The Electronic Journal of Linear Algebra. A publication of the International Linear Algebra Society. Volume 7, pp. 1-20, February 2000. ELA

ISSN 1081-3810. http://math.technion.ac.il/iic/ela

THE GENERAL TOTALLY POSITIVE MATRIX COMPLETION



PROBLEM WITH FEW UNSPECIFIED ENTRIES

y z x

SHAUN M. FALLAT , CHARLES R. JOHNSON , AND RONALD L. SMITH

Abstract. For m-by-n partial totally p ositive matrices with exactly one unsp eci ed entry,

the set of p ositions for that entry that guarantee completability to a totally p ositive matrix are

characterized. They are the p ositions i; j , i + j  4 and the p ositions i; j , i + j  m + n 2. In

each case, the set of completing entries is an op en and in nite in case i = j =1 or i = m, j = n

interval. In the pro cess some new structural results ab out totally p ositive matrices are develop ed.

In addition, the pairs of p ositions that guarantee completability in partial totally p ositive matrices

with two unsp eci ed entries are characterized in low dimensions.

Key words. totally p ositive matrices, matrix completion problems, partial matrix.

AMS sub ject classi cations. 15A48

1. Intro duction. An m-by-n matrix A is called total ly positive, TP, total ly

nonnegative, TN, if every minor of A is p ositive nonnegative. A partial matrix is a

rectangular array in which some entries are sp eci ed, while the remaining, unsp eci ed,

entries are free to b e chosen from an indicated set e.g., the real eld. A completion

of a partial matrix is a choice of allowed values for the unsp eci ed entries, resulting

in a conventional matrix, and a matrix completion problem asks which partial ma-

trices have a completion with a particular desired prop erty. The TP, TN matrices

arise in a varietyofways in geometry, combinatorics, algebra, di erential equations,

function theory, etc., and have attracted considerable attention recently [1,7,9,17].

There is also now considerable literature on a variety of matrix completion problems

[3, 4, 5, 6, 8, 12, 13, 14, 15]. Since total p ositivity and total nonnegativity are in-

herited by arbitrary submatrices, in order for a partial matrix to have a TP TN

completion, it must b e partial ly TP TN, i.e., every fully sp eci ed submatrix must b e

TP TN. Study of the TN completion problem was b egun in [14], where the square,

combinatorially symmetric patterns for which every partially TN matrix has a TN

completion were characterized under a regularity condition that has b een removed

in [6].

Here, we b egin the study of the TP completion problem, in the rectangular case



Received by the editors on 13 July 1999. Accepted for publication on 10 January 2000. Handling

Editor: Ludwig Elsner.

y

Department of Mathematics and Statistics, University of Regina, Regina, Saskatchewan, S4S

0A2, CANADA. The work of this author was b egun while as a Ph.D candidate at the College of

William and Mary [email protected].

z

Department of Mathematics, College of William and Mary, Williamsburg, VA. 23187-8795, USA

[email protected]. Research supp orted in part by the Oce of Navel Research contract

N00014-90-J-1739 and NSF grant 92-00899.

x

Department of Mathematics, UniversityofTennessee at Chattano oga, Chattano oga, TN 37403-

2598, USA [email protected]. The work of this author was b egun while on sabbatical at the

College of William and Mary, supp orted by the University of Chattano oga Foundation. The research

was completed under a Summer Research grant from the UniversityofTennessee at Chattano oga's

Center of Excellence in Computer Applications. 1

ELA

2 S.M. Fallat, C.R.. Johnson and R.L. Smith

without any combinatorial symmetry assumption. It should b e noted that there are

imp ortant di erences b etween the TP and TN completion problems. In the former,

there are stronger requirements on the data, but also much stronger requirements on

the completion. In practice, this di erence is considerable, and for example, results

analogous to those in [14] are not yet clear in the TP case. Here in the TP case we

again fo cus on patterns, and ask for which patterns do es every partial TP matrix have

a TP completion. In general, this app ears to b e a substantial and subtle problem. We

give a complete answer for m-by-n patterns with just one unsp eci ed entry, and even

here the answer is rather surprising and nontrivial. If min fm; ng 4, then only 12

individual p ositions ensure completability: the upp er left and lower right triangles of 6

entries each. If minfm; ng =1; 2 or 3, then any single p osition ensures completability.

In order to prove these results several classical and some new e.g., ab out TP linear

systems facts ab out TP matrices are used. Patterns with two unsp eci ed entries are

also considered. An inventory of completable pairs when m; n  4, and some general

observations are given. However, the general question of two unsp eci ed entries is

shown to b e subtle: examples are given to show that the answer do es not dep end just

on the \relative" p ositions of the twoentries.

The submatrix of an m-by-n matrix A lying in rows and columns , M =

f1; 2 :::;mg, N = f1; 2 :::;ng, is denoted by A[ ; ]. Similarly, A ;  denotes

the submatrix obtained from A by deleting the rows indexed by and columns indexed

by . When m = n, the principal submatrix A[ ; ] is abbreviated to A[ ], and the

complementary principal submatrix is denoted by A . As usual, for N we let

c

denote the complementof relativeto N . An m-by-n matrix is TN resp. TP 

k k

for 1  k  minm; n, if all minors of size at most k are nonnegative resp. p ositive.

It is well known that if A =[a ]isanm-by-n TP TN  matrix, then the m-by-n

ij k k

~

matrix de ned by A =[a ] is also a TP TN  matrix. Wemay refer to

mi+1;nj +1 k k

~

the matrix A =[a ] as b eing obtained from A by reversing the indices of

mi+1;nj +1

A.

Wenow present some identities, discovered by Sylvester see [10], for complete-

ness and clarity of comp osition. Let A be an n-by-n matrix, N , and supp ose

c

j j = k . De ne the n k -by-n k  matrix B =[b ], with i; j 2 , by setting

ij

c

b = det A[ [fig; [fj g], for every i; j 2 . Then Sylvester's identity states that

ij

c

for each ;  , with j j = j j = l ,

l1

1 detB [; ] = det A[ ] detA[ [ ; [ ]:

nk 1

Observe that a sp ecial case of 1 is detB = det A[ ] detA. Another very useful

sp ecial case which is employed throughout is the following. Let A be an n-by-n matrix

partitioned as follows

3 2

T

a a a

11 13

12

5 4

a A a

; A =

21 22 23

T

a a a

31 33

32

where A is n 2-by-n 2 and a ;a are scalars. De ne the matrices

22 11 33

ELA

Totally Positive Completions 3

   

T T

a a a a

11 13

12 12

B = ; C = ;

a A A a

21 22 22 23

   

a A A a

21 22 22 23

D = ; E = :

T T

a a a a

31 33

32 32

~ ~

If we let b = det B ,~c = det C , d = detD , ande ~ = det E , then by 1 it follows that

 

~

b c~

= detA detA: det

22

~

d e~

Hence, provided det A 6=0,wehave

22

detB det E det C detD

detA = 2 :

detA

22

Wenow state an imp ortant result due to Fekete see, e.g., [1], who givesavery

useful and app ealing criterion or characterization of totally p ositive matrices. First

we need to de ne the notion of the disp ersion of an index set. For = fi ;i ;:::;i g

1 2 k

N , with i

1 2 k

k 1

X

i i 1 = i i k 1; with the convention that d  = 0 when is

j +1 j k 1

j =1

a singleton. The disp ersion of a set represents a measure of the \gaps" in the set

. In particular, observe that d  = 0 whenever is a contiguous i.e., an index set

based on consecutive indices subset of N .

