Determinantal inequalities for block triangular matrices
aa r X i v : . [ m a t h . F A ] O c t DETERMINANTAL INEQUALITIES FOR BLOCKTRIANGULAR MATRICES ♦ MINGHUA LIN
Abstract.
Let T = (cid:20) X Y Z (cid:21) be an n -square matrix, where X, Z are r -square and ( n − r )-square, respectively. Among other determinantalinequalities, it is proveddet ( I n + T ∗ T ) ≥ det ( I r + X ∗ X ) · det ( I n − r + Z ∗ Z )with equality holds if and only if Y = 0. Introduction
The well known Fischer inequality [4, p. 506] states that if A = (cid:20) A A A ∗ A (cid:21) is positive semidefinite, thendet A ≤ det A · det A . (1.1)As any positive semidefinite matrix A can be factorized as A = T ∗ T with T = (cid:20) X Y Z (cid:21) being comformally partitioned as A , inequality (1.1) canbe written asdet X ∗ X · det Z ∗ Z = det T ∗ T ≤ det X ∗ X · det( Y ∗ Y + Z ∗ Z ) . (1.2)This paper presents some results that complement (1.2). We believe ourresults are of new pattern concerning determinantal inequalities. Let us fixsome notation. The matrices considered here have entries from the fieldof complex numbers. X ′ , X, X ∗ stands for transpose, (entrywise)conjugate,conjugate transpose of X , respectively. For two n -square Hermitian matrices X, Y , we write X ≥ Y to mean X − Y is positive semidefinite (so X ≥ X is positive semidefinite). The n -square identity matrix is denotedby I n . The Frobenius norm of X is denoted by k X k F . It is known that k X k F = √ tr X ∗ X , where tr denotes the trace. If X = (cid:20) X X X X (cid:21) is an n -square matrix with X nonsingular, then the Schur complement of X in X is defined by X/X = X − X ∗ X − X . A well known property of the Mathematics Subject Classification.
Key words and phrases. determinantal inequality, block triangular matrices. ♦ Dedicated to Stephen Drury, whose beautiful works on matrix analysis inspire theauthor.
Schur complement is det X = det X · det( X/X ). Finally, for an n -squarematrix, we denote by λ j ( X ) and σ j ( X ), j = 1 , . . . , n , the eigenvalues andsingular values of X , respectively, such that | λ ( X ) | ≥ · · · ≥ | λ n ( X ) | and σ ( X ) ≥ · · · ≥ σ n ( X ). 2. Main results
We present the following result, showing that when more matrices aresummed, the identity in (1.2) becomes an inequality.
Theorem 2.1.
Let T k = (cid:20) X k Y k Z k (cid:21) , k = 1 , . . . , m , be n -square conformallypartitioned matrices. Then det m X k =1 T ∗ k T k ! ≥ det m X k =1 X ∗ k X k ! · det m X k =1 Z ∗ k Z k ! . (2.1) Proof.
By a standard continuity argument, we may assume X ∗ k X k is non-singular for k = 1 , . . . , m . As (cid:20) X ∗ k X k X ∗ k Y k Y ∗ k X k Y ∗ k Y k (cid:21) = (cid:2) X k Y k (cid:3) ∗ (cid:2) X k Y k (cid:3) ≥ , summing for k from 1 to m gives (cid:20)P mk =1 X ∗ k X k P mk =1 X ∗ k Y k P mk =1 Y ∗ k X k P mk =1 Y ∗ k Y k (cid:21) ≥ , Hence, m X k =1 Y ∗ k Y k − m X k =1 Y ∗ k X k ! m X k =1 X ∗ k X k ! − m X k =1 X ∗ k Y k ! ≥ . On the other hand, T ∗ k T k = (cid:20) X ∗ k X k X ∗ k Y k Y ∗ k X k Y ∗ k Y k + Z ∗ k Z k (cid:21) . Thus m X k =1 T ∗ k T k ! . m X k =1 X ∗ k X k ! = m X k =1 ( Y ∗ k Y k + Z ∗ k Z k ) − m X k =1 Y ∗ k X k ! m X k =1 X ∗ k X k ! − m X k =1 X ∗ k Y k ! ≥ m X k =1 Z ∗ k Z k ≥ . Applying the determinant on both sides gives the desired inequality. (cid:3)
Remark 2.2.
