The General Totally Positive Matrix Completion Problem with Few

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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.

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 mbyn partial totally positive matrices with exactly one unsp ecied entry

the set of p ositions for that entry that guarantee completabilitytoa totally p ositive matrix are

characterized They are the p ositions i j i j and the p ositions i j i j m n In

each case the set of completing entries is an op en and innite in case i j 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 ecied entries are characterized in low dimensions

Key words totally p ositive matrices matrix completion problems partial matrix

AMS sub ject classications A

Intro duction An mbyn 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 ecied while the remaining unsp ecied

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

of a partial matrix is a choice of allowed values for the unsp ecied 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 dierential equations

function theory etc and have attracted considerable attention recently

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

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 ie every fully sp ecied submatrix must b e

TP TN Study of the TN completion problem was b egun in 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

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

Received by the editors on July Accepted for publication on January Handling

Editor Ludwig Elsner

Department of Mathematics and Statistics University of Regina Regina Saskatchewan SS

A CANADA The work of this author was b egun while as a PhD candidate at the College of

William and Mary sfallatmathureginaca

Department of Mathematics College of William and Mary Williamsburg VA USA

crjohnsomathwmedu Research supp orted in part by the Oce of Navel Research contract

NJ and NSF grant

Department of Mathematics UniversityofTennessee at Chattano oga Chattano oga TN

USA rsmithcecasunutcedu 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 ogas

Center of Excellence in Computer Applications

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SM Fallat CR Johnson and RL Smith

without anycombinatorial symmetry assumption It should b e noted that there are

imp ortant dierences 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 dierence is considerable and for example results

analogous to those in 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 mbyn patterns with just one unsp ecied entryandeven

here the answer is rather surprising and nontrivial If min fm ng then only

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

entries each If minfm ng or then any single p osition ensures completability

In order to prove these results several classical and some new eg ab out TP linear

systems facts ab out TP matrices are used Patterns with two unsp ecied entries are

also considered An inventory of completable pairs when m n and some general

observations are given However the general question of two unsp ecied 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 mbyn matrix A lying in rows and columns M

f mg N f 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 welet

denote the complementof relativeto N An mbyn matrix is TN resp TP

k k

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

It is well known that if A a isanmbyn TP TN matrix then the mbyn

ij k k

matrix dened by A a isalsoaTP TN matrix Wemay refer to

minj k k

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

minj

Wenow present some identities discovered by Sylvester see for complete

ness and clarity of comp osition Let A be an nbyn matrix N and supp ose

jj k Dene the n k byn k matrix B b with i j by setting

b det A fig fj g for every i j Then Sylvesters identity states that

for each with j j j j l

detB det A detA

Observe that a sp ecial case of is detB detA detA Another very useful

sp ecial case which is employed throughout is the following Let A be an nbyn matrix

partitioned as follows

a a a

a A a

a a a

where A is n byn and a a are scalars Dene the matrices

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T T

a a a a

B C

a A A a

D E

T T

a a a a

If welet b det B c detC d detD ande detE thenby it follows that

b c

detA detA det

d e

Hence provided det A wehave

detB det E det C detD

detA

Wenow state an imp ortant result due to Fekete see eg who givesavery

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

we need to dene the notion of the disp ersion of an index set For fi i i g

N with i i i the dispersion of denoted by d is dened to be

i i i i k with the convention that d when is

j j k

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

In particular observethat d whenever is a contiguous ie an index set

based on consecutive indices subset of N

Theorem Feketes Criterion An mbyn matrix A is total ly positive if and

only if det A for al l f mg and f ng with jj j j

and dd In other words a matrix is totally p ositive if and only if the

determinant of every square submatrix based on contiguous rowand 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 Let A a a a be an n byn TP matrix Then for

k n

a y a

k i i

has a unique solution y for which

sg n if k is o dd

sg ny

sg n if k is even

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SM Fallat CR Johnson and RL Smith

Proof If k has solution

y a a a a

det a a a a

deta a a

deta a a a

and sg ny sg n If k then has solution

y a a a a a a

k k n k

deta a a a a a

k k k n

deta a a a a a

k k k n

deta a a a a a

k k k n

deta a a a

k k n

deta a a a a a

k k k n

deta a a a a a a

k k k k n

deta a a a a a

k k n k

and we see that if k is odd sg ny sg n while if k is even sg ny

i i

sg n

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

mbyn patterns with just one unsp ecied entry We b egin with a preliminary lemma

which is used throughout the remainder of the pap er

Lemma Let

x a a

a a a

n n nn

n bepartial TP in which x is the only unspecied entry Then if x is chosen so

that det A det Af kg k n In particular x

Proof The pro of follows inductively using Sylvesters identity for determi

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

example

detAdetAf n g

detAf n gdetAf ng

detAf n g fngdetAfng f n g

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and since A is partial TP wehavethatdetAf n g If A is replaced by