Theorem 1.1. Fekete's Criterion An m-by-n matrix A is total ly positive if and

only if det A[ ; ] > 0, for al l f1; 2;:::;mg and f1; 2;:::;ng, with j j = j j

and d =d =0. In other words a matrix is totally p ositive if and only if the

determinant of every square submatrix based on contiguous row and column index

sets is p ositive.

The next lemma whichmay b e of indep endentinterest proves to b e very useful

for certain completion problems for TP matrices, and is no doubt useful for other

issues with TP matrices.

Lemma 1.2. Let A =[a ;a


;a ] be an n 1-by-n TP matrix. Then, for

1 2 n

k =1; 2;


;n;

n

X

a = 3 y a ;

k i i

i=1

16=k

has a unique solution y for which



i

sg n1 ; if k is o dd

sg ny =

i

i1

sg n1 ; if k is even:

ELA

4 S.M. Fallat, C.R.. Johnson and R.L. Smith

Proof. If k = 1, 3 has solution

1

y =[a ;a ;


;a ] a

2 3 n 1

2 3

det [a ;a ;a ;


;a ]

1 3 4 n

6 7

det [a ;a ;a ;


;a ]

2 1 4 n

1

6 7

=

6 7

.

.

det[a ;a ;


;a ] 4 5

2 3 n

.

det[a ;a ;


;a ;a ]

2 3 n1 1

i

and sg ny =sg n1 . If k>1, then 3 has solution

i

1

y =[a ;a ;


;a ;a ;


;a ] a

1 2 k 1 k +1 n k

3 2

det[a ;a ;a ;


;a ;a ;


;a ]

k 2 3 k 1 k +1 n

7 6

det[a ;a ;a ;


;a ;a ;


;a ]

1 k 3 k 1 k +1 n

7 6

7 6 .

.

7 6

.

7 6

7 6

.

.

7 6

.

1

7 6

=

7 6

det[a ;a ;


;a ;a ;a ;


;a ]

1 2 k 2 k k +1 n

7 6

det[a ;


;a ;a ;


;a ]

1 k 1 k +1 n

7 6

det[a ;a ;


;a ;a ;a ;


;a ]

1 2 k 1 k k +2 n

7 6

7 6

det[a ;a ;


;a ;a ;a ;a ;


;a ]

1 2 k 1 k +1 k k +3 n

7 6

7 6

.

.

5 4

.

det[a ;a ;


;a ;a ;


;a ;a ]

1 2 k 1 k +1 n1 k

i

and we see that if k is o dd, sg ny  = sg n1 while if k is even, sg ny  =

i i

i1

sg n1 :

2. Main Results. In this section we consider the TP completion problem for

m-by-n patterns with just one unsp eci ed entry. We b egin with a preliminary lemma

which is used throughout the remainder of the pap er.

Lemma 2.1. Let

2 3

x a


a

12 1n

6 7

a a


a

21 22 2n

6 7

A = ;

6 7

. . .

. . .

4 5

. . .

a a


a

n1 n2 nn

n  2; bepartial TP in which x is the only unspeci ed entry. Then, if x is chosen so

that det A  0, det A[f1;


;kg] > 0, k =1;


;n 1: In particular, x>0.

Proof. The pro of follows inductively using Sylvester's identity 2 for determi-

nants applied to the leading principal minors of A in decreasing order of size. For

example,

0  detAdetA[f2; 3;:::;n 1g]

= detA[f1; 2;:::;n 1g]detA[f2; 3;:::;ng]

detA[f1; 2;:::;n 1g; f2;:::;ng]detA[f2;:::;ng; f1; 2;:::;n 1g];

ELA

Totally Positive Completions 5

and since A is partial TP wehave that detA[f1; 2;:::;n 1g] > 0. If A is replaced by

A[f1; 2;:::;n 1g] in the ab ove identity, then it follows that detA[f1; 2;:::;n 2g] > 0,

etc.

Theorem 2.2. Let A be an n-by-n partial TP matrix, n  3, with x, the only

unspeci ed entry, lying in the s; t position. If s + t  4 or s + t  2n 2, then A is

completable to a TP matrix.

Proof. Throughout the pro of A = Ax=A x will denote an n-by-n partial TP

n

matrix with exactly one unsp eci ed entry x in the s; t p osition. For k =1;


;n 1;

let

A x=A[f1;


;kg];

k

B x=A[f1;


;kg; f2;


;k +1g];

k

C x=A[f2;


;k +1g; f1;


;kg]; and

k

D x=A[f2;


;k +1g].

k

Case 1 s; t=1; 1: Every minor which do es not involve A[1; 1] is p ositiveby

assumption. Furthermore A[1; 1] enters p ositively into every minor in which it is

involved. Hence, by making A[1; 1] large enough, we will obtain a TP completion of

A.

Case 2 s; t=1; 2: Let F = A[f2;


;n 1g; f2;


;ng]=[f ;f ;


;f ]:

1 2 n1

P

n1

i

By Lemma 1.2 f = y f in which sg ny  = sg n1 . Let B x  =

1 i i i n1 B

i=2

P

n1

[b ;b ;


;b ] in which x is chosen so that b = y b : Thus, det B x =

1 2 n1 B 1 i i n1 B

i=2

0 and, by Lemma 2.1, det B x  > 0;k =1; 2;


;n 2; in particular, x > 0. So

k B B

det A x  = det[b ;b ;b ;


;b ]

n1 B 0 1 2 n2

n1

X

= det[b ; y b ; b ;


;b ]

0 i i 2 n2

i=2

= det[b ;y b ;b ;


;b ]

0 n1 n1 2 n2

n3

=1 y det [b ;b ;


;b ;b ]

n1 0 2 n2 n1

= jy j det[b ;b ;


;b ;b ]

n1 0 2 n2 n1

> 0:

Applying Sylvester's identity 2 for determinants, we obtain det A>0 and we can

continue to apply this identity to obtain:

det A x  > 0; k =2;


;n 2:

k B

We can then increase x so as to make det B x > 0 and obtain a TP completion

B n1

of A.

