
+-,
a21al3
a23
4 =
-az2
+-,
42923
d, = -a33
+F,
a13
the matrix A + diag(dl, d2, d3) has rank one and hence does not belong to cs. For Suppose that the statement holds for n > 3 and let A E C,,,. the first proof we assume that by simultaneous row and column permutations A has been brought into the form
aoT
=
b2,...,
al,+d,
alj #
0,
j =
2,...,m+
1,
(4)
TRIDIAGONAL
where
MATRICES
89
no row of A has less than m nonzero
the square matrices For any dr # -
off-diagonal
aI1 the matrix
^
of B,,
may assume that dr # nonzero. matrix
Bll
= A,,
elements.
B is an irreducible,
By irreducibility,
are at least m nonzero
off-diagonal
This implies that m < 2. at least m nonzero,
m = 2, for otherwise
A;
elements step;
off-diagonal
in one of the first m rows m 3 n -
n
x
n
with an edge between
matrix vertex
elimination
(3) is obtained
A, that
is, the graph j exactly
[2] on changes
step.
In terms
matrix. structure
G(A) on n if i # j and
in the structure
of our elimination
graph G(B) of the matrix
from G(A) by deleting
vertex
1 together
B
with all
with it and by adding an edge between any vertices i and i,
i # i, which were both adjacent The characterization path.
the case
irreducible
of the (undirected)
i and vertex
We use a result of Parter
due to a Gaussian
edges incident
This excludes
tridiagonal,
step (3), this result states that the structure in Eq.
2 > 0 rows of A
by choice of m, they
at best one row of B has less than two such elements.
of a symmetric
aij = aj, # 0.
2 rows with two, nonzero
elements.
The second proof is based on the notion vertices
of B1, is
permuted tridiagonal
therefore,
Thus, m = 1 and A is indeed a permuted graph
T.
is not zero, and hence there
The last n -
were not changed in the elimination contain
ccaOa,,
can be zero only for a single value of dr, we
and thus has two rows with one, and n -
of B.
graph
-
all was chosen so that every element
By induction hypothesis,
off-diagonal
and
m, respectively.
B in Eq. (3) then has the form
,
Since an element
elements,
and AZ2 have order m and n -
A,,
theorem
to vertex states
We proceed again by induction
1 in G(A).
that for A E C,, G(A) must be a
starting
from the earlier observation
that the result holds for n = 2, 3. Moreover, we use once more that for any A E Gn+r a symmetric produces a submatrix
Gaussian
elimination
B for which-by
step with any diagonal
induction
hypothesis-G(B)
pivot is a
path. Observe degree
first that
greater
than
for A E Cn+l, 12 > 3, no vertex two.
vertices i, j, 1 were adjacent
In fact,
some d, # -
a ,,-contains
hence is not a path. Since the irreducibility
that
in G(A)
three
to the vertex k. Then the elimination
aRk + d, # 0 as pivot produces -for
suppose
of G(A) can have
a resulting
structure
a three-cycle
Thus, all vertices
graph
distinct step with
G(B) which
with vertices
i, j, 1, and
of G(A) have degree at most two.
of A is equivalent
to the connectedness
of G(A),
WERNER
90
C. RHEINBOLDT
AND ROGER
this implies that G(A) is either a path or an (n + 1) cycle. case an elimination structure
step on any vertex would produce
graph G(B) which contradicts
B E C,.
A. SHEPHERD
In the latter
an n-cycle
as the
Thus, G(A) must be a
path and the proof is once more complete. REFERENCES 1 M. Fiedler, A characterization
of tridiagonal
matrices, Lin. Alg. and A@Z. 2(1969),
191-197. 2 S. Parter, Use of linear graphs in Gauss elimination, Received November,
197 1
SIAM
Rev. 3(1961), 119-130.