By a simple induction, Theorem 2.1 can be extended to the p × p ( p >
2) block upper triangular case.
ETERMINANTAL INEQUALITIES 3
A full characterization of the equality case in (2.1) is nasty, so we do notinclude it here. We extract a special case of Theorem 2.1 with m = 2 forlater use. Moreover, the equality case is concise. Corollary 2.3.
Let T = (cid:20) X Y Z (cid:21) be an n -square matrix, where X, Z are r -square and ( n − r ) -square, respectively. Then det ( I n + T ∗ T ) ≥ det ( I r + X ∗ X ) · det ( I n − r + Z ∗ Z ) . (2.2) Equality holds in (2.2) if and only if Y = 0 .Proof. It suffices to show the equality case. If Y = 0, clearly (2.2) becomesan equality. Conversely, if the equality in (2.2) holds, thendet (cid:16) I n − r + Z ∗ Z + Y ∗ Y − Y ∗ X ( I r + X ∗ X ) − X ∗ Y (cid:17) = det( I n − r + Z ∗ Z ) . (2.3)As Y ∗ Y − Y ∗ X ( I r + X ∗ X ) − X ∗ Y ≥
0, the equality (2.3) holds only if Y ∗ Y − Y ∗ X ( I r + X ∗ X ) − X ∗ Y = 0, i.e, Y ∗ (cid:16) I r − X ( I r + X ∗ X ) − X ∗ (cid:17) Y = 0 . Now for any j = 1 , . . . , r , λ j (cid:16) I r − X ( I r + X ∗ X ) − X ∗ (cid:17) = 1 − λ r − j +1 ( X ∗ X )1 + λ r − j +1 ( X ∗ X )= 11 + λ r − j +1 ( X ∗ X ) > . so I r − X ( I r + X ∗ X ) − X ∗ is positive definite, forcing Y = 0. (cid:3) Corollary 2.4.
Let T k = (cid:20) X k Y k Z k (cid:21) , k = 1 , . . . , m , be n -square conformallypartitioned matrices. If X k , Z k are all normal matrices, then det m X k =1 T ∗ k T k ! ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) det m X k =1 X k X k !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) det m X k =1 Z k Z k !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (2.4) Proof.
In view of Theorem 2.1, it suffices to showdet m X k =1 X ∗ k X k ! ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) det m X k =1 X k X k !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . As (cid:20)P mk =1 X k X ′ k P mk =1 X k X k P mk =1 X ∗ k X ′ k P mk =1 X ∗ k X k (cid:21) = m X k =1 (cid:2) X ′ k X k (cid:3) ∗ (cid:2) X ′ k X k (cid:3) ≥ , this yields (cid:20) det (cid:0)P mk =1 X k X ′ k (cid:1) det (cid:0)P mk =1 X k X k (cid:1) det ( P mk =1 X ∗ k X ′ k ) det ( P mk =1 X ∗ k X k ) (cid:21) ≥ , M. LIN and sodet m X k =1 X k X ′ k ! · det m X k =1 X ∗ k X k ! ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) det m X k =1 X k X k !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . But X k , k = 1 , . . . , m , are normal, and sodet m X k =1 X k X ′ k ! = det m X k =1 X k X ′ k ! ′ = det m X k =1 X k X ∗ k ! = det m X k =1 X ∗ k X k ! , as desired. (cid:3) The author does not know if there is a simple characterization for theequality case in (2.4). The following example shows that (2.4) may failwithout the normality assumption.
Example 2.5.
Taking T = − − − −
100 0 − , T = −
16 3 3 − −
160 0 − − , a simple calculation using Matlab givesdet ( T ∗ T + T ∗ T ) = 1 . × < (cid:12)(cid:12) det (cid:0) X + X (cid:1)(cid:12)(cid:12) · (cid:12)(cid:12) det (cid:0) Z + Z (cid:1)(cid:12)(cid:12) = 9 . × . Our next result says that, to some extent, Corollary 2.4 can be improved.
