ON AN INEQUALITY OF MARCUS
MINGHUA LIN
Abstract. We give a simple proof of a determinantal inequality due
to Marcus (1976).
1. Introduction
About 40 years ago, Marcus [1] proved the following result.
Theorem 1.1. [1, Theorem 4] Let A be an m × n complex matrix of rank
m, H = AA∗ and assume 1 ≤ k ≤ m ≤ n. Let X and R be n × p and
m × p matrices respectively, satisfying AX = R, and let κ1 ≥ · · · ≥ κn be
the eigenvalues of XX ∗ . Partition A and R into
A1
R1
A=
,
R=
,
A2
R2
where A1 is k × n, A2 is (m − k) × n, R1 is k × p, R2 is (m − k) × p and let
H11 H12
,
H=
∗
H22
H12
−1
∗
where H11 is k × k. If M = R2 − H12
H11
R1 , then
m−k
m−k
det H Y
det H Y
∗
κn−j+1 ≤ det M M ≤
κj .
det H11 j=1
det H11 j=1
Among others, a converse of the Fischer inequality follows from Theorem
1.1.
Theorem 1.2. [1, Theorem 2] Assume H is an m × m positive definite
Hermitian matrix and H = AA∗ . Partition H into
H11 H12
H=
,
∗
H12
H22
where H11 is k × k, and partition the m × m matrix A in the same way,
A11 A12
A=
.
A21 A22
Then
(1.1)
−1
∗
det H
| det(A22 − H12
H11
A12 )|2
≥
.
det H11 det H22
det H22
2010 Mathematics Subject Classification. 15A45.
Key words and phrases. Determinantal inequality, singular value inequality .
1
2
M. LIN
Equality holds if and only if A12 = 0.
Marcus’ main weapon is the Grassmann algebra and is somewhat sophisticated. In this letter, we make use of a classical inequality for singular
values of matrix product to produce a more transparent proof of Theorem
1.1. An independent simple proof of Theorem 1.2 is also included.
For an m × n complex matrix A, we use σj (A), j = 1, . . . , min{m, n},
to denote the j-th largest singular value of A. By convention, σj (A) = 0
for
n}. The n × n identity matrix is denoted by In . If X =
j > min{m,
X11 X12
with X11 nonsingular, then the Schur complement of X11 in X
X21 X22
−1
is given by X/X11 := X22 − X21 X11
X12 .
2. Proofs
Our proof of Theorem 1.1 relies on the following singular value inequality;
see e.g., [2, p. 364-365].
Lemma 2.1. Let S, T be m × n, n × p complex matrices, respectively. Then
m
m
m
Y
Y
Y
σj (S)σn−j+1 (T ) ≤
σj (ST ) ≤
σj (S)σj (T ).
j=1
j=1
j=1
Proof of Theorem 1.1. Clearly, A1 X = R1 and A2 X = R2 . Thus we may
rewrite M as
M = A2 X − A2 A∗1 (A1 A∗1 )−1 A1 X = ST,
where S = A2 (In − A∗1 (A1 A∗1 )−1 A1 ), T = X.
Note that In − A∗1 (A1 A∗1 )−1 A1 is an orthogonal projection, thus
σj2 (S) = σj (SS ∗ ) = σj A2 (In − A∗1 (A1 A∗1 )−1 A1 )A∗2
∗
∗
∗ −1
∗
= σj A2 A2 − A2 A1 (A1 A1 ) A1 A2
j = 1, . . . , m − k.
= σj (H/H11 ),
Now by Lemma 2.1,
m−k
Y
j=1
2
σj (H/H11 )σn−j+1
(X)
≤
m−k
Y
σj2 (M )
j=1
The desired result follows by noting
≤
m−k
Y
σj (H/H11 )σj2 (X).
j=1
m−k
Y
σj (H/H11 ) =
det H
.
det H11
j=1
H11 A12
Proof of Theorem 1.2. Let K =
. We note that (1.1) is equivalent
∗
H12
A22
to
(2.1)
det H det H11 ≥ | det K|2 .
ON AN INEQUALITY OF MARCUS
3
Compute
det K =
=
=
=
=
H11 A12
det
∗
H12
A22
A11 A∗11 + A12 A∗12 A12
det
A21 A∗11 + A22 A∗12 A22
A11 A∗11 A12
det
A21 A∗11 A22
0
A11 A12 A∗11
det
0 Im−k
A21 A22
∗
det A det A11 .
Thus (2.1) is equivalent to det(A11 A∗11 + A12 A∗12 ) ≥ det A11 A∗11 , which is
obviously true and the equality holds if and only if A12 = 0.
References
[1] M. Marcus, Converses of the Fischer inequality and an inequality of
A. Ostrowski, Linear Multilinear Algebra, 4 (1976) 137-148.
[2] F. Zhang, Matrix Theory: Basic Results and Techniques, Springer, New
York, 2nd ed., 2011.
Department of Mathematics and Statistics, University of Victoria, Victoria, BC, Canada, V8W 2Y2.
E-mail address: [email protected]