Case 3 s; t=1; 3: Notice that it suces to show that there is x>0 such that

eachof

A x; k =3;


;n;

k

B x; k =2;


;n 1; and

k

G x=A[f1;


;kg; f3;


;k +2g], k =1;


;n 2; k

ELA

6 S.M. Fallat, C.R.. Johnson and R.L. Smith

has p ositive determinant since the determinants of these matrices comprise all of

the contiguous minors of A involving x. By Theorem 4.7 in the app endix there

exists an x > 0 such that B x 2 TP is singular. Therefore, since B x,

n1 n2 k

k =2;


;n 2, and G x, k =1;


;n 2, are submatrices of B x wehave

k n1

det B x > 0, k =2;


;n 2; and det G x > 0, k =1;


;n 2: Let

k k

F = A[f2;


;n 1g; f2;


;ng]= [f ;f ;


;f ]

1 2 n1

n1

X

y f in which sg ny  = and B x = [b ;b ;


;b ]: By Lemma 1.2, f =

i i i n1 1 2 n1 2

i=1

i6=2

i1

sg n1 and, since B x is singular and b ;b ;


;b are linearly indep en-

n1 1 3 n1

n1

X

dent, b = y b : Thus, if A x=[b ;b ;b ;


;b ];

2 i i n1 0 1 2 n2

i=1

i6=2

n1

X

det A x = det [b ;b ; y b ;b ;


;b ]

n1 0 1 i i 3 n2

i=1

i6=2

= det [b ;b ;y b ;b ;


;b ]

0 1 n1 n1 3 n2

n4

=1 y det[b ;b ;b ;


;b ;b ]

n2 0 1 3 n2 n1

= jy j det [b ;b ;b ;


;b ;b ]

n2 0 1 3 n2 n1

> 0:

By Sylvester's identity 2, it follows that det A = det A x > 0 and det A x > 0,

n k

k = n 2;


; 3: By decreasing x, det B x b ecomes p ositive and we obtain a TP

n1

completion of A.

Case 4 s; t=2; 2: Notice that it suces to show that there is an x>0 such

that eachof

A x, k =2;


;n;

k

B x, k =2;


;n 1;

k

C x, k =2;


;n 1; and

k

D x, k =2;


;n 1;

k

has p ositive determinant since the determinants of each of these submatrices comprise

all of the contiguous minors involving x. By Theorem 4.7, there exists an x > 0

B

resp. x > 0 such that B x  resp. C x  is singular and lies in TP .

C n1 B n1 C n2

Without loss of generality assume x = minfx ;x g. Thus, det B x  > 0,

B B C k B

k =2;


;n 2; and det C x   0. Since

n1 B

 

 

B x = ;

n1 B

D x  

n2 B

wehave that det D x  > 0 and hence, by Lemma 2.1, det D x  > 0,

n2 B k B

k =1;


;n 2. Now

   

C x    

k 1 B

C x = = ;

k B

  D x  

k 1 B

ELA

Totally Positive Completions 7

k = n 1;


; 3: So we can apply Sylvester's identity 2 rep eatedly to obtain

det C x  > 0, k =2;


;n 2.

k B

We next show that det A x  > 0, k =2;


;n. Let

k B

F = A[f1; 3;


;n 1g; f2;


;ng]= [f ;f ;


;f ] and let

1 2 n1

B x =Ax [f1; 2;:::;n 1g; f2; 3;:::;ng]=[b ;b ;


;b ]:

n1 B B 1 2 n1

P

n1

i

y f in which sg ny =sg n1 and since B x is By Lemma 1.2, f =

i i i n1 B 1

i=2

P

n1

singular and b ;b ;


;b are linearly indep endent, b = y b : Since

2 3 n1 1 i i

i=2

A x =[b ;b ;


;b ]wehave

n1 B 0 1 n2

n1

X

det A x  = det[b ; y b ;b ;


;b ]

n1 B 0 i i 2 n2

i=2

= det[b ;y b ;b ;


b ]

0 n1 n1 2 n2

n3

=1 y det [b ;b ;


;b ;b ]

n1 0 2 n2 n1

= jy j det[b ;b ;


;b ;b ]

n1 0 2 n2 n1

> 0:

It then follows from Sylvester's identity 2 that det Ax  = det A x  > 0

B n B

and since, for k = n 1;


; 3,

       

A   B    

k 1 k 1

A = = = = ;

k

    C   D

k 1 k 1

we can rep etitively apply Sylvester's identity 2 to obtain det A x  > 0,

k B

k =2;


;n 1.

To show that det D x  > 0, pro ceed as follows. Let

n1 B

2 3

H

1

6 7

H

2

6 7

H = A[f1;


;n 1g; f3;


;ng]= :

6 7

.

.

4 5

.

H

n1

n1

X

i1

By the rowversion of Lemma 1.2, H = y H in which sg ny  = sg n1 .

2 i i i

i=1

i6=2

2 3

1

6 7

2

6 7

Since B x = and ; ;


; are linearly indep endent,

6 7

.

n1 B 1 3 n1

.

4 5

.

n1

n1

X

= y . Now

2 i i

i=1

i6=2

ELA

8 S.M. Fallat, C.R.. Johnson and R.L. Smith

2 3

2

6 7

3

6 7

6 7

.

.

D x =

6 7

n1 B

.

6 7

4 5

n1

n

and hence

2 3

P

3 2

n1

3 2

y i

i=1

y

i

1 1

i6=2

6 7

1

7 6

6 7

3

7 7 6 6 3

6 7

3

7 7 6 6

.

6 7

.

.

> 0: = y det det D x  = det = det

7 7 6 6

.

1 n1 B

.

.

6 7

.

.

7 6

5 4

.

6 7

5 4

4 5

n1

n1

n

n

n

Nowby decreasing x det B x and det C x b ecome p ositive and we obtain a

B n1 n1

TP completion of A. If s; t=2; 1 or 3,1, then A is completable to a TP matrix

by reversing the indices. Thus, A is completable to a TP matrix if s + t  4. And by

reversing the indices A is completable to a TP matrix if s + t  2n 2.

Lemma 2.3. Let A beanm-by-n TP matrix. Then there exists a positive vector

n

x 2 R such that the augmented matrix [Ajx] is an m-by-n +1 TP matrix. Similarly,

h i

 

A w

there exist positive vectors u; v ; w such that [ujA], , and are al l TP.

v A

n

T

Proof. Consider a vector x = x ;


;x  2 R and the augmented matrix

1 n

[Ajx]. The idea of the pro of is to cho ose the x 's sequentially so that

i

[Ajx][f1;


;ig; f1;


;n +1g] is TP. Cho ose x > 0 arbitrarily. Cho ose x > 0

1 2

large enough so that [Ajx][f1; 2g; f1;


;n+1g]isTP. This is p ossible since x enters

2

p ositively into each minor that contains x . Continue this argument inductively. The

2

latter part follows by analogous arguments.

Lemma 2.4. Let A be an m-by-n partial TP matrix. Then there exist positive

h i

 

w A

areallpartial TP. vectors x; u; v ; w such that [Ajx], [ujA], , and

A v

Proof. As b efore the idea is to cho ose the x 's sequentially. First cho ose x > 0

i 1

arbitrarily. Cho ose x > 0 large enough so that all fully sp eci ed minors that involve

2

x are p ositive. Observe that suchanx exists by Lemma 2.3. Continue this argument

2 2

i h

 

w A

, and follow analogously. inductively. The cases [ujA],

A v

Theorem 2.5. Let A beanm-by-n partial TP matrix with 3  m  n and with

x, the only unspeci ed entry, lying in the s; t position. If s + t  4 or s + t  m + n 2,

then A is completable to a TP matrix.