Af n g in the ab ove identity then it follows that detAf n g

etc

Theorem Let A be an nbyn partial TP matrix n with x the only

unspecied entry lying in the s t position If s t or s t n then A is

completable to a TP matrix

Proof Throughout the pro of A AxA x willdenoteannbyn partial TP

matrix with exactly one unsp ecied entry x in the s t p osition For k n

let

A xAf kg

B xAf kg f k g

C xAf k g f kg and

D xAf k g

Case s t Every minor which do es not involve A is p ositiveby

assumption Furthermore A enters p ositively into every minor in which it is

involved Hence bymakingA large enough we will obtain a TP completion of

Case s t Let F Af n g f ngf f f

By Lemma f y f in which sg ny sg n Let B x

i i i n B

b b b in which x is chosen so that b y b Thus det B x

n B i i n B

and by Lemma det B x k n in particular x So

k B B

det A x detb b b b

n B n

detb y b b b

i i n

detb y b b b

n n n

y det b b b b

n n n

jy j detb b b b

n n n

Applying Sylvesters identity for determinants we obtain det Aandwecan

continue to apply this identity to obtain

det A x k n

k B

We can then increase x so as to makedetB x and obtain a TP completion

B n

of A

Case s t Notice that it suces to show that there is xsuchthat

eachof

A x k n

B x k n and

G xAf kg f k g k n k

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SM Fallat CR Johnson and RL Smith

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

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

exists an x such that B x TP is singular Therefore since B x

n n k

k n and G x k n are submatrices of B x wehave

k n

det B x k n and det G x k n Let

k k

F Af n g f ng f f f

y f in which sg ny and B x b b b By Lemma f

i i i n n

sg n and since B x is singular and b b b are linearly indep en

n n

dent b y b Thus if A xb b b b

i i n n

det A x det b b y b b b

n i i n

det b b y b b b

n n n

y detb b b b b

n n n

jy j det b b b b b

n n n

By Sylvesters identity it follows that det A detA x anddetA x

n k

k n By decreasing x det B x b ecomes p ositiveandwe obtain a TP

completion of A

Case s t Notice that it suces to show that there is an xsuch

that eachof

A x k n

B x k n

C x k n and

D x k n

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

all of the contiguous minors involving x By Theorem there exists an x

resp x suchthatB x resp C x is singular and lies in TP

C n B n C n

Without loss of generality assume x minfx x g Thus det B x

B B C k B

k n and det C x Since

n B

B x

n B

D x

n B

wehave that det D x and hence by Lemma det D x

n B k B

k n Now

C x

k B

C x

k B

D x

k B

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k n So we can apply Sylvesters identity rep eatedly to obtain

det C x k n

k B

We next show that det A x k n Let

k B

F Af n g f ng f f f andlet

B x Ax f n g f ngb b b

n B B n

y f in which sg ny sg n and since B x is By Lemma f

i i i n B

singular and b b b are linearly indep endent b y b Since

n i i

A x b b b wehave

n B n

det A x detb y b b b

n B i i n

detb y b b b

n n n

y det b b b b

n n n

jy j detb b b b

n n n

It then follows from Sylvesters identity that det Ax det A x

B n B

and since for k n

A B

k k

C D

k k

we can rep etitively apply Sylvesters identity to obtain det A x

k B

k n

Toshow that det D x pro ceed as follows Let

n B

H Af n g f ng

By the rowversion of Lemma H y H in which sg ny sg n

i i i

Since B x and are linearly indep endent

n B n

y Now

i i

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SM Fallat CR Johnson and RL Smith

