Download On Singular Value Inequalities for the Sum of

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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
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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
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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