Proof. If s + t  4, we can apply Lemma 2.4 and obtain an n-by-n partial TP

i h

A

in which the unsp eci ed entry x lies in the s; t p osition where matrix A =

1

B

s + t  4. If s + t  m + n 2, we can apply Lemma 2.4 and obtain an n-by-n partial

i h

C

TP matrix A = in which the unsp eci ed entry x lies in the s; t p osition where

2 A

ELA

Totally Positive Completions 9

s + t  2n 2: In either case we can apply Theorem 2.2 to obtain a completion of A

1

or A and hence one of A itself.

2

Theorem 2.6. Let A bea 2-by-n partial TP matrix with exactly one unspeci ed

entry. Then A is completable to a TP matrix.

Proof. By reversing the indices wemay assume without loss of generality that

 

a a


a x a


a

1 2 k 1 k +1 n

: A =

b b


b b b


b

1 2 k 1 k k +1 n

a

a j

i

Since A is partial TP it follows that > for i< j and distinct from k . Then, if

b b

i j

wecho ose x such that a =b > x=b >a =b , the resulting completion of A

k 1 k 1 k k +1 k +1

is TP.

Lemma 2.7. Let A be a 3-by-4 partial TP matrix with exactly one unspeci ed

entry. Then A is completable to a TP matrix.

This result simply follows from Theorem 2.5.

Theorem 2.8. Let A bea 3-by-n partial TP matrix with exactly one unspeci ed

entry, n  3. Then A is completable to a TP matrix.

Proof. By Fekete's criterion see Theorem 1.1 it suces to prove the theorem

for n = 5: To see this, supp ose the unsp eci ed entry in A is in column k . If a

value for the unsp eci ed entry can be chosen so that A[f1; 2; 3g; fk 2;k 1;kg],

A[f1; 2; 3g; fk 1;k;k +1g], and A[f1; 2; 3g; fk; k +1;k +2g] if k =1; 2;n 1, or

n, then consider only the submatrices ab ove that apply are all TP, then it follows

that every contiguous minor of A is p ositive. Thus, byFekete's criterion, A is TP.

Observe that in this case it was enough to complete the submatrix of A, namely

A[f1; 2; 3g; fk 2;k 1;k;k +1;k +2g], consisting of at most 5 columns. Hence,

we can assume n = 5. So supp ose A is a 3-by-5 partial TP matrix with exactly

one unsp eci ed entry x. By Theorem 2.5 and Lemma 2.7, we need only consider the

case in which x lies in the 2,3 p osition. In this case there is x > 0 such that

m

Ax [f1; 2; 3g; f2; 3; 4g] is singular and TP by decreasing its 2,2 entry whichis

m 2

the 2,3 entry of A, see Theorem 4.3. Let F = A[f1; 3g; f2; 3; 4g] = [f ;f ;f ].

1 2 3

Then f = y f + y f where sg ny  = sg ny  is p ositive by Lemma 1.2. Thus,

2 1 1 3 3 1 3

Ax [f1; 2; 3g; f2; 3; 4g]= [g ;g ;g ] in which g = y g + y g and so

m 1 2 3 2 1 1 3 3

det Ax [f1; 2; 3g] = det[g ;g ;g ] = det[g ;g ;y g + y g ]

m 0 1 2 0 1 1 1 3 3

= y det [g ;g ;g ] > 0:

3 0 1 3

Similarly,

det Ax [f3; 4; 5g] = det[g ;g ;g ]=y det [g ;g ;g ] > 0:

m 2 3 4 1 1 3 4

Since these two 3-by-3 minors and all 2-by-2 minors are p ositive, we can increase x

m

and obtain a TP completion of A.

Lemma 2.9. Thereisa4-by-4 partial TP matrix with the 1; 4 or 4; 1 entry

unspeci ed that has no TP completion.

Proof. Consider the partial TP matrix

ELA

10 S.M. Fallat, C.R.. Johnson and R.L. Smith

2 3

100 100 40 x

6 7

40 100 100 40

6 7

A = :

4 5

20 80 100 100

3 20 40 100

In order for the determinant itself to b e p ositive, wemust have x<1144=14. Since

wemust also have x>0, we see that A is not completable to a TP matrix.

Lemma 2.10. Thereisa4-by-4 partial TP matrix with the 2; 3 or 3; 2 entry

unspeci ed that has no TP completion.

Proof. Consider the partial TP matrix

3 2

1000 10 10 10

7 6

20 9 x 10

7 6

A = :

5 4

2 1 9 10

1 2 40 1000

For det A[f1; 2g; f3; 4g] to b e p ositivewemust have x<1. But for

det A[f2; 3; 4g; f1; 2; 3g] to b e p ositivewemust have x>199=3. Hence A has no TP

completion.

Theorem 2.11. Main Result Let A beanm-by-n partial TP matrix in which

4  m  n and in which the only unspeci ed entry lies in the s,t position. Any such

A has a TP completion if and only if s + t  4 or s + t  m + n 2.

Proof. Suciency follows from Theorem 2.5. To prove necessity assume that

5  s + t  m + n 3. Notice that if an m-by-n matrix has this prop erty, then

there is a 4-by-4 submatrix such that the s; t p osition of the matrix corresp onds to

the 1,4, 4,1, 2,3, or 3,2 p osition of that submatrix. Without loss of generality,

assume that the s; t p osition of the matrix corresp onds to the 1,4 p osition of the

submatrix. By Lemma 2.9 there is a 4-by-4 partial TP matrix with its only unsp eci ed

entry in the 1,4 p osition that has no TP completion. Using Lemma 2.4 and by the

preceding remarks, we can augment this matrix with p ositiverows/columns so that

we obtain an m-by-n partial TP matrix A whose only unsp eci ed entry lies in the

s; t p osition. Since the original matrix is a partial submatrix of A that has no TP

completion, A itself has no TP completion.

3. Further Discussion. For 3  m; n  4 we consider partial m-by-n TP

matrices with exactly two unsp eci ed entries. We note that completing a 2-by-n

partial TP matrix with two unsp eci ed entries follows easily from Theorem 2.6. In

the case when min fm; ng = 3 all such completable patterns have b een characterized

in [3]. For example, all partial 3-by-3 TP matrices with exactly two unsp eci ed entries

are completable except for the following four patterns:

2 3 2 3 2 3 2 3

x ? x x x x x ? x x x x

4 5 4 5 4 5 4 5

? x x x x ? x x ? ? x x

; ; ; and :

x x x x ? x x x x x ? x

ELA

Totally Positive Completions 11

Similarly, when m = 3 and n = 4, Table 1 contains the p ositions of the pairs of

unsp eci ed entries for which there is no completion in general. Observe that it is

sucient to only consider the six entries 1,1 through 2,2 since the conclusions for

the remaining six entries follows by reversal of indices; see also [3].