D x

n B

and hence

y i

y det det D x det det

n B

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

B n n

TP completion of A If s t or then A is completable to a TP matrix

byreversing the indices Thus A is completable to a TP matrix if s t And by

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

Lemma Let A beanmbyn TP matrix Then thereexistsapositive vector

x R such that the augmented matrix Ajx is an mbyn TP matrix Similarly

h i

A w

there exist positive vectors u v w such that ujA and areallTP

v A

Proof Consider a vector x x x R and the augmented matrix

Ajx The idea of the pro of is to cho ose the x s sequentiallysothat

Ajxf ig f n g is TP Cho ose x arbitrarily Cho ose x

large enough so that Ajxf g f ngisTP This is p ossible since x enters

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

latter part follows by analogous arguments

Lemma Let A be an mbyn 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

arbitrarily Cho ose x large enough so that all fully sp ecied minors that involve

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

i h

w A

and follow analogously inductively The cases ujA

A v

Theorem Let A beanmbyn partial TP matrix with m n and with

x the only unspecied entry lying in the s t position If s t or s t m n

then A is completable to a TP matrix

Proof If s t wecan apply Lemma and obtain an nbyn partial TP

i h

in which the unsp ecied entry x lies in the s t p osition where matrix A

s t If s t m n we can apply Lemma and obtain an nbyn partial

i h

TP matrix A in which the unsp ecied entry x lies in the s t p osition where

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

or A and hence one of A itself

Theorem Let A bea byn partial TP matrix with exactly one unspecied

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

k k n

b b b b b b

k k k n

a j

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

b b

i j

wecho ose x suchthat a b xb a b the resulting completion of A

k k k k k

is TP

Lemma Let A be a by partial TP matrix with exactly one unspecied

entry Then A is completable to a TP matrix

This result simply follows from Theorem

Theorem Let A bea byn partial TP matrix with exactly one unspecied

entry n Then A is completable to a TP matrix

Proof By Feketes criterion see Theorem it suces to provethe theorem

for n To see this supp ose the unsp ecied entry in A is in column k If a

value for the unsp ecied entry can be chosen so that Af g fk k kg

Af g fk kk g and Af g fk k k gif k n or

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

that every contiguous minor of A is p ositive Thus by Feketes criterion A is TP

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

Af g fk k kk k g consisting of at most columns Hence

we can assume n So supp ose A is a by partial TP matrix with exactly

one unsp ecied entry x By Theorem and Lemma we need only consider the

case in which x lies in the p osition In this case there is x such that

Ax f g f g is singular and TP by decreasing its entry whichis

the entry of A see Theorem Let F Af g f g f f f

Then f y f y f where sg ny sg ny is p ositive by Lemma Thus

Ax f g f g g g g in which g y g y g and so

det Ax f g detg g g detg g y g y g

y det g g g

Similarly

det Ax f g detg g g y det g g g

Since these twoby minors and all by minors are p ositive we can increase x

and obtain a TP completion of A

Lemma Thereisaby partial TP matrix with the or entry

unspecied that has no TP completion

Proof Consider the partial TP matrix

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SM Fallat CR Johnson and RL Smith

In order for the determinant itself to b e p ositive wemust have x Since

wemust also have x we see that A is not completable to a TP matrix

Lemma Thereisaby partial TP matrix with the or entry

unspecied that has no TP completion

Proof Consider the partial TP matrix

For det Af g f gtobepositivewemust have x But for

det Af g f gtobepositivewemust have x Hence A has no TP

completion

Theorem Main Result Let A beanmbyn partial TP matrix in which

m n and in which the only unspecied entry lies in the st position Any such

A has a TP completion if and only if s t or s t m n

Proof Suciency follows from Theorem To prove necessity assume that

s t m n Notice that if an mbyn matrix has this prop erty then

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

the or p osition of that submatrix Without loss of generality

assume that the s t p osition of the matrix corresp onds to the p osition of the

submatrix By Lemma there is a by partial TP matrix with its only unsp ecied

entry in the p osition that has no TP completion Using Lemma and bythe

preceding remarks we can augment this matrix with p ositiverowscolumns so that

we obtain an mbyn partial TP matrix A whose only unsp ecied 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

Further Discussion For m n we consider partial mbyn TP

matrices with exactly two unsp ecied entries We note that completing a byn

partial TP matrix with two unsp ecied entries follows easily from Theorem In

the case when min fm ng all such completable patterns havebeencharacterized

in For example all partial by TP matrices with exactly two unsp ecied entries

are completable except for the following four patterns

x x x x x x x x x x

x x x x x x x x

and

x x x x x x x x x x

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Similarly when m and n Table contains the p ositions of the pairs of

unsp ecied entries for which there is no completion in general Observe that it is

sucient to only consider the six entries through since the conclusions for

the remaining six entries follows byreversal of indices see also

None

Table

We b egin the by case with some observations that aid in the classication of

the completable and hence noncompletable pairs of unsp ecied entries Supp ose

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

s t The rst observation concerns the case when either p s or q t ie the

unsp ecied 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

we know that the only unsp ecied p ositions in a by partial TP matrix that do

not guarantee a TP completion are and 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 by partial TP matrix to a TP matrix whichfollows from Theorem

