Download THE INVERSE MEAN PROBLEM OF GEOMETRIC AND

THE INVERSE MEAN PROBLEM OF GEOMETRIC AND
CONTRAHARMONIC MEANS
YONGDO LIM
Abstract. In this paper we solve the inverse mean problem of contraharmonic
and geometric means of positive definite matrices (proposed in [2])

A = X + Y − 2(X −1 + Y −1 )−1
B = X#Y
by proving its equivalence to the well-known nonlinear matrix equation X = T −
BX −1 B where T = 21 (A+A#(A+8BA−1 B)) is the unique positive definite solution
of X = A + 2BX −1 B. The inverse mean problem is solvable if and only if B ≤ A.
1. Introduction
In [2] Anderson and coauthors have introduced a natural matrix generalization of
the contraharmonic mean of scalars (a2 + b2 )/(a + b) and studied related fixed point
and least squares problems on positive semidefinite matrices. The contraharmonic
mean C(A, B) of positive definite matrices A and B is defined by
C(A, B) = A + B − 2(A−1 + B −1 )−1 .
Inverse mean problems involving the contraharmonic mean are considered and answered for the problem of contraharmonic and arithmetric means in [2] where the
inverse mean problem of contraharmonic and geometric means is proposed even for
the scalar version. The geometric mean A#B of positive definite matrices A and
B is defined by A#B = A1/2 (A−1/2 BA−1/2 )1/2 A1/2 and is well-known in theory of
matrix inequalities (cf. [4], [5], [6], [15]) and appears in semidefinite programming
(scaling point, [20]) and in geometry (geodesic middle, [16], [8], [18]). Although the
inverse mean problem occurs on positive semidefinite matrices A and B, we restrict
our attention to the positive definite case. Similar questions on geometric, arithmetic
and harmonic means are answered in [3].
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2
YONGDO LIM
The inverse mean problem of contraharmonic and geometric means is to find positive definite matrices X and Y for the system of nonlinear matrix equations:

A
B
= C(X, Y )
= X#Y
(1.1)
where A and B are given positive definite matrices of same size. The main result
of this paper is that the inverse mean problem (1.1) is equivalent to a system of the
well-known nonlinear matrix equations (Theorem 2.3):

X
Y
= A + 2BX −1 B
= X − BY −1 B.
(1.2)
The matrix equations X = Q + A∗ X −1 A and X = Q − A∗ X −1 A with Q positive
definite have been studied recently by several authors (see [1], [9], [10], [11], [12], [14],
[17], [19], [21], [22]). For the application areas where the equations arise, see the
references given in [1] and [12].
In section 3, we compute an explicit solution T of the first equation
X = A + 2BX −1 B
of (1.2) and then assert that the inverse mean problem is equivalent to the matrix
equation
X = T − BX −1 B.
Solving directly the matrix equation X = T − BX −1 B we conclude that the inverse
mean problem is solvable if and only if 2B ≤ T, equivalently B ≤ A (Theorem 3.3).
Extreme solutions of the inverse mean problem are also discussed.
2. The inverse mean problem
The following characterization and properties of the geometric mean are wellknown. See [4], [5], [13], [18], and [7] for multivariable geometric means.
Lemma 2.1. (The Riccati Lemma) The geometric mean A#B of positive definite
matrices A and B is the unique positive definite solution of the Riccati equation
XA−1 X = B.
INVERSE MEAN PROBLEM
3
Furthermore, the geometric mean operation has the following properties:
(i) A#B = B#A,
(ii) (A#B)−1 = A−1 #B −1 ,
(iii) M (A#B)M ∗ = (M AM ∗ )#(M BM ∗ ) for any nonsingular matrix M,
(iv) A0 #B 0 ≤ A#B whenever A0 ≤ A and B 0 ≤ B where A ≤ B means that B − A
is positive semidefinite.
For given positive definite matrices A and B, the inverse mean problem (1.1) of
contraharmonic and geometric means is said to be solvable if there exist positive
definite matrices X and Y satisfying the equations of the problem, and in this case
the pair (X, Y ) is called a solution of the problem. One immediately sees that solutions
of the inverse mean problem do not depend on the order from the commutativity of
the contraharmonic and geometric means. That is, if (X, Y ) is a solution of (1.1)
then so is (Y, X). By the Riccati Lemma, the second equation of (1.1) is solved in
terms of X:
B = X#Y
implies Y = BX −1 B.
Thus the inverse mean problem (1.1) can be alternatively written as

