Sufficiency of Signed Principal Minors for Semidefiniteness: A Relatively Easy Proof

Transcription

1 Sufficiency of Signed Principal Minors for Semidefiniteness: A Relatively Easy Proof David M. Mandy Department of Economics University of Missouri 118 Professional Building Columbia, MO USA mandyd@missouri.edu September 23, 2016 Abstract. Extant proofs that signed principal minors is sufficient for semidefiniteness of a Hermitian matrix are relatively inaccessible to economics audiences. The paper presents a new proof that relies only on facts familiar to economics graduate students. Keywords. Semidefinite, Hermitian, Matrix, Principal Minors. EL Code. C02. 1

2 1 Introduction Characterizing semidefiniteness of a matrix is a frequent need in econometrics and economic theory. Sometimes the matrix is definite and this is established with relative ease by studying the signs of leading principal minors. However some contexts require that the semidefinite case be considered explicitly. For example, duality for price-taking economic actors requires that profit (cost/expenditure) be a linear homogeneous and convex (concave) function of prices. Linear homogeneity implies the Hessian is singular (when it exists), whence the matrix under study in these settings cannot be definite. Another example is necessary second order conditions for an extremum. The necessary, but not sufficient, conditions are implied by maximizing behavior so semidefiniteness is all that can be inferred from the postulated behavior. Although it is well-known that all principal minors must be signed to establish semidefiniteness, proofs in the economics literature are scarce. Standard mathematical economics books including Chiang [2, pp ], de la Fuente [4, p. 270], Lancaster [8, pp ], Simon and Blume [9, Theorem 16.2], Sydsæter et al. [10, Theorem 1.7.1] and Takayama [11, Theorem 1.E.11] do not prove the semidefinite case. Extant proofs typically work directly with the quadratic form or characteristic equation and use a limit argument [3] [1, p. 223] [5, p. 307], making them relatively inaccessible to many economists. This short note provides an alternate proof that signing all of the principal minors is sufficient for semidefiniteness of a Hermitian matrix. The method of proof enables relatively easy consideration of many situations with a singular matrix. The proof relies on six facts about matrices which, for clarity of the background needed to understand the proof, are listed at the outset. These six facts have long been familiar to economics graduate students. For example, they are all succinctly proven in Chapter 4 of ohnston [7]. 1 With these six facts in hand, the proof herein uses nothing more than simple observations about the entries in the matrix. 2 Notation and Reminders x s = y means the real numbers x and y have the same sign. ᾱ denotes the complex conjugate of α C 1. ᾱα is real and nonnegative, and is zero if and only if α = 0. Given a matrix A with complex entries a ij, A denotes the conjugate transpose of A (the (j, i) entry of A is ā ij ). If A C n (a vector) then A A is real and nonnegative, and is zero if and only if A = 0. Given an n n complex matrix A, let {1,..., n} denote an index set of the rows and columns of A ( ). A denotes the principal submatrix of A consisting of the rows 1 ohnston works only with real matrices. Everything herein is stated for complex matrices because doing so adds little complication. 2

3 and columns in, and # denotes the number of elements in (the order of the principal submatrix). For the special cases = {1,..., i} for i = 1,..., n, A is denoted A i (i.e., A i is the i i leading principal submatrix of A, and A n = A). A 0 1 for notational convenience. The determinant A is a principal minor of order #. Enumerating the elements of in ascending order as = {j 1,..., j # } yields a permutation matrix P with entries p ij = 0 for i, j = 1,..., n except p iji = 1 for i = 1,..., # and p ii = 1 for i = # + 1,..., n. When symmetrically applied to A, P makes A a leading principal submatrix. As with all permutation matrices, P P = I n. A is Hermitian (symmetric, in the real case) if A = A. A Hermitian matrix is square and has real diagonal entries. An n n matrix A is positive (negative) definite if x Ax > (<) 0 for every nonzero vector x C n. If the defining inequality is weak then A is positive (negative) semidefinite. Implicit in this definition is that the product x Ax is real for every nonzero x C n. 2 3 Six Familiar Facts about Complex Matrices 1. (AB) = B A for conformable matrices A and B. 2. αa = α n A for α C 1 and an n n matrix A. 3. If a square matrix A is partitioned as: [ ] A11 A A = 12, A 21 A 22 where A 11 is square (i.e., a leading principal submatrix) and nonsingular, then: A = A 11 A 22 A 21 A 1 11 A Every positive (semi) definite (Hermitian) matrix has a unique positive (semi) definite (Hermitian) square root matrix. 5. (The definite case; necessity) Every principal minor of a positive definite (Hermitian) matrix is positive. 6. (The definite case; sufficiency) If every leading principal minor of a Hermitian matrix is positive then the matrix is positive definite. 2 A is Hermitian if and only if x Ax is real for every x C n [6, Theorem 4.1.4]. We do not use this fact directly but it shows that semidefinite matrices are always Hermitian. 3

