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The General Totally Positive Matrix Completion Problem

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The General Totally Positive Matrix Completion Problem 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 <i < < i , the dispersion of , denoted by d , is de ned to be 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.
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