Observe that now A is a partial TP matrix with only one unsp ecied entry

Since this remaining unsp ecied entry is in a go o d p osition it follows that wecan

complete A to a TP matrix

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

the or p osition and the other unsp ecied entry is in a go o d p osition

ie satises one of the conditions in Theorem Since the or 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 ecied entry Then A may be completed to TP matrix since the remaining

unsp ecied 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 mbyn partial TP matrix with

exactly two unsp ecied entries with one in the or m n p osition and the other

in a go o d p osition as dened by Theorem then A is completabletoaTP

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

Finally supp ose the unsp ecied entries o ccur in p ositions p q and s twhere

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

the case when p q and s t or viceversa First assume that

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SM Fallat CR Johnson and RL Smith

p q and s t and consider the following example Let

Ax y

seealso Then it is easy to checkthat every fully sp ecied minor of Ax y is

p ositive However

det Ax y x y xy

Hence detAx y for x and y p ositive On the other hand assume that p q

or s t Then since either jp sj or jq tj it follows that at

least one of the unsp ecied entries is in a go o d p osition Without loss of generality

assume that q t and that q t Hence q and t Now delete the column

of the unsp ecied entry that is in a go o d p osition By Theorem wemay complete

the remaining by matrix to a TP matrix Now consider the by matrix A with

only one unsp ecied entry If we can showthat A is a partial TP matrix then we

are done since the remaining unsp ecied entry is in a go o d p osition The reason

that A with only one unsp ecied entry is partial TP is b ecause since jq tj all

of the fully sp ecied minors of A based on contiguous index sets are p ositive Hence

byFeketes criterion A is a partial TP matrix

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

not ruled out are contained in Table

Table

As b efore weonlyhave to consider the six entries

and since the class TP is closed under transp osition and reversal of indices In

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

from others by transp osition andor reversal of indices see Table

As we shall see all but one of the ab ove namelythepairf g pairs are

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

pairs of noncompletable entries in the by and by cases For example supp ose

the unsp ecied entries are in the and p ositions There exists a by

TP matrix with unsp ecied entries in those p ositions that has no TP completion

eg We can b order this by partial TP matrix by Lemma to

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Table

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

be applied using b oth the by and by results to rule out the pairs f

g f g f g f g f g f g f

g f g f g f g For the remaining cases we consider

examples of by partial TP matrices

Let

Then A is a partial TP matrix However detAf g f g ifand

only if y But for y detAf g f g Hence A has no TP

completion

Let

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

detAf g x and det Af g f g x

Let

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

detAf g y and detAf g f g y

Let

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SM Fallat CR Johnson and RL Smith

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

detAf g f g y and detAf g f g

Let

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

detAf g f g y and detAf g f g

Let

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

detAf g y and det Af g f g y

The only pair remaining is f g 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 entry in a by matrix whichwenow state and prove

