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1 On Singular Value Inequalities for the Sum of Two Matrices arXiv:1507.06630v1 [math.GM] 8 Jul 2015 Sergey Loyka Abstract A counter-example to lower bounds for the singular values of the sum of two matrices in [1] and [2] is given. Correct forms of the bounds are pointed out. AMS Classification: 15A18, 15A42 Keywords: Singular value; Inequality; Matrix The following lower bound for singular values of the sum of two n × n matrices A and B is reported in [1] (see G.3.b on p. 334): k X σi (A + B) ≥ i=1 k X i=1 σi (A) − k X σn−i+1 (B), 1 ≤ k ≤ n, (1) i=1 where singular values σi are in decreasing order, σ1 ≥ σ2 ≥ .... This lower bound is also claimed in [2] (see Problem 4 in Sec. 10.5, p. 361). For k = 1, this reduces to σ1 (A + B) ≥ σ1 (A) − σn (B) (2) To see that this does not hold in general, consider the following example: A = diag{1, 0}, B = diag{−1, 0}, which results in 0 = σ1 (A + B) ≥ σ1 (A) − σn (B) = 1 (3) This example also disproves the lower bound given in Theorem 8.13 of [2], namely σi (A + B) ≥ σi (A) + σn (B) (4) S. Loyka is with the School of Electrical Engineering and Computer Science, University of Ottawa, Ontario, Canada, K1N 6N5, e-mail: [email protected]. July 24, 2015 DRAFT 2 The correct form of this bound is σi (A + B) ≥ σi (A) − σ1 (B) (5) which follows from e.g. Theorem 3.3.16(c) in [3]. An inspection of the ”proof” in [1] reveals the following incorrect step min Re trUBV+ = − k X σn−i+1 (B) (6) i=1 where min is over all semi-unitary matrices U, V such that UU+ = VV+ = Ik and Ik is k × k identity matrix. The correct form of this step is as follows: min Re trUBV+ = − max Re trU(−B)V+ = − k X σi (−B) = − i=1 k X σi (B) (7) i=1 where 2nd equality is due to e.g. Theorem 3.4.1 in [3], and the last equality is due to σi (−B) = σi (B), which results in the following lower bound: k X σi (A + B) ≥ k X σi (A) − k X σi (B) (8) i=1 i=1 i=1 This lower bound is not the tightest one in general. A tighter lower bound can be established using e.g. Theorem 3.4.5. in [3]: k X σi (A − B) ≥ i=1 k X s[i] (9) i=1 where si = |σi (A) − σi (B)| and s[i] are in decreasing order (note that while σi are in decreasing order, si do not need to be so), from which it follows that k X σi (A + B) ≥ i=1 k X s[i] i=1 = ≥ ≥ max 1≤i1 <...<ik ≤n max 1≤i1 <...<ik ≤n k X k X |σil (A) − σil (B)| l=1 k X (σil (A) − σil (B)) l=1 (σi (A) − σi (B)) (10) i=1 where 1st inequality follows from (9) via the substitution B → −B. It should be noted that (7)-(10) hold for rectangular matrices as well. July 24, 2015 DRAFT 3 R EFERENCES [1] A.W. Marshall, I. Olkin, B.C. Arnold, Inequalities: Theory of Majorization and Its Applications, Springer, New York, 2011. [2] F. Zhang, Matrix Theory, Springer, New York, 2011. [3] R.A. Horn, C.R. Johnson, Topics in Matrix Analysis, Cambridge University Press, 1991. July 24, 2015 DRAFT