4
I \ JOURNAL OF RESEARCH of the Noti onal Bureau of Standards - B. Ma th emat i ca l Sciences Vol. 75B, No s. 1 a nd 2, January-June 1971 An Improvement of the Fischer Inequality Russe ll Mer ri s* Inst itu te for Basic Sta ndards, National Bureau of Standards , Washington, D.C. 20234 (October 2 1, 1970) An improvement of the classica l Fi sc her inequality for the de terminant of a positive defin ite he rmitian mat rix is proved. It is use d to analyze th e Hadamard de terminant theor e m. Key words: Fi sc her in e qua lit y; Hadamard determinant theorem; permanent; positive de fin ite hermitian matrix . 1. Introduction Let H = (hij) be an n-squar e po sitive se midefinite hermitian matrix , n 2. Th e ju s tly famous Hadamard dete rminant theorem [3; 1, p. 64; 5, p. 114]1 states that 11 IT hll det H. 1=1 Two of the many ge neralizations of this theorem ar e of int er es t to us. Partition H, so that H22 is k s quar e. E. Fi sc her [2; 5, p. 117- 118] proved that det HII det H 22 det H. One obtains Hadamard 's theorem from Fis c her's by induction . (1) In 1918, I. Sc hur [7] e xtended Fischer 's inequal ity. He defined certain matrix func tions: Let C be a subgroup of the sy mmetri c group 51/. Suppose A is a c hara cter of de gree s on C. Then /I d(H) = 1; A(U) D h((T(t). If C = 51/ and A = sgn, d = det. If C = 51/ and A = 1, d = per (per manent) . Finally, if C = {l} th en /I d (H) = s IT hit. Schur proved that (=1 d(H) s det (H). AMS Subject. Classification: Primary 15A1 5, Secondary 15A42, 15A57. *Thi s work wa s don e while the author was a National Acad e my of Sciences-National Research Council Postdoctoral Resea rch Associate at the Nat ional Burea u of Standards. Wash ington, D.C. 20234. 1 Fig ures in brac kets indicate th e literature refere nces at the end of this paper. 77

Improvement of Fischer inequality - NIST Page · 2011-05-17 · Fischer [2; 5, p. 117-118] proved that det HII det H 22 ~ det H. One obtains Hadamard's theorem from Fischer's by induction

  • Upload
    others

  • View
    1

  • Download
    0

Embed Size (px)

Citation preview

Page 1: Improvement of Fischer inequality - NIST Page · 2011-05-17 · Fischer [2; 5, p. 117-118] proved that det HII det H 22 ~ det H. One obtains Hadamard's theorem from Fischer's by induction

I

\

JOURNAL OF RESEARCH of the Notional Bureau of Standards - B. Mathematica l Sciences Vol. 75B, Nos. 1 a nd 2, January-June 1971

An Improvement of the Fischer Inequality

Russell Merri s*

Institute for Basic Sta ndards, National Bureau of Standards, Washington, D.C. 20234

(October 2 1, 1970)

An improve me nt of the c lass ical Fische r inequality for the determinant of a pos itive defin ite hermitian matrix is proved. It is used to analyze the Hadamard determinant theore m.

Key words : Fi scher in equa lity; Hadamard d e term ina nt theore m; permanent; positive de fin ite he rmitian matrix.

1. Introduction

Let H = (hij) be an n-square positive semidefinite hermitian matrix , n ~ 2. The jus tly famous Hadamard de terminant theorem [3; 1, p. 64; 5 , p. 114]1 states that

11

IT hll ~ det H. 1=1

Two of the many generalizations of this theorem are of interest to us. Partition H,

so that H22 is k square. E. Fischer [2; 5, p. 117- 118] proved that

de t HII de t H 22 ~ det H.

One obtains Hadamard 's theore m from Fischer's by induction.

(1)

In 1918, I. Schur [7] extended Fischer's inequality. He defined certain matrix func tions: Let C be a subgroup of the symmetric group 51/. Suppose A is a charac ter of degree s on C. Then

/I

d(H) =1; A(U) D h((T(t).

If C = 51/ and A = sgn, d = det. If C = 51/ and A = 1, d = per (permanent). Finally, if C = {l} then /I

d (H) = s IT hit. Schur proved that (=1

d(H) ~ s det (H ) .

AMS Subject. Classification: Primary 15A1 5, Secondary 15A42, 15A57.

*Thi s work was done while the author was a National Acade my of Sciences-National Research Council Postdoctoral Research Associate a t the Nat ional Burea u of S tand ards. Wash ington, D.C. 20234.

1 Figures in brackets ind icate the literature refe rences at the end of th is paper.

77

Page 2: Improvement of Fischer inequality - NIST Page · 2011-05-17 · Fischer [2; 5, p. 117-118] proved that det HII det H 22 ~ det H. One obtains Hadamard's theorem from Fischer's by induction

If S~. is the symmetric group on {n-k+1, . .. , n}, G=SII_kXS~" and A.=sgn, one obtains

Fischer's inequality. In this note an inequality improving Fischer's is derived. Though the methods are simple,

the result allows a sharp analysis of Hadamard's theorem.

2. Results

THEOREM 1: Let H be a positive definite hermitian matrix partitioned as in (1). Then

with equality when k = 1. When n = 2k, the result becomes

det HII det H22 -det Hi2 det HI2 ~ det H,

a special case of Schur's theorem.