Lemma Let A beapartial TP matrix with the or entry unspec

ied and let Ax denote the matrix where x is a value that is specied for the

or entry of A Then A has a TP completion if and only if det A

Proof Observe that det Ax det A xdetAf g f g Thus it

is clear that A has no TP completion if detA To verify that A can be

completed to a TP matrix in the case when detA itisenoughtoshowthat

there exists an x such that detAx det Axf g f g and

detAxf g f g byFeketes criterion Note that suchanx exists if

detAf g f g a a det A

min

det Af g a detAf g f g

Now observe that by Sylvesters identity

detAf g f g

detAf g f gdetAf g f g det Af ga a

detAf ga a

a a detAf g f g

detAf g a

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Similar arguments show

det A detAf g f g

detAf g detAf g f g

This completes the pro of

We are nowin a p osition to prove that the pair f g is a completable

pair

Proposition Let A be a by partial TP matrix with the and

entries unspecied Then A may becompleted to a TP matrix

Proof Let

a a a x

a a a a

a y a a

a a a a

First cho ose y suchthatAf g f g is singular but TP suchay exists by

Theorem and is unique Write Af g f g a a a a where a

is the i column of Af g f gi Then by applying Lemma

to Af g f g and Af g f gwehavethat a a a where

We claim that all minors of Af g f g are p ositive except of

course Af g f g The by minors are easily veried to b e p ositive For

the by minors consider Af g f g for example Observe that

det Af g f g deta a a deta a a deta a a

det a a a

Similarly all the minors of Af g f g are p ositive except for

Af g f g Since detAf g f g and A is partial TP it follows

that for anyvalue of x detA Moreover wecanincrease y a small amountso

that detAf g f g all other fully sp ecied minors of A are p ositive and

detAwhen x Hence by Lemma we can complete A to a TP matrix

Table contains a complete list of all noncompletable pairs for a by partial

TP matrix

Table

For larger values of m and n a complete classication of all p ossible pairs of

completable entries seems to b e dicult and remains op en

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SM Fallat CR Johnson and RL Smith

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

a TP completion Recall that the and m n can b e increased without b ound

in a TP matrix and preserve the prop erty of being TP This fact proves that the

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

unsp ecied 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

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 ecied entry large enough so that

the fully sp ecied 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 ecied entry since it o ccurs in the p osition of the

submatrix that is to b e completed Finally in itis shown that one can always

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

App endix Single Entry Perturbations of Totally Positive matrices

It is not dicult to provethatifwe increase the or the m nentry of an mbyn

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 ie 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 nbyn p ositive semidenite

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

p ositive semidenite M P or inverse M matrix resp ectively see Recall

that E denotes the nbyn i j standard basis matrix ie the matrix whose

i j entry is and whose remaining entries are zero Supp ose A is an nbyn

matrix Then det A tE detA tdetAfg Therefore if det Afg then

detA tE when t detAdetAfg Consider the following lemma

Lemma Let A beannbyn total ly nonnegative matrix with detAfg

Then A xE is total ly nonnegative for al l x detAdetAfg

Proof Firstly observe that for every value x detAdetAfg det A

xE Recall that A admits a ULfactorization see into totally nonnegative

matrices Partition A as follows

a a

a A

where a is by and A Afg Partition L and U conformally with A Then

a a

A UL

a A

u u l

U l L

T T

u l u l u L

U l U L

ELA

Totally Positive Completions

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

x is degenerate and x is the only allowed value for x The desired result is

trivial Thus we assume that l Consider the matrix A xE with x

detAdetAfg Then

T T

u l u l x u L

A xE

U l U L

u u

U L

l L

Toshow that A xE is totally nonnegative it is enough to verify that u xl

Since if this was the case it follows that U is totally nonnegativeandas L is totally

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

ative Since l anddetAfg it follows that L and U are nonsingular

Hence detA xE u xl detU det L from which it follows that

u xl

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

the previous result for the class TP as follows

Theorem Let A be an nbyn total ly positive matrix Then A tE is a

TP matrix for al l t detAdetAfg

Proof Following the pro of of Lemma we can write

u u

U L A tE

l L

where U U E and b oth U L are triangular TP matrices see Observe

that u l det AdetAfg If u tl then U and L are triangular

TP matrices and hence A tE is TP So consider the case u tl or

equivalently det A tE or t detAdetAfg Let B A tE Ob

serve that B f ng f ng and B f ng f ng are TP ma

trices since A is TPThus the only contiguous minors left to verify are the leading

contiguous minors of B Consider the submatrices B f kg for k n

Then detB f kg det Af kg tdetAfkg This minor is p os

itive if and only if detAf kg tdetAfkg which is equivalent to

detAf kgdetAfg detAdet Afkg an example of a Koteljanskii

inequality see 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 ie assume there exist two index sets

and suchthatdetA det A detAdet A For simplicitywemay

assume that N otherwise replace A by A in the following By Jacobis

c c c c

identity see wehavedetA det A detA detA

Let B SA S forS diag Then B is TP and the ab ove equation

c c

implies detB detB detB Byaresultin B is reducible which is nonsense

since B is TP Thus A tE is TP by Feketes criterion This completes the

n pro of

ELA

SM Fallat CR Johnson and RL Smith

An obvious next question is what other entries can b e increaseddecreased to the

point of singularity so that the matrix is TP As it turns out decreasing the

entry of a TP matrix results in a TP matrix

Theorem Let A be an nbyn total ly positive matrix Then A tE is

TP for al l t detAdetAfg

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

submatrix B A tE f ng f ng ToshowthatB is TP we need

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

detC tdet C fg fg detA where det A is some minor of

A Thus detC is a p ositivelinearcombination of minors of A and hence is p ositive