1,1 None

1,2 2,1, 2,3, 3,1, 3,3

1,3 2,2, 2,4

1,4 2,3, 3,3

2,1 1,2, 3,2

2,2 1,3, 3,1, 3,3

Table 1

We b egin the 4-by-4 case with some observations that aid in the classi cation of

the completable and hence non-completable pairs of unsp eci ed entries. Supp ose

A is a partial TP matrix with exactly two unsp eci ed entries in p ositions p; q  and

s; t. The rst observation concerns the case when either p = s or q = t, i.e., the

unsp eci ed entries o ccur in the same row or the same column. In this case A is always

completable to a TP matrix. The idea of the pro of is as follows. By Theorem 2.11,

we know that the only unsp eci ed p ositions in a 4-by-4 partial TP matrix that do

not guarantee a TP completion are 1,4, 2,3, 3,2, and 4,1. Since either p = s or

q = t at least one of the p ositions p; q ors; t is not among the list of four \bad"

p ositions listed ab ove. Without loss of generality assume that p = s. Then delete

the column that corresp onds to the entry that is not in the ab ove list, and complete

the remaining 4-by-3 partial TP matrix to a TP matrix which follows from Theorem

2.8. Observe that now A is a partial TP matrix with only one unsp eci ed entry.

Since this remaining unsp eci ed entry is in a \go o d" p osition, it follows that we can

complete A to a TP matrix.

The next observation deals with the case when one of the unsp eci ed entries is in

the 1,1 or 4,4 p osition and the other unsp eci ed entry is in a \go o d" p osition,

i.e., satis es one of the conditions in Theorem 2.11. Since the 1,1 or 4,4 entry

enters p ositively into every minor that it is contained in, it follows that we can sp ecify

a large enough value for this entry so that A is a partial TP matrix with only one

unsp eci ed entry. Then A may be completed to TP matrix since the remaining

unsp eci ed entry is in a \go o d" p osition. We note here that this observation can b e

extended in the general case as follows. If A is an m-by-n partial TP matrix with

exactly two unsp eci ed entries with one in the 1,1 or m; n p osition and the other

in a \go o d" p osition as de ned by Theorem 2.11, then A is completable to a TP

matrix. The pro of in this general case is identical to the pro of in the 4-by-4 case.

Finally, supp ose the unsp eci ed entries o ccur in p ositions p; q  and s; t where

either jp sj =3, or jq tj =3. Then A is completable to a TP matrix except in

the case when p; q  = 1; 4 and s; t = 4; 1 or vice-versa. First assume that

ELA

12 S.M. Fallat, C.R.. Johnson and R.L. Smith

p; q =1; 4 and s; t=4; 1, and consider the following example. Let

3 2

1 1 :4 x

7 6

:4 1 1 :4

7 6

; Ax; y =

5 4

:2 :8 1 1

y :2 :4 1

see also [14]. Then it is easy to check that every fully sp eci ed minor of Ax; y  is

p ositive. However,

det Ax; y =:0016 :008x :328y :2xy ;

Hence detAx; y  < 0 for x and y p ositive. On the other hand assume that p; q  6=

1; 4 or s; t 6=4; 1. Then since either jp sj =3,or jq tj = 3 it follows that at

least one of the unsp eci ed entries is in a \go o d" p osition. Without loss of generality

assume that q  t and that q t =3. Hence q = 4 and t =1. Now delete the column

of the unsp eci ed entry that is in a go o d p osition. By Theorem 2.8, wemay complete

the remaining 4-by-3 matrix to a TP matrix. Now consider the 4-by-4 matrix A with

only one unsp eci ed entry. If we can show that A is a partial TP matrix, then we

are done, since the remaining unsp eci ed entry is in a \go o d" p osition. The reason

that A with only one unsp eci ed entry is partial TP is b ecause since jq tj = 3 all

of the fully sp eci ed minors of A based on contiguous index sets are p ositive. Hence,

byFekete's criterion, A is a partial TP matrix.

Using the ab ove three observations the pairs of p ositions for the unsp eci ed entries

not ruled out are contained in Table 2.

1,1 2,3, 3,2

1,2 2,1, 2,3, 2,4, 3,1, 3,3, 3,4

1,3 2,1, 2,2, 2,4, 3,1, 3,2, 3,4

1,4 2,2, 2,3, 3,2, 3,3

2,2 1,3, 1,4, 3,1, 3,3, 3,4, 4,1, 4,3

2,3 1,1, 1,2, 1,4, 3,1, 3,2, 3,4, 4,1, 4,2, 4,4

Table 2

As b efore we only have to consider the six entries: 1,1, 1,2, 1,3, 1,4, 2,2,

and 2,3, since the class TP is closed under transp osition and reversal of indices. In

fact the list in Table 2 can b e even further re ned as some of the ab ove cases follow

from others by transp osition and/or reversal of indices; see Table 3.

As we shall see all but one of the ab ove namely, the pair f1,4, 3,2g pairs are

not in general completable. We b egin by rst ruling out the pairs that follow from the

pairs of non-completable entries in the 3-by-3 and 3-by-4 cases. For example, supp ose

the unsp eci ed entries are in the 1,2 and 2,1 p ositions. There exists a 3-by-3

TP matrix with unsp eci ed entries in those p ositions that has no TP completion,

2 3

3 x 2

4 5

y 3 1

e.g., . We can b order this 3-by-3 partial TP matrix by Lemma 2.4 to

3 4 3

ELA

Totally Positive Completions 13

1,1 2,3

1,2 2,1, 2,3, 2,4, 3,1, 3,3, 3,4

1,3 2,2, 2,4, 3,1, 3,2,

1,4 2,2, 2,3, 3,2, 3,3

2,2 3,3

2,3 3,2, 3,4

Table 3

obtain a 4-by-4 partial TP matrix that has no TP completion. Similar arguments can

be applied using b oth the 3-by-3 and 3-by-4 results to rule out the pairs: f1,2,

2,3g, f1,2, 3,1g, f1,2, 3,3g, f1,3, 2,2g, f1,3, 2,4g, f1,4, 2,3g, f1,4,

3,3g, f2,2, 3,3g, f2,3, 3,2g, f2,3, 3,4g. For the remaining cases we consider

examples of 4-by-4 partial TP matrices.

1. 1,1, 2,3: Let

3 2

x 1 1 1

7 6

2 :9 y 1

7 6

: A =

5 4

:2 :1 :9 1

:2 :2 4 100

Then A is a partial TP matrix. However, detA[f2; 3; 4g; f1; 2; 3g] > 0 if and

only if y>6:3. But for y>6:3, detA[f1; 2g; f3; 4g] < 0. Hence A has no TP

completion.

2. 1,2, 2,4: Let

2 3

3:21 x 2 1

6 7

2:3 2:1 2 y

6 7

A = :

4 5

2:1 2 2 2

1 1 2 3

Then A is a partial TP matrix. However, A has no TP completion since

detA[f1; 2g] > 0 , x<2:931, and det A[f1; 3; 4g; f2; 3; 4g] > 0 , x>3.