Before proving Theorem 1, we introduce the Schur complement of H I I in H as

LEMMA (known): If H is positive definite hermitian, so is (H/H II); and det H = det H II det (H/HII ).

PROOF: Haynsworth [4] has noticed that

Hence, (H/ H II) is POSItIve definite because the nonsingular congruence, Q* HQ , preserves positive definiteness. The determinant formula (due originally to Schur) follows because det Q = 1. I

PROOF of THEOREM 1: It is a well known and elementary fact that if A and B are positive ( semidefinite hermitian matrices of the same size then

det (A + B) ~ det A + det B.

(In fact, a much stronger result is known, namely the Minkowski determinant theorem [5 , p. 115].) Using this fact, the definition of (H/HI d, and the lemma, we see

(Clearly, (2) is equality when k = 1.) Multiplication of (2) by det H II yields the result. We proceed to use Theorem 1 to analyze Hadamard's theorem. THEOREM 2: Let H be positive definite hermitian. Suppose Ut is the row vector (h t, 1, ... , ht,H),

and H t is the leading tXt principal sub matrix ofH. Then

and

/I

hll IT (htr-utHt~lun =det H. 1= 2

78

Page 3: Improvement of Fischer inequality - NIST Page · 2011-05-17 · Fischer [2; 5, p. 117-118] proved that det HII det H 22 ~ det H. One obtains Hadamard's theorem from Fischer's by induction

l

In a certain sense, Theorem 2 gives the final word on the Hadamard theorem. It shows just what is being thrown away to obtain the inequality.

PROOF: The proof is by induction on n. From Theorem 1 (with k = 1),

But, det H II _I, det H > O. Hence (h llll - UIIH-;;~ I u~) > O. By induction,

/I - I

detHII _I = h ll IT (hll-UtH(.lluj) (= 2

and (h ll - UHf_II un > O. (Alternatively, one can prove, using the Laplace expansion theorem [5, p. 14] , that

detHt det H t - I

Using different methods, Marcus and Soul es [6] obtain ed Theorem 1 for the case k= 1. They also obtained a n analog of it for the permanent. They used their result to prove Theorems 3 and 4 (below) but failed to notice Theorems 2 and 5.

THEOREM 3: Let H be positive definite hermitian. Let U t and H t be as in Theorem 2. Then

II IT hlt-det H =(det HII'_ I )UIIH -;;~ IU II / = 1

The proof is an easy induction on n. THEOREM 4: Let H be positive definite hermitian with minimum eigenvalue 11-. Then

/ = 1 i<j

PROOF: Let pn - k be the largest eigenvalue of H n - k. Cauchy's inequalities [5 , pp. 119] yield

II

(i) IT hu ~ 11- ,," - 1, / = n - k +2

(ii) det Hn - k ~ Pn _ kl1-" - k - l ,

The result follows from (i), (i i), (iii), and Theorem 3. THEOREM 5: Let H be positive definite hermitian with largest eigenvalue p. Then

n det H IT hu - det H ~ -2- L Ihijl2. /=1 P i<j

(3)

PROOF: For 1 ~ k ~ n, let H(k) denote the submatrix of H obtained by deleting row and column k. Then

79

Page 4: Improvement of Fischer inequality - NIST Page · 2011-05-17 · Fischer [2; 5, p. 117-118] proved that det HII det H 22 ~ det H. One obtains Hadamard's theorem from Fischer's by induction

(i) (=nU+2 htt ) det H n - k ~ det H(n - k + 1) ,

(ii) det H(n - k + 1) ~ p - I det H,

( ... ) H I * III 112 III U n - k + l n - kUn - k + l ~ p - U n -k+ l .

Number (i) follows by induction on Fischer's inequality or from Schur's theorem. Numbers (ii) and (iii) follow from Cauchy's inequalities. The result follows from (i), (ii), (iii), and Theorem 3.

Theorem 5 can be restated as THEOREM 5': Let 11.1 , ., An be the eigenvalues of the positive definite matrix H.

Then

n (htt) 1 n II ~ - 1 ~ 22 L (AT - hit) ~ o. t = 1 P t = 1

PROOF: Divide (3) by det H = Al ... An and observe that

i <j i = 1 i = 1

The author is pleased to acknowledge helpful discussions with Alan Goldman. Note Added in Proof: It has been pointed out to me that theorem 1 was proved by Frederic 1'.

Metcalf, A Bessel·Schwarz inequality for Gramiams and related bounds for determinants, Ann. Mat. Pura ed Appl. 68,201-232 (1965), corollary 10.2.

3. References

1] Beckenback, E., and Bellman, R., Inequalities (Springer·Verlag, Berlin , 1961).

2] Fischer, E. , Uber den Hadamardschen Determinantensatz , Archiv d. Math. u. Phys. (3), 13,32-40 (1907).

3] Hadamard , J., Resolution d'une question relative aux determi· nants, Bull. Sci. Math. 2, 240-248 (1893).

4] Haynsworth , E., Determination of the inertia of a partitioned hermitian matrix , Lin. Alg. App!. 1,73-81 (1968).

[5] Marcus , M. , and Minc , H., A Survey of Matrix Theory and Matrix Inequalities (Allyn and Bacon, Boston , 1964).

[6] Marcus, M., and Soules, G. , Some inequalities for combinatorial matrix fun!<tions , 1. Comb. Theory 2,145-163 (1967).

[7] Schur, I., Uber endliche Gruppen und Hermitesche Formen, Math. Z. 1, 184-207 (1918).

(Paper 75B1&2-346)

80

(