Matrix Note 12
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Proof. Without loss of generality, assume that S = {1, . . . , k } and P = A B where A = P (S ) ∈ Hk . Hence A − BC −1 B ∗ 0 and B∗ C P −1 = I 0 ∗ −1∗ B A I A−1 0 ∗ −1 0 (C − B A B )−1 I A−1 B . 0 I
n n
f (αI + (1 − α)Λ) =
i=1
log(1 + (1 − α)λi ) ≥
i=1 n
[log 1 + (1 − α) log λi ]
= (1 − α)
i=1
log λi = (1 − α)f (Λ)
with equality holds if λi = 1, i.e. Λ = I and B = A. Theorem 5. Let A, B 0 and let 0 ≤ α ≤ 1. Then
Proof. It is known that (c) and (d) are equivalent. (a) implies (b) is trivial. (b) implies (c) by putting x = A−1/2 BC −1/2 y . (c) implies (a) by showing that |x∗ (A−1/2 BC −1/2 y )|2 ≤ (x∗ x)(y ∗ C −1/2 B ∗ A−1 BC −1/2 y ) ≤ (x∗ x)(y ∗ y ) and then replace x by A1/2 x and y by C 1/2 y . Theorem 11. Let P 0 and S ⊆ {1, . . . , n}, then P −1 (S ) P (S )−1 .
=
A 0 . 0 C − B ∗ A−1 B
Theorem 10. Let A ∈ Hn and C ∈ Hm be positive definite and let B ∈ Mn,m (C). The following are equivalent: 1. (x∗ Ax)(y ∗ Cy ) ≥ |x∗ By |2 for all x ∈ Cn and y ∈ Cm . 2. x∗ Ax + y ∗ Cy ≥ 2|x∗ By | for all x ∈ Cn and y ∈ Cm . 3. ρ(B ∗ A−1 BC −1 ) = ρ(C −1/2 B ∗ A−1 BC −1/2 ) ≤ 1. 4. A B B∗ C 0
det(αA + (1 − α)B ) ≥ (det A)α (det B )1−α with equality if andΒιβλιοθήκη Baiduonly if A = B .
3
2
Positive semidefinite ordering
B if A − B 0. Similarly,
Definition 1. Let A, B ∈ Hn . We write A A B if A − B 0. Examples: 6 4 4 3 • 2 3 4 1 • A2 + I • A+B
Then P −1 (S ) = (A − BC −1 B ∗ )−1 and P (S )−1 = A−1 . Since C 0, BC −1 B ∗ 0 and thus A A − BC −1 B ∗ P −1 (S ) P (S )−1 . 5
0, we have
A1/2 BA1/2 and A1/2 BA1/2 is similar to
Theorem 8. Suppose A, B 1. A 2. If A
0, then A−1 ;
B if and only if B −1
B then det A ≥ det B and tr A ≥ tr B ;
3. λk (A) ≥ λk (B ) for k = 1, 2, . . . , n. Proof. (c) is just Weyl’s theorem.
2
The change of variable is therefore resulting formulae become
z w
= U ∗ Du =
x−y . The 2x
5x2 − 2xy + y 2 = z 2 + w2 and 1 x2 + 2xy − y 2 = −z 2 + w2 . 2
Theorem 4. The function f (A) = log det A is a strictly concave function on the convex set of positive definite Hermitian matrices in Mn (C). Proof. Take A, B 0, we want to show f (αA + (1 − α)B ) ≥ αf (A) + (1 − α)f (B ). Write A = CC ∗ and B = CλΛC ∗ for some nonsingular C and Λ = diag (λ1 , . . . , λn ). Then f (αA + (1 − α)B ) = f (CC ∗ ) + f (αI + (1 − α)Λ) = f (A) + f (αI + (1 − α)Λ) and αf (A)+(1 − α)f (B ) = αf (A)+(1 − α)(f (CC ∗)+ f (Λ)) = f (A)+(1 − α)f (Λ). Now
Matrix Analysis Lecture 12. Positive definite matrices (2)
7 November 2015
In this lecture, we will continue to discuss the positive definite matrices. We will study the congruence relation and the positive definite matrix ordering.
1
1
Congruence
Theorem 1. Let A, B ∈ Hn and A 0. Then AB is a diagonalizable matrix with real eigenvalues, with the same number of positive, negative and zero eigenvalues as B . Furthermore, every diagonalizable matrices with real eigenvalues is the product of a positive definite matrix and a Hermitian matrix. Proof. A−1/2 ABA1/2 = A1/2 BA1/2, therefore AB is similar to a Hermitian matrix which is ∗-congruence to B . For the last assertion, if C = SDS −1 then C = SS ∗ (S −1∗ DS −1 ). Theorem 2. Let A, B ∈ Hn and there exists s, t ∈ R such that sA + tB 0. Then there exists a nonsingular C ∈ Mn (C) such that both C ∗ AC and C ∗ BC are diagonal. Proof. As sA + tB 0, it is ∗-congruent to In . There exists nonsingular C1 ∗ ∗ such that sC1 AC1 + tC1 BC1 = I . Note that one of s and t is nonzero, and so we may assume that s = 0. Take a unitary matrix such that (C1 U )∗ B (C1 U ) is diagonal, then (C1 U )∗ A(C1 U ) is diagonal as well. Theorem 3. Let A, B ∈ Hn and A 0. Then there exists a nonsingular C ∈ Mn (C) such that A = CC ∗ and B = C ΛC ∗ where Λ are diagonal. Indeed if D = A1/2 and D−1 BD−1 = U ΛU ∗ then C can be chosen to be DU . Examples: Find a single change of variables such that both 5x2 − 2xy + y 2 and 2 x + 2xy − y 2 are weighted sums of squares. Solution: Let u = (x, y )t , then 5x2 − 2xy + y 2 = ut Au where A = 5 −1 1 1 and x2 + 2xy − y 2 = ut Bu where B = . −1 1 1 −1 1 7 −1 D = A1/2 = √ − 1 3 10 and 1 −1 0 −1 3 D−1 BD−1 = U ∗ U where U = √ . 0 1/2 10 3 1
2 3 4 1
4 1 1 3
3 1 2 4
1 2 4 0 3 1 3 1 −2 1 −2 0
2A for all A ∈ Hn . 2A1/2 B 1/2 for all A, B ∈ Hn with AB = BA.
Theorem 6. Suppose A, B ∈ Hn and T ∈ Mm,n (C). If A B then T AT ∗ T BT ∗ . If A B then T AT ∗ T BT ∗ . Theorem 7. If A 0 and B 0, then A and A B if and only if ρ(BA−1 ) < 1. Proof. A BA−1 . B if and only if I B if and only if ρ(BA−1 ) ≤ 1
4
Theorem 9. The Hermitian matrix C
A B 0 if and only if A B∗ C B ∗ A−1 B or equivalently ρ(B ∗ A−1 BC −1 ) < 1. I 0 −B A−1∗ I
∗
0 and
Proof.
A B B∗ C
I −A−1 B 0 I