3. 1,2, 3,4: Let

3 2

3 x 2 1

7 6

2 2:1 2 2

7 6

: A =

5 4

1 2 2 y

:4 1 2 4

Then A is a partial TP matrix. However, A has no TP completion since

detA[f2; 3; 4g] > 0 , y<2:18, and detA[f1; 2; 3g; f1; 3; 4g] > 0 , y>3.

4. 1,3, 3,1: Let

2 3

2:5 2 x 1

6 7

2 2 1:21 2

6 7

A = :

4 5

y 2 1:5 2:5

:3 2 3 10

ELA

14 S.M. Fallat, C.R.. Johnson and R.L. Smith

Then A is a partial TP matrix. However, A has no TP completion since

detA[f1; 2; 3g; f1; 2; 4g] > 0 , y>1:75, and detA[f2; 3; 4g; f1; 2; 3g] > 0 ,

y<1:725.

5. 1,3, 3,2: Let

2 3

3 2 x 1

6 7

2 2 1:1 2

6 7

A = :

4 5

:3 y 1 2

1 5 4 10

Then A is a partial TP matrix. However, A has no TP completion since

detA[f2; 3; 4g; f1; 2; 3g] > 0 , y>1:268, and detA[f1; 2; 3g; f1; 2; 4g] > 0 ,

y<1:15.

6. 1,4, 2,2: Let

3 2

10 2 3 x

7 6

3:3 y 2 1

7 6

: A =

5 4

2:1 2 2 2

1 1 2 3

Then A is a partial TP matrix. However, A has no TP completion since

detA[f2; 3; 4g] > 0 , y>3, and det A[f1; 2g; f2; 3g] > 0 , y<4=3.

The only pair remaining is f1,4, 3,2g. As we shall see, this pair is actually a

completable pair. However, the argument presented b elow is a little long and requires

a lemma which provides a necessary and sucient condition for the completability

of the 1,4 entry in a 4-by-4 matrix whichwenow state and prove.

Lemma 3.1. Let A beapartial TP matrix with the 1,4 or 4,1 entry unspec-

i ed, and let Ax denote the matrix where x is a value that is speci ed for the 1,4

or 4,1 entry of A. Then A hasaTPcompletion if and only if det A0 > 0.

Proof. Observe that det Ax = det A0 xdetA[f2; 3; 4g; f1; 2; 3g]. Thus it

is clear that A has no TP completion if detA0  0. To verify that A can be

completed to a TP matrix in the case when detA0 > 0 it is enough to show that

there exists an x > 0 such that detAx > 0, det Ax[f1; 2; 3g; f2; 3; 4g] > 0, and

detAx[f1; 2g; f3; 4g] > 0, byFekete's criterion. Note that suchanx>0 exists if

 

detA0[f1; 2; 3g; f2; 3; 4g] a a det A0

13 24

< min ; :

det A[f2; 3g] a detA[f2; 3; 4g; f1; 2; 3g]

23

Now observe that by Sylvester's identity 2

detA0[f1; 2; 3g; f2; 3; 4g]=

detA[f1; 2g; f2; 3g]detA[f2; 3g; f3; 4g] det A[f2; 3g]a a

13 24

=

a

23

detA[f2; 3g]a a

13 24

> ;

a

23

or

a a detA0[f1; 2; 3g; f2; 3; 4g]

13 24

< :

detA[f2; 3g] a 23

ELA

Totally Positive Completions 15

Similar arguments show

detA0[f1; 2; 3g; f2; 3; 4g] det A0

< :

detA[f2; 3g] detA[f2; 3; 4g; f1; 2; 3g]

This completes the pro of.

We are nowin a p osition to prove that the pair f1,4, 3,2g is a completable

pair.

Proposition 3.2. Let A be a 4-by-4 partial TP matrix with the 1,4 and 3,2

entries unspeci ed. Then A may becompleted to a TP matrix.

Proof. Let

2 3

a a a x

11 12 13

6 7

a a a a

21 22 23 24

6 7

: A =

4 5

a y a a

31 33 34

a a a a

41 42 43 44

First cho ose y such that A[f2; 3; 4g; f1; 2; 3g] is singular but TP suchay exists by

2

Theorem 4.3 and is unique. Write A[f2; 3; 4g; f1; 2; 3; 4g]= [a ;a ;a ;a ], where a

1 2 3 4 i

th

is the i column of A[f2; 3; 4g; f1; 2; 3; 4g]i =1; 2; 3; 4. Then by applying Lemma

1.2 to A[f2; 3g; f1; 2; 3g] and A[f3; 4g; f1; 2; 3g]wehave that a = a + a , where

2 1 1 3 3

; > 0. We claim that all minors of A[f2; 3; 4g; f1; 2; 3; 4g] are p ositive except, of

1 3

course, A[f2; 3; 4g; f1; 2; 3g]. The 2-by-2 minors are easily veri ed to b e p ositive. For

the 3-by-3 minors consider A[f2; 3; 4g; f1; 2; 4g], for example. Observe, that

det A[f2; 3; 4g; f1; 2; 4g] = det [a ;a ;a ]= det[a ;a ;a ]+ det[a ;a ;a ]

1 2 4 1 1 1 4 3 1 3 4

= det [a ;a ;a ] > 0:

3 1 3 4

Similarly, all the minors of A[f1; 2; 3; 4g; f1; 2; 3g] are p ositive except for

A[f2; 3; 4g; f1; 2; 3g]. Since detA[f2; 3; 4g; f1; 2; 3g] = 0 and A is partial TP it follows

that for anyvalue of x, detA>0. Moreover, we can increase y a small amountso

that detA[f2; 3; 4g; f1; 2; 3g] > 0, all other fully sp eci ed minors of A are p ositive, and

detA>0 when x =0. Hence by Lemma 3.1 we can complete A to a TP matrix.

Table 4 contains a complete list of all non-completable pairs for a 4-by-4 partial

TP matrix.

2,3 1,1

1,2 2,1, 2,3, 2,4, 3,1, 3,3, 3,4

1,3 2,2, 2,4, 3,1, 3,2,

2,2, 2,3, 3,3 1,4

2,2 3,3

2,3 3,2, 3,4

Table 4

For larger values of m and n, a complete classi cation of all p ossible pairs of

completable entries seems to b e dicult and remains op en.