A
Y
The matrix (X
−1
+Y
−1 −1
)
= C(X, Y )
= BX
−1
B.
(2.3)
in the first equation of (1.1) or in the contraharmonic
mean C(X, Y ) is known as the parallel addition of X and Y , and has an alternative
expression useful for optimal control theory and algebraic Riccati equations (cf. [2]):
(X −1 + Y −1 )−1 = X − X(X + Y )−1 X.
But, the following expression looks new and will be crucial for our approach.
Lemma 2.2. For positive definite matrices X and Y,
(X −1 + Y −1 )−1 = (X#Y )(X + Y )−1 (X#Y ).
Proof. The Riccati Lemma implies that
X −1 = (Y −1 #X −1 )Y (Y −1 #X −1 )
Y −1 = (X −1 #Y −1 )X(X −1 #Y −1 ).
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YONGDO LIM
It then follows from the commutative and inversion properties of the geometric mean
(Lemma 2.1)
X −1 = (X#Y )−1 Y (X#Y )−1
Y −1 = (X#Y )−1 X(X#Y )−1
and therefore (X −1 + Y −1 ) = (X#Y )−1 (X + Y )(X#Y )−1 . Taking the inverses of the
both sides leads
(X −1 + Y −1 )−1 = (X#Y )(X + Y )−1 (X#Y ).
¤
Theorem 2.3. The inverse mean problem (1.1) is equivalent to the system of nonlinear matrix equations

