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After substituting into f and simple manipulations, we get 1 min f (x) = − bT A+ b. 10) x∈Rn In particular, if A is positive deﬁnite, then 1 min f (x) = − bT A−1 b. 11) x∈Rn The formulae for the minimum of the unconstrained minimization problems can be used to develop useful estimates. 37) to get 1 1 1 f (x) ≥ − bT A+ b = − bT A† b ≥ − A† 2 2 2 b 2 = − b 2 /(2λmin ), where A† denotes the Moore–Penrose generalized inverse and λmin denotes the least nonzero eigenvalue of A. In particular, it follows that if A is positive deﬁnite and λmin denotes the least eigenvalue of A, then for any x ∈ Rn 1 1 f (x) ≥ − bT A−1 b ≥ − A−1 2 2 b 2 = − b 2 /(2λmin ).
20) both λ1 λ2 > 0 and λ1 + λ2 > 0 for suﬃciently large values of . Since the latter implies that at least one of the eigenvalues of H is positive for suﬃciently large , it follows from λ1 λ2 > 0 that λ1 > 0 and λ2 > 0 provided is suﬃciently large. 40). We often use bounds on the spectrum of some matrix expressions with penalized matrices that are based on the following lemma. 4. Let m < n be given positive integers, let A ∈ Rn×n denote a symmetric positive deﬁnite matrix, and let B ∈ Rm×n . 42) μi = βi /(1 + βi ), i = 1, .
Let m < n be positive integers, let A ∈ Rn×n denote a symmetric positive deﬁnite matrix, let B ∈ Rm×n , and let r denote the rank of the matrix B. 43) ImB = ImBA−1 = ImBA−1 BT and the eigenvalues β i of BA−1 BT |ImB of the restriction of BA−1 BT to ImB are related to the positive eigenvalues β1 ≥ β2 ≥ · · · ≥ βr of BA−1 BT by β i = βi /(1 + βi ), i = 1, . . , r. 44) Proof. First observe that if C ∈ Rn×n is nonsingular and B ∈ Rm×n , then Bx = BC(C−1 x), so that ImB = ImBC. 30) to get ImB = ImBA−1 = ImBA−1/2 = ImBA−1/2 (BA−1/2 )T = ImBA−1 BT .