By Garrett Birkhoff

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**Example text**

Therefore, cT b + d TB−1 d ≥ 0. 11 following the order (1) =⇒ (2) =⇒ (3) =⇒ (1). Proof. 11 holds. 15) that ∀x0 = 0, U(x0 ) > 0 when cT x0 > 0. 4 Several Equivalent SANC for Negative Deﬁnite Derivatives 51 where m is a constant. Thus, cT x¯ = cT B−1 d − cT B−1 d − m T −1 c B c = m. cT B−1 c So for all m ∈ (0, +∞), U(x) ¯ > 0. However, on the other hand, U(x) ¯ = x¯T B x¯ − 2 d T x¯ − cT b, cT B−1 d − m T −1 T c (B ) cT B−1 c cT B−1 d − m −1 B c × B B−1 d − cT B−1 c cT B−1 d − m −1 B c − cTb. 21) Because B is a symmetric positive deﬁnite matrix, so is B−1 , and (B−1 )T = B−1 .

X are ⎧ T −1 ⎨ c B d ≤ 0, 1 ⎩ (cT B−1 d)2 − d T B−1 d − cT b ≥ 0. 4) to prove the following equivalent result. 11. t. x. 50 3 Sufﬁcient Conditions of Absolute Stability: Classical Methods (1) U(x) = xT B x − 2 d Tx − cT b ≥ 0 (cT x = 0), 1. 7; 2. cT B−1 d + cT B−1 c (d T B−1 d + cTb) ≤ 0; ⎧ T −1 2 ⎨ (1) cT b + η TB−1 d − (c B d) ≤ 0, cT B−1 c where d = 12 (AT c + 2 Pb). 3. 11, we need the following lemma. 12. For any symmetric matrix B, we have cT b + d TB−1 d ≥ 0, where d= 1 T (A c + 2 Pb), 2 Proof.

T. 13) where d = P b + 12 AT c. Proof. Since B is a symmetric positive deﬁnite, so is B−1 , and there exists matrix H such that B = H T H, B−1 = H −1 (H −1 )T . 14) ⎪ ⎩ 2 U f (σ ) when f (σ ) = 0, where U= H x − (H −1 )d f (σ ) T H x − (H −1)d − (d T B−1 d + cTb). 15) 48 3 Sufﬁcient Conditions of Absolute Stability: Classical Methods 1 we only need to prove that for any cT x ≥ , we have U > 0. To achieve this, k introduce the nonsingular transform y = H x − (H −1)T d under which U becomes U = yT y − ρ , where ρ = d T B−1 d + cTb.

### Lattice theory (Colloquium publications - American Mathematical Society) by Garrett Birkhoff

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