Therefore B is TP Similar arguments showthatA tE fng f ng

is TP This completes the pro of

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

next result follows directly from Theorems and

Corollary If A is an nbyn total ly positive matrix and i f n ng

then A tE is TP for al l t detAdetAfig

ii n

Corollary Let n If A is an nbyn total ly positive matrix and

i nthen A tE is TP for al l t detAdetAfig

ii n

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

onal entry in general of a TP matrix and stayTP

Example Consider the following matrix

Then A is a totally p ositive matrix with detAdetAfg However

detAf g f g

det Af g f g

Thus for t detA tE f g f g and hence A tE is not

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 for

example

Up to this p ointwehave only considered decreasing a single diagonal entryand

obvious next step is to consider increasing or decreasing odiagonal entriesinaTP

matrix We b egin our study of p erturbing odiagonal entries by considering the

entry

Theorem Let A be an nbyn total ly positive matrix Then A tE is

TP for al l t detAdetAfg fg

ELA

Totally Positive Completions

Proof Since the entry of A enters negatively into detA we increase a to

a detAdetAfg fg so that det A tE where t detAdetAfg fg

Observe that the submatrix A tE fng f ng is equal to

Afng f ng and hence is TP Moreover

A tE f ng fng is TP since wehave increased the entry of

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

of A tE Observe that for t detAdetAfg fg

detA tE detAf n g

detA tE f n gdetAf ng

detA tE f n g fngdetAfng f n g

follows bySylvesters identity Hence detA tE f n g Re

placing A tE byA tE f n gintheaboveidentity yields detA

tE f n g and so on This completes the pro of

Corollary If A is an nbyn total ly positive matrix and i j

n nor n n thenA tE is TP for al l t detAdetAfig fj g

ij n

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

decreasing odiagonal entries Consider the following example

Example Let

Then detAdetAfg fg and det Af g f ga Thus for

t detA tE f g f g Thus A tE is not TP

For the entry of a TP matrix consider the following example Let

Then detAdetAfg fg and det Af g f ga Thus for

t detA tE f g f g Thus A tE is not TP

Finally of the case of the entry wehave the following example Let

Then detAdetAfg fg and det Af g f ga Thus for

t detA tE f g f g Thus A tE is not TP

ELA

SM Fallat CR Johnson and RL 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 to show that in general there are no other

odiagonal entries that can b e increaseddecreased until the matrix is singular and

the resulting matrix lie in the class TP

REFERENCES

T Ando Totally Positive Matrices Linear Algebra and its Applications

D Carlson Weakly SignSymmetric Matrices and Some Determinantal Inequalities Col lo

quium Mathematicum

KE DiMarco and CR 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

JH Drew and CR Johnson The completely p ositive and doubly nonnegative completion

problems Linear and Multilinear Algebra

JH Drew CR Johnson KL Karlof and KL Shuman The cycle completable graphs for the

completely p ositive and doubly nonnegative completion problems manuscript

EB Dryden and CR Johnson Total ly Nonnegative 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

FR Gantmacher and MG Krein Oszil lationsmatrizen Oszil lationskerne und kleine

Schwingungen Mechanischer SystemeAkademieVerlag Berlin

R Grone CR Johnson E Sa and H Wolkowicz Positive denite completions of partial

hermitian matrices Linear Algebra and its Applications

M Gasca and CA Micchelli Total Positivity and its Applications Mathematics and its Ap

plications Vol Kluwer Academic

R A Horn and C R Johnson Matrix AnalysisCambridge University Press New York

R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press New

York

CR Johnson Matrix completion problems A Survey Proceedings of Symposia in Applied

Mathematics

CR Johnson and BK Kroschel The combinatorially symmetric P matrix completion prob

lem Electronic Journal of Linear Algebra

CR Johnson BK Kroschel and M Lundquist The totally nonnegative completion problem

Fields Institute Communications American Mathematical So ciety Providence RI

CR Johnson and RL Smith The completion problem for Mmatrices and inverse Mmatrices

Linear algebra and its Applications

CR Johnson and RL Smith Insertions in Totally Positive Matrices in preparation

S Karlin Total PositivityVol I Stanford University Press Stanford