Theorem 2.6.
Let T = (cid:20) X Y Z (cid:21) be an n -square matrix, where X, Z are r -square and ( n − r ) -square, respectively. Then det ( I n + T ∗ T ) ≥ det (cid:0) I r + XX (cid:1) · det (cid:0) I n − r + ZZ (cid:1) . (2.5) Equality holds in (2.5) if and only if Y = 0 and X, Z are symmetric.
Remark 2.7.
Compared with (2.4), we don’t use absolute value on theright hand side of (2.5). This is because det (cid:0) I r + XX (cid:1) ≥
0, an observationby Djokovi´c [1, 2]. However, it is possible that det (cid:0)P mk =1 X k X k (cid:1) < X = X ′ = (cid:20) (cid:21) , a simple calculation givesdet( X X + X X ) = det (cid:20) (cid:21) = − < . ETERMINANTAL INEQUALITIES 5
We need a lemma, which plays a key role in establishing the equalitycase in Theorem 2.6.
Lemma 2.8.
Let X be an n -square matrix. Then det( I n + X ∗ X ) ≥ det (cid:0) I n + XX (cid:1) . (2.6) Equality holds in (2.6) if and only if X is symmetric.Proof. From the proof of Corollary 2.4, we havedet( I n + XX ∗ ) · det( I n + X ∗ X ) ≥ det (cid:0) I n + XX (cid:1) . Note thatdet( I n + XX ′ ) = det( I n + XX ′ ) ′ = det( I n + XX ∗ ) = det( I n + X ∗ X ) . This proves (2.6).If X is symmetric, then X = X ∗ and so det (cid:0) I n + XX (cid:1) = det( I n + X ∗ X ).We show the other way around. It is clear thatdet (cid:0) I n + XX (cid:1) = n Y j =1 (cid:16) λ j ( XX ) (cid:17) ≤ n Y j =1 (cid:16) | λ j ( XX ) | (cid:17) with the second inequality becoming an equality only if λ j ( XX ) ≥ j . By Weyl’s inequality [5, p. 317], for k = 1 , . . . , n , k Y j =1 | λ j ( XX ) | ≤ k Y j =1 σ j ( XX ) ≤ k Y j =1 σ j ( X ) σ j ( X ) = k Y j =1 σ j ( X ) = k Y j =1 σ j ( X ∗ X ) , where equality holds when k = n . The strict convexity of the function f ( t ) = log(1 + e t ) ([5, p. 156]) implies that n Y j =1 (cid:16) | λ j ( XX ) | (cid:17) ≤ n Y j =1 (cid:16) σ j ( X ∗ X ) (cid:17) = det( I n + X ∗ X )with the first inequality becoming an equality only if XX is normal.Thus, if det (cid:0) I n + XX (cid:1) = det( I n + X ∗ X ), then XX ≥ λ j ( XX ) = σ j ( X ∗ X ) for all j . In particular, tr XX = tr X ∗ X . We shall show the latterimplies that X is symmetric. Compute k X − X ′ k F = tr( X − X ′ ) ∗ ( X − X ′ )= tr X ∗ X − tr XX − tr( XX ) ∗ + tr XX ′ = 2 (cid:16) tr X ∗ X − tr XX (cid:17) = 0 , and so X = X ′ , as required. (cid:3) M. LIN
Proof of Theorem 2.6.
The inequality (2.5) follows from (2.2) and (2.6).If Y = 0 and X, Z are symmetric, thendet ( I n + T ∗ T ) = det( I r + X ∗ X ) · det( I n − r + Z ∗ Z )= det (cid:0) I r + XX (cid:1) · det (cid:0) I n − r + ZZ (cid:1) . Conversely, if the equality holds in (2.5), then actually we havedet ( I n + T ∗ T ) = det( I r + X ∗ X ) · det( I n − r + Z ∗ Z )= det (cid:0) I r + XX (cid:1) · det (cid:0) I n − r + ZZ (cid:1) . In view of Corollary 2.3, the first equality gives Y = 0. By Lemma 2.8, thesecond equality implies that X, Z are symmetric. (cid:3)
An immediate consequence of Theorem 2.6 is the following, which is dueto Drury [3, Lemma 4].