ELA

16 S.M. Fallat, C.R.. Johnson and R.L. Smith

We close this section we some nal remarks on other general patterns that yield

a TP completion. Recall that the 1,1 and m; n can b e increased without b ound

in a TP matrix and preserve the prop erty of b eing TP. This fact proves that the

following pattern always has a TP completion. Let A b e a partial TP matrix whose

unsp eci ed entries have the following prop erty: if a =?, then either a =?, for s  i

ij st

and t  j or a =?, for s  i and t  j . For example, supp ose A is a partial TP

st

matrix such that if a =?, then a =?, for s  i and t  j , Then b egin completing A

ij st

bycho osing a value for the b ottom rightmost unsp eci ed entry large enough so that

the fully sp eci ed submatrix b elow and to the rightisTP, and continue this pro cess

bymoving to the left until the entire row has b een completed. Rep eat this pro cess by

moving one row up and continue. Observe that at each stage in this pro cess there is

always a choice for the unsp eci ed entry since it o ccurs in the \1,1" p osition of the

submatrix that is to b e completed. Finally, in [16]itisshown that one can always

insert a row or a column into a TP matrix and remain TP.

4. App endix: Single Entry Perturbations of Totally Positive matrices.

It is not dicult to prove that if we increase the 1,1 or the m; nentry of an m-by-n

totally nonnegative p ositive matrix, then the resulting matrix is totally nonnegative

p ositive. We wish to investigate further which entries of a totally nonnegative

p ositive matrix may b e p erturb ed i.e., increased or decreased so that the result is

a totally nonnegative p ositive matrix. These issues have already b een addressed for

other p ositivity classes of matrices, for example, if A is an n-by-n p ositive semide nite,

M -, P -, or inverse M -matrix, then A + D D a nonnegative diagonal matrix is a

p ositive semide nite, M -, P -, or inverse M -matrix, resp ectively; see [10,11]. Recall

th

that E denotes the n-by-n i; j  standard basis matrix, i.e., the matrix whose

ij

th

i; j  entry is 1 and whose remaining entries are zero. Supp ose A is an n-by-n

matrix. Then det A tE  = det A tdetAf1g. Therefore, if det Af1g 6= 0, then

11

detA tE  = 0, when t = detA=detAf1g. Consider the following lemma.

11

Lemma 4.1. Let A beann-by-n total ly nonnegative matrix with detAf1g 6=0.

Then A xE is total ly nonnegative for al l x 2 [0; detA=detAf1g].

11

Proof. Firstly, observe that for every value x 2 [0; detA=detAf1g], det A

xE   0. Recall that A admits a UL-factorization see [1] into totally nonnegative

11

matrices. Partition A as follows,

 

T

a a

11

12

A = ;

a A

21 22

where a is 1-by-1 and A = Af1g. Partition L and U conformally with A. Then

11 22

 

T

a a

11

12

A = = UL

a A

21 22

  

T

u u l 0

11 11

12

=

0 U l L

22 21 22

 

T T

u l + u l u L

11 11 21 22

12 12

= :

U l U L

22 21 22 22

ELA

Totally Positive Completions 17

Note that if l =0, then L, and hence A, is singular. In this case the interval for

11

x is degenerate, and x = 0 is the only allowed value for x. The desired result is

trivial. Thus we assume that l > 0. Consider the matrix A xE , with x 2

11 11

[0; detA=detAf1g]. Then

 

T T

u l + u l x u L

11 11 21 22

12 12

A xE =

11

U l U L

22 21 22 22

  

x

T

u u

l 0

11

11

12 0

l

11

= = U L:

l L

0 U

21 22

22

To show that A xE is totally nonnegative it is enough to verify that u x=l  0.

11 11 11

0

Since if this was the case it follows that U is totally nonnegative and as L is totally

nonnegativeby assumption, wehave that their pro duct, A xE is totally nonneg-

11

ative. Since l > 0 and detAf1g > 0, it follows that L and U are nonsingular.

11 22

Hence 0  detA xE  = u x=l detU det L, from which it follows that

11 11 11 22

u x=l  0.

11 11

Note that a similar result holds for decreasing the n; nentry. We can extend

the previous result for the class TP as follows.

Theorem 4.2. Let A be an n-by-n total ly positive matrix. Then A tE is a

11

TP matrix, for al l t 2 [0; detA=detAf1g].

n1

Proof. Following the pro of of Lemma 4.1 we can write

  

t

T

u u

l 0

11

11

0

12

l

11

= U L; A tE =

11

l L

0 U

21 22

22

t

0

where U = U  E , and b oth U; L are triangular TP matrices; see [1]. Observe

11

l

11

0

that u l = det A=detAf1g. If u t=l > 0, then U and L are triangular

11 11 11 11

TP matrices and hence A tE is TP. So consider the case u t=l = 0, or

11 11 11

equivalently, det A tE  = 0, or t = detA=detAf1g. Let B = A tE . Ob-

11 11

serve that B [f1; 2;:::;ng; f2; 3;:::;ng] and B [f2; 3 :::;ng; f1; 2;:::;ng] are TP ma-

trices since A is TP.Thus the only contiguous minors left to verify are the leading

contiguous minors of B . Consider the submatrices B [f1; 2;:::;kg] for 1  k < n.

Then detB [f1; 2;:::;kg] = det A[f1; 2;:::;kg] tdetA[f2;:::;kg]: This minor is p os-

itive if and only if detA[f1; 2;:::;kg] > tdetA[f2;:::;kg], which is equivalent to

detA[f1; 2;:::;kg]detAf1g > detAdet A[f2;:::;kg], an example of a Koteljanskii

inequality see [11]. The only issue left to settle is whether or not equality holds for

the ab ove Koteljanskii inequality. We claim here that for a TP matrix every Koteljan-

skii inequality is strict. Supp ose to the contrary, i.e., assume there exist two index sets

and such that detA[ [ ]det A[ \ ] = detA[ ]det A[ ]. For simplicity,wemay

assume that [ = N , otherwise replace A by A[ [ ] in the following. By Jacobi's

1 c 1 c 1 c 1 c

identity see [10] wehave det A [ [  ]det A [ \  ] = detA [ ]detA [ ].

1

Let B = SA S , for S = diag1; 1;


; 1. Then B is TP and the ab ove equation

c c

implies detB = detB [ ]detB [ ]. By a result in [2], B is reducible, which is nonsense

since B is TP. Thus A tE is TP , by Fekete's criterion. This completes the

11 n1 pro of.

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18 S.M. Fallat, C.R.. Johnson and R.L. Smith

An obvious next question is what other entries can b e increased/decreased to the

p oint of singularity so that the matrix is TP . As it turns out decreasing the 2,2

n1

entry of a TP matrix results in a TP matrix.

n1

Theorem 4.3. Let A be an n-by-n total ly positive matrix. Then A tE is

22

TP for al l t 2 [0; detA=detAf2g].

n1

Proof. Using the fact that all Koteljanskii inequalities are strict it follows that

all of the leading prop er principal minors of A tE are p ositive. Consider the

22

submatrix B =A tE [f1; 2;:::;ng; f2; 3;:::;ng]. To show that B is TP we need

22

only consider the contiguous minors of B that involve the rst and second row and rst

column, all other minors are p ositiveby assumption. Let C denote such a submatrix

of B . To compute detC , expand the determinant along the second rowof C . Then

1+2

detC =1 tdet C f2g; f1g + detA[ ; ]; where det A[ ; ] is some minor of

A. Thus detC is a p ositive linear combination of minors of A, and hence is p ositive.

Therefore B is TP. Similar arguments show that A tE [f2;:::;ng; f1; 2;:::;ng]

22

is TP. This completes the pro of.