4 4 Main Theorem Theorem 1. If the n n Hermitian matrix A has a principal submatrix A satisfying: 1. A is positive (negative) definite, 2. No higher order principal submatrix of A is positive (negative) definite, and 3. A K 0 (has sign ( 1) #K or is zero) for every K satisfying K {1,..., n}; then A is positive (negative) semidefinite. Proof. Begin with the positive semidefinite case. If A = A then A is trivially positive semidefinite, so assume A excludes some rows and (the same) columns. Then: [ ] P AP A B = B C for some matrices B and C (C is Hermitian). Consider an arbitrary column b i from B and the corresponding diagonal entry c ii from C. Using A > 0 from item 1 (Fact 5) and Fact 3 yields: A b i = A c ii b i A 1 b i = s c ii b i A 1 b i. b i c ii This is nonnegative due to item 3. If it is positive then the principal submatrix under consideration is positive definite (Facts 5 and 6), contradicting item 2, so c ii b i A 1 b i = 0. Now consider any two columns from B and the corresponding 2 2 submatrix from C: A b i b j [ ] [ ] b i c ii c ij b j c ij c jj = A cii c ij b c i ij c jj b A 1 [ ] bi b j j s = 0 c ij b i A 1 b j c ij b j A 1 b i 0 = (c ij b i A 1 b j)(c ij b ja 1 b i) = (c ij b i A 1 b j)(c ij b i A 1 b j). The last equality uses Fact 1. This is again nonnegative due to item 3, which implies c ij b i A 1 b j = 0 [ for every i, j ] (including i = j). Therefore C = B A 1 B. Now use Fact 4 to define F = A 1/2 B and using Fact 1 note that: A 1/2 [ ] F A 1/2 [ ] [ ] [ ] F = B A 1/2 A 1/2 A 1/2 A B A B B = B B A 1 B = B. C 4

5 For any nonzero x C n, let y(x) = F P x. Then, again using Fact 1: x Ax = x P F F P x = y(x) y(x) 0. Hence A is positive semidefinite. For the negative semidefinite case, replace A in the above argument by A and A by A, noting that (1) A is positive definite, (2) no higher order principal submatrix of A is positive definite, and (3) A K = ( 1) #K A K 0 for every K satisfying K {1,..., n} (Fact 2). The conclusion is A is positive semidefinite, or A is negative semidefinite. 5 Applications Theorem 1 readily gives the sufficient condition for semidefiniteness. Corollary 1. If every principal minor of a Hermitian matrix A is nonnegative (of order # has sign ( 1) # or is zero) then A is positive (negative) semidefinite. Proof. If every diagonal entry of A is zero then: 0 a ii a ij = a ijā ij for every i, j = 1,..., n. ā ij a jj This implies a ij = 0 for every i, j. That is, A is a matrix of zeros, and is therefore trivially positive (negative) semidefinite. So assume there is a nonzero diagonal entry. That entry is a positive (negative) definite principal submatrix, so A has a highest order positive (negative) definite principal submatrix A satisfying the assumptions of Theorem 1. In most duality applications even though the matrix under consideration is singular there is an (n 1) dimensional submatrix that is definite. That definiteness is verified by checking only (n 1) leading principal minors, and is therefore much easier than attempting to verify semidefiniteness directly with Corollary 1. Theorem 1 shows that doing so suffices to establish semidefiniteness of the original matrix. Corollary 2. Assume A is an n n singular Hermitian matrix. If A has an (n 1) st order positive (negative) definite principal submatrix A then A is positive (negative) semidefinite. Proof. The only principal submatrix of higher order than A is A, and A = 0. Apply Theorem 1. Acknowledgements. I am grateful to onathan Hamilton and. Isaac Miller for helpful discussions. 5

6 References [1] Abadir, KM and R Magnus. Matrix Algebra. Cambridge University Press, Cambridge (2005). [2] Chiang, A. Fundamental Methods of Mathematical Economics (third ed.). Mcgraw-Hill, New York (1984). [3] Debreu, G. Definite and Semidefinite Quadratic Forms. Econometrica 20 (1952) pp [4] de la Fuente, A. Mathematical Methods and Models for Economists. Cambridge University Press, Cambridge (2000). [5] Gantmacher, FR. The Theory of Matrices, Vol. 1. Chelsea, New York (1959). [6] Horn, RA and CR ohnson. Matrix Analysis (second ed.). Cambridge University Press, Cambridge (2013). [7] ohnston,. Econometric Methods (third ed.). McGraw-Hill, New York (1984). [8] Lancaster, K. Mathematical Economics. Macmillian, London (1970). [9] Simon, CP and L. Blume. Mathematics for Economists. Norton, New York (1994). [10] Sydæeter, K, P Hammond, A Seierstad and A Strøm. Further Mathematics for Economic Analysis (second ed.). Prentice-Hall, London (2008). [11] Takayama, A. Mathematical Economics (second ed.). Cambridge University Press, Cambridge (1985). 6