X
Y
= A + 2BX −1 B
= X − BY −1 B.
(2.4)
More precisely, if we denote S and T by the solution sets of the system (1.1) and
(2.4), respectively, then the maps
Φ : S → T , Φ(X, Y ) = (X + Y, X)
Ψ : T → S, Ψ(X, Y ) = (Y, X − Y )
are bijective and S −1 = T .
Proof. Suppose that (X, Y ) ∈ S. That is, X and Y are positive definite matrices so
that
A = X + Y − 2(X −1 + Y −1 )−1 , B = X#Y.
Setting S := X and T := X + Y, we have
(2.3)
S = X = T − Y = T − BX −1 B = T − BS −1 B.
INVERSE MEAN PROBLEM
5
The equation A = C(X, Y ) = X + Y − 2(X −1 + Y −1 )−1 of (1.1) leads to
T
X + Y = A + 2(X −1 + Y −1 )−1
=
Lemma(2.2)
=
A + 2(X#Y )(X + Y )−1 (X#Y )
=
A + 2BT −1 B.
Therefore, (T, S) = (X + Y, X) is a solution of (2.4).
Conversely, suppose that (X, Y ) is a solution of (2.4). That is, X and Y are positive
definite matrices so that
X = A + 2BX −1 B
Y
and
= X − BY −1 B.
(2.5)
(2.6)
Put T := Y and S := X − Y. Then
X = S + Y = S + T.
(2.7)
It then follows from (2.6) that S = X − Y = BY −1 B > 0 and then by the Riccati
Lemma
B = S#Y = S#T.
(2.8)
The equation (2.5) X = A + 2BX −1 B leads to
A
=
(2.7),(2.8)
=
Lemma(2.2)
X − 2BX −1 B
(T + S) − 2(S#T )(T + S)−1 (S#T )
=
(T + S) − 2(T −1 + S −1 )−1
=
C(S, T ).
Therefore, (T, S) = (Y, X − Y ) is a solution of the inverse mean problem (1.1).
It is obvious to see that the maps Φ and Ψ are bijective and S −1 = T .
¤
Remark 2.4. If (X, Y ) is a solution of (2.4) then (Y, BY −1 B) is a solution of (1.1)
and (Y, BY −1 B) = (Y, X − Y ). Indeed, Y = X − BY −1 B implies that X − Y =
BY −1 B.
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YONGDO LIM
3. Solvability
In this section we give a complete answer for the inverse mean problem (1.1) by
solving the equivalent system of matrix equations (2.4) in Theorem 2.3. First, we
solve the first equation X = A + 2BX −1 B in (2.4). The uniqueness and existence
of a positive definite solution of the matrix equation X = A + BX −1 B appear in
(Proposition 4.1, [12]). However, we compute it explicitly in terms of geometric
means.
Proposition 3.1. Let A and B be positive definite matrices. Then the nonlinear
matrix equation
X = A + BX −1 B
has the unique positive definite solution X = 12 (A + A#(A + 4BA−1 B)).
Proof. By setting Y = A−1/2 XA−1/2 and C = A−1/2 BA−1/2 , we observe that the
matrix equation X = A + BX −1 B is equivalent to
Y = I + CY −1 C.
(3.9)
By Lemma 2.1 the equation (3.9) means that C = Y #(Y − I) = (Y 2 − Y )1/2 and
hence Y 2 − Y − C 2 = 0. This equation has a unique positive definite solution defined
as
1
1
Y = (I + (I + 4C 2 )1/2 ) = (I + I#(I + 4C 2 )).
2
2
Therefore again by Lemma 2.1 the equation (3.9) has a unique positive definite solution
1
X = (A + A#(A + 4BA−1 B)).
2
¤
Proposition 3.2. The inverse mean problem (1.1) is equivalent to the nonlinear
matrix equation
X = T − BX −1 B
where T = 21 (A+A#(A+8BA−1 B)). In particular, if X is a positive definite solution
of X = T −BX −1 B then (X, BX −1 B) = (X, T −X) is a solution of the inverse mean
problem (1.1).
INVERSE MEAN PROBLEM
7
Proof. By Proposition 3.1, the nonlinear equation X = A + 2BX −1 B of the system
(2.4) has the unique positive definite solution T = 21 (A + A#(A + 8BA−1 B)) and
therefore (2.4) and hence (1.1) is equivalent to the single matrix equation
X = T − BX −1 B.
If X is a positive solution of X = T − BX −1 B then (T, X) is a solution of (2.4)
and hence Ψ(T, X) = (X, T − X) = (X, BX −1 B) is a solution of the inverse mean
problem (1.1) by Theorem 2.3 and Remark 2.4.
¤
The solvability criterion of the matrix equation
X = T − BX −1 B
(3.10)
in Proposition 3.2 appeared in Anderson-Morley-Trapp (Theorem 4.2, [1]) and Engwerda [10]: (3.10) has a positive definite solution if and only if
Ã
!
T /2 B
≥ 0,
B T /2
and it is well-known that this
is equivalent to T /2 ≥ B. Indeed for
à requirement
!
A B
positive definite A and B,
≥ 0 if and only if A ≥ BA−1 B (Lemma 3.3,
B A
[18]) if and only if B −1/2 AB −1/2 ≥ B 1/2 A−1 B 1/2 if and only if B −1/2 AB −1/2 ≥ I if
and only if A ≥ B.
We also have a direct proof of solvability of (3.10) as follows. Positive definite
solvability of (3.10) is equivalent to solvability of the equation (T −X)#X = B with a
side condition T > X > 0. With transforms Z = T −1/2 XT −1/2 and D = T −1/2 BT −1/2
this is equivalent to solvability of the equation, with a side condition I > Z > 0,
(I − Z)#Z = (Z − Z 2 )1/2 = D
or Z 2 − Z + D2 = 0.
(3.11)
It is immediate that the solvability in the last form is I − 2D ≥ 0. Therefore we can
conclude that positive definite solvability of the original equation is T ≥ 2B.
Theorem 3.3. The inverse mean problem (1.1) is solvable if and only if B ≤ A. In
this case (X+ , Y+ ), defined as
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YONGDO LIM
1
(T + T #(T − 4BT −1 B)) =
2
1
≡
(T − T #(T − 4BT −1 B)) =
2
X+ ≡
Y+
1
(T + (T + 2B)#(T − 2B)),
2
1
(T − (T + 2B)#(T − 2B))
2
where
1
T = (A + A#(A + 8BA−1 B)),
2
is the extreme solution in the sense that X ≤ X+ and Y+ ≤ Y for all solutions (X, Y )
of (1.1).
Proof. By Theorem 2.3 and Proposition 3.1 we have to consider the solvability of the
matrix equation
1
T = (A + A#(A + 8BA−1 B)).
2
But this is the case if and only if 2B ≤ T , equivalently 4B ≤ A + A#(A + 8BA−1 B).
Y = T − BY −1 B
with
Let C = B −1/2 AB −1/2 . Then by Lemma 2.1 the above inequality becomes
4I ≤ C + C#(C + 8C −1 ) = C + (C 2 + 8I)1/2 .
(3.12)
It is straightforward to see
4≤λ+
√
λ2 + 8 (λ > 0) ⇐⇒ λ ≥ 1.
Hence the inequality (3.12) is equivalent to I ≤ C, hence to B ≤ A.
Now the equation
X = T − BX −1 B
(3.13)
is equivalent to the equation (3.11)
Z 2 − Z + D2 = 0
(3.14)
where Z = T −1/2 XT −1/2 and D = T −1/2 BT −1/2 . Since 2B ≤ T is equivalent to
4D2 ≤ I, the maximum and the minimum solution of (3.14) are respectively
1
(I + (I − 4D2 )1/2 ) =
2
1
=
(I − (I − 4D2 )1/2 ) =
2
Z+ =
Z−
1
(I + (I + 2D)#(I − 2D)) and
2
1
(I − (I + 2D)#(I − 2D)).
2
INVERSE MEAN PROBLEM
9
Therefore again by Lemma 2.1 the maximum and the minimum solutions of (3.13)
are respectively
1
1
X+ ≡
(T + T #(T − 4BT −1 B)) = (T + (T + 2B)#(T − 2B)) and
2
2
1
1
X− ≡
(T − T #(T − 4BT −1 B)) = (T − (T + 2B)#(T − 2B)).
2
2
Since X− = T − X+ , the assertion on the extremity follows from Theorem 2.3.
¤
Remark 3.4. The simple condition A ≥ B is very natural in the sense that it corresponds to the fact that twice of the arithmetic mean is larger than the sum of the
geometric and harmonic means: If (X, Y ) is a solution of (1.1) then
A − B = 2A(X, Y ) − H(X, Y ) − (X#Y )
= (A(X, Y ) − X#Y ) + (A(X, Y ) − H(X, Y ))
where A(X, Y ) and H(X, Y ) are the arithmetic and harmonic means of X and Y.
The arithmetic-geometric-harmonic mean inequality
A(X, Y ) ≥ X#Y ≥ H(X, Y )
implies that A ≥ B.
Acknowledgements. The author is very grateful to the referees for many points
to improve the paper, particularly on Theorem 3.3 and Remark 3.4.
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Department of Mathematics, Kyungpook National University, Taegu 702-701, Korea
E-mail address: [email protected]