A similar fact holds for decreasing the n 1;n 1 entry of a TP matrix. The

next result follows directly from Theorems 4.2 and 4.3.

Corollary 4.4. If A is an n-by-n total ly positive matrix and i 2f1; 2;n 1;ng,

then A tE is TP for al l t 2 [0; detA=detAfig].

ii n1

Corollary 4.5. Let n  4. If A is an n-by-n total ly positive matrix and

1  i  n, then A tE is TP for al l t 2 [0; detA=detAfig].

ii n1

According to the next example we cannot decrease any other interior main diag-

onal entry in general of a TP matrix and stayTP .

n1

Example 4.6. Consider the following matrix.

3 2

100 10 7=5 2 1

7 6

22 5 2 3 2

7 6

7 6

3 1 1:01 2 3

A = :

7 6

5 4

1 1 2 5 12

1=2 2 5 15 50

Then A is a totally p ositive matrix with detA=detAf3g  :03. However,

detA[f1; 2; 3g; f3; 4; 5g]

= :01:

det A[f1; 2g; f4; 5g]

Thus for t 2 :01;:03], detA tE [f1; 2; 3g; f3; 4; 5g] < 0, and hence A tE is not

33 33

TP .

4

Note that this example can be emb edded into a larger example by b ordering

the matrix ab ove so as to preserve the prop erty of b eing TP using Lemma 2.3, for

example.

Up to this p ointwehave only considered decreasing a single diagonal entry, and

obvious next step is to consider increasing or decreasing o -diagonal entries in a TP

matrix. We b egin our study of p erturbing o -diagonal entries by considering the 1,2

entry.

Theorem 4.7. Let A be an n-by-n total ly positive matrix. Then A + tE is

12

TP for al l t 2 [0; detA=detAf1g; f2g].

n1

ELA

Totally Positive Completions 19

Proof. Since the 1,2 entry of A enters negatively into detA we increase a to

12

a + detA=detAf1g; f2g so that det A + tE  = 0, where t = detA=detAf1g; f2g.

12 12

Observe that the submatrix A + tE [f2;:::;ng; f1; 2;:::;ng] is equal to

12

A[f2;:::;ng; f1; 2;:::;ng] and hence is TP. Moreover,

A + tE [f1; 2;:::;ng; f2;:::;ng] is TP since wehave increased the \1,1" entry of

12

a TP matrix. The only remaining minors to verify are the leading principal minors

of A + tE . Observe that for t 2 [0; detA=detAf1g; f2g];

12

0  detA + tE detA[f2; 3;:::;n 1g]

12

= detA + tE [f1; 2;:::;n 1g]detA[f2; 3;:::;ng]

12

detA + tE [f1; 2;:::;n 1g; f2;:::;ng]detA[f2;:::;ng; f1; 2;:::;n 1g];

12

follows by Sylvester's identity 2. Hence detA + tE [f1; 2;:::;n 1g] > 0. Re-

12

placing A + tE byA + tE [f1; 2;:::;n 1g] in the ab ove identity yields detA +

12 12

tE [f1; 2;:::;n 2g] > 0 and so on. This completes the pro of.

12

Corollary 4.8. If A is an n-by-n total ly positive matrix and i; j  = 2; 1,

n 1;n,or n; n 1, then A + tE is TP for al l t 2 [0; detA=detAfig; fj g].

ij n1

Unfortunately, this is all that can be said p ositively concerning increasing or

decreasing o -diagonal entries. Consider the following example.

Example 4.9. Let

3 2

13 33 31 10

7 6

132 383 371 120

7 6

: A =

5 4

13 38 37 12

1 3 3 1

Then detA=detAf1g; f3g = 1 and det A[f1; 2g; f3; 4g]=a =1=12: Thus for

24

t 2 1=12; 1], detA tE [f1; 2g; f3; 4g] < 0. Thus A tE is not TP .

13 13 3

For the 2,3 entry of a TP matrix consider the following example. Let

2 3

1 12 111 100

6 7

3 37 344 310

6 7

A = :

4 5

3 37 356 321

1 13 123 112

Then detA=detAf2g; f3g= 1=3 and det A[f1; 2g; f3; 4g]=a =1=10: Thus for

44

t 2 1=10; 1=3], detA + tE [f1; 2g; f3; 4g] < 0. Thus A + tE is not TP .

23 23 3

Finally, of the case of the 1,4 entry wehave the following example. Let

2 3

1 2 3 1

6 7

3 7 11 4

6 7

A = :

4 5

3 8 14 6

1 3 6 4

Then detA=detAf1g; f4g = 1 and det A[f1; 2g; f3; 4g]=a =1=11: Thus for

23

t 2 1=11; 1], detA + tE [f1; 2g; f3; 4g] < 0. Thus A + tE is not TP .

14 14 3

ELA

20 S.M. Fallat, C.R.. Johnson and R.L. Smith

As in the case of the diagonal entries these three examples ab ove can b e emb edded

into a larger TP matrix using Lemma 2.3 to show that in general there are no other

o -diagonal entries that can b e increased/decreased until the matrix is singular and

the resulting matrix lie in the class TP .

n1

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[1] T. Ando, Totally Positive Matrices, Linear Algebra and its Applications. 90:165-219, 1987.

[2] D. Carlson, Weakly, Sign-Symmetric Matrices and Some Determinantal Inequalities, Col lo-

quium Mathematicum 17:123-129, 1967.

[3] K.E. DiMarco and C.R. Johnson, Total ly Positive Pattern Completions, An unpublished pap er

from the National Science Foundation Research Exp eriences for Undergraduates program

held at the College of William and Mary in the summer of 1998.

[4] J.H. Drew and C.R. Johnson, The completely p ositive and doubly nonnegative completion

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[7] F.R. Gantmacher and M.G. Krein, Oszil lationsmatrizen, Oszil lationskerne und kleine

Schwingungen Mechanischer Systeme,Akademie-Verlag, Berlin, 1960.

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[10] R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press, New York, 1985.

[11] R. A. Horn and C. R. Johnson, Topics in Matrix Analysis, Cambridge University Press, New

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[13] C.R. Johnson and B.K. Kroschel, The combinatorially symmetric P -matrix completion prob-

lem, Electronic Journal of Linear Algebra 1:59-63, 1996.

[14] C.R. Johnson, B.K. Kroschel, and M. Lundquist The totally nonnegative completion problem,

Fields Institute Communications, American Mathematical So ciety, Providence, RI, 18:97-

108, 1998.

[15] C.R. Johnson and R.L. Smith, The completion problem for M-matrices and inverse M-matrices,

Linear algebra and its Applications. 241-243:655-667, 1996.

[16] C.R. Johnson and R.L. Smith, Insertions in Totally Positive Matrices, in preparation.

[17] S. Karlin, Total Positivity,Vol. I, Stanford University Press, Stanford, 1968.