Corollary 2.9. (Drury’s inequality)
Let T = [ t ij ] be an n -square uppertriangular matrix. Then det ( I n + T ∗ T ) ≥ n Y j =1 (1 + | t jj | ) . Equality holds if and only if T is diagonal. More results
The absolute value of a matrix X is defined to be the matrix | X | =( X ∗ X ) / , the unique positive semidefinite square root of X ∗ X . The in-equality (2.1) can be rewritten asdet m X k =1 | T k | ! ≥ det m X k =1 | X k | ! · det m X k =1 | Z k | ! . (3.1)The following result is an extension of Corollary 2.3. Theorem 3.1.
Let T = (cid:20) X Y Z (cid:21) be an n -square matrix, where X, Z are r -square and ( n − r ) -square, respectively. Then for any p ≥ I n + | T | p ) ≥ det ( I r + | X | p ) · det ( I n − r + | Z | p ) . (3.2) Equality holds in (3.2) if and only if Y = 0 .Proof. The proof is similar to the one given in [3, Lemma 4]. Let X = U | X | , Z = U | Z | be the polar decomposition of X, Z , respectively. Taking U = (cid:20) U U (cid:21) (so U is unitary) gives U ∗ A = (cid:20) | X | U ∗ Y | Z | (cid:21) . We have by Weyl’s ETERMINANTAL INEQUALITIES 7 inequality [5, p. 317] k Y j =1 | λ j ( U ∗ T ) | ≤ k Y j =1 σ j ( U ∗ T ) = k Y j =1 σ j ( T ) , k = 1 , . . . , n, i.e., k Y j =1 λ j (cid:18)(cid:20) | X | | Z | (cid:21)(cid:19) ≤ k Y j =1 σ j ( T ) , k = 1 , . . . , n, where equality holds when k = n .By the convexity of the function f ( t ) = log(1 + e pt ) for p ≥
1, we obtainfrom [5, p. 156] thatdet( I r + | X | p ) · det( I n − r + | Z | p ) = det (cid:18) I n + (cid:20) | X | p | Z | p (cid:21)(cid:19) ≤ det( I n + | T | p ) . This proves (3.2).If Y = 0, then clearlydet ( I n + | T | p ) = det ( I r + | X | p ) · det ( I n − r + | Z | p ) . Conversely, ifdet ( I n + | T | p ) = det ( I r + | X | p ) · det ( I n − r + | Z | p ) , the strict convexity of f ( t ) = log(1 + e pt ), p ≥
1, gives that k Y j =1 λ j (cid:18)(cid:20) | X | U Y | Z | (cid:21)(cid:19) = k Y j =1 σ j (cid:18)(cid:20) | X | U Y | Z | (cid:21)(cid:19) , k = 1 , . . . , n. And so (cid:20) | X | U Y | Z | (cid:21) is normal, which is the case only if U Y = 0 andtherefore Y = 0. (cid:3) Nevertheless, (3.1) does not have such an analogue. We show by anexample that, in general, it is not true thatdet ( | T | + | T | ) ≥ det ( | X | + | X | ) · det ( | Z | + | Z | ) , (3.3)where T , T , are as in Theorem 2.1. Example 3.2.
Taking T = − − − −
190 0 0 −
20 0 − , T = −
12 12 100 0 14 −
20 0 23 − , a simple calculation using Matlab givesdet ( | T | + | T | ) = 5193 . < det ( | X | + | X | ) · det ( | Z | + | Z | ) = 20248 . M. LIN
References [1] D. Z. Djokovi´c, On some representations of matrices, Linear Multilin-ear Algebra 4 (1976) 33-40.[2] D. Z. Djokovi´c, O. P. Lossers, Problem E2525, det (cid:0) I + AA (cid:1) ≥ × Department of Mathematics and Statistics, University of Victoria, Vic-toria, BC, Canada, V8W 3R4.
E-mail address ::