Volume 2010, Article ID 897279,11pages doi:10.1155/2010/897279
Research Article
On The Frobenius Condition Number of
Positive Definite Matrices
Ramazan T ¨urkmen and Z ¨ubeyde Uluk ¨ok
Department of Mathematics, Science Faculty, Selc¸uk University, 42003 Konya, Turkey
Correspondence should be addressed to Ramazan T ¨urkmen,[email protected]
Received 19 February 2010; Revised 4 May 2010; Accepted 15 June 2010
Academic Editor: S. S. Dragomir
Copyrightq2010 R. T ¨urkmen and Z. Uluk ¨ok. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We present some lower bounds for the Frobenius condition number of a positive definite matrix depending on trace, determinant, and Frobenius norm of a positive definite matrix and compare these results with other results. Also, we give a relation for the cosine of the angle between two given real matrices.
1. Introduction and Preliminaries
The quantity
κA
⎧ ⎨ ⎩
AA−1 ifA is nonsingular,
∞ ifAis singular 1.1
is called the condition number for matrix inversion with respect to the matrix norm·. Notice thatκA A−1A ≥ A−1AI ≥1 for any matrix normsee, e.g.,1, page 336. The
condition numberκA AA−1of a nonsingular matrixAplays an important role in the
numerical solution of linear systems since it measures the sensitivity of the solution of linear systemsAxbto the perturbations onAandb. There are several methods that allow to find good approximations of the condition number of a general square matrix.
whereA∗denotes the conjugate transpose ofA. A Hermitian matrixAis said to be positive semidefinite or nonnegative definite, written asA≥0, ifsee, e. g.,2, p.159
x∗Ax≥0, ∀x∈Cn, 1.2
Ais further called positive definite, symbolizedA >0, if the strict inequality in1.2holds for all nonzerox∈Cn. An equivalent condition forA∈Cn×nto be positive definite is thatA is Hermitian and all eigenvalues ofAare positive real numbers.
The trace of a square matrixAthe sum of its main diagonal entries, or, equivalently, the sum of its eigenvalues is denoted by trA. Let Abe any m×n matrix. The Frobenius
Euclideannorm of matrixAis
AF
⎛ ⎝m
i1 n
j1
aij 2
⎞ ⎠
1/2
. 1.3
It is also equal to the square root of the matrix trace ofAA∗, that is,
AF √trAA∗. 1.4
The Frobenius condition number is defined byκFA AFA−1F. InRn×nthe Frobenius
inner product is defined by
A, BF trATB 1.5
for which we have the associated norm that satisfiesA2F A, AF. The Frobenius inner product allows us to define the cosine of the angle between two given realn×nmatrices as
cosA, B A, BF
AFBF. 1.6
The cosine of the angle between two real n×nmatrices depends on the Frobenius inner product and the Frobenius norms of given matrices. Then, the inequalities in inner product spaces are expandable to matrices by using the inner product between two matrices.
Buzano in3obtained the following extension of the celebrated Schwarz inequality in a real or complex inner product spaceH; ·,·:
| a, x x, b| ≤ 1
2ab| a, b|x
2, 1.7
for any a, b, x ∈ H. It is clear that for a b, the above inequality becomes the standard Schwarz inequality
with equality if and only if there exists a scalarλ ∈K K RorCsuch thatx λa. Also Dragomir in4has stated the following inequality:
a, x x, b x2 −
a, b
2
≤ ab
2 , 1.9
wherea, b, x∈H, x /0. Furthermore, Dragomir4has given the following inequality, which is mentioned by Precupanu in5, has been showed independently of Buzano, by Richard in
6:
1
2 a, b − abx
2≤ a, x x, b ≤ 1
2 a, babx
2. 1.10
As a consequence, in next section, we give some bounds for the Frobenius condition numbers and the cosine of the angle between two positive definite matrices by considering inequalities given for inner product space in this section.
2. Main Results
Theorem 2.1. LetAbe positive definite real matrix. Then
2 trA
detA1/n −n≤κFA, 2.1
whereκFAis the Frobenius condition number.
Proof. We can extend inequality1.9given in the previous section to matrices by using the Frobenius inner product as follows: LetA, B, X∈Rn×n. Then we write
A, XF X, BF
X2 F
− A, BF
2
≤ AFBF
2 , 2.2
where A, XF trATX,and · Fdenotes the Frobenius norm of matrix. Then we get
trATXtrXTB
X2 F
−tr
ATB
2
≤ AFBF
2 . 2.3
In particular, in inequality2.3, if we takeBA−1, then we have
trATXtrXTA−1
X2 F
−tr
ATA−1
2
≤ AFA−1F
Also, ifXandAare positive definite real matrices, then we get
trAXtrXA−1
X2 F
−n
2
≤ AFA−1F
2
κFA
2 , 2.5
whereκFAis the Frobenius condition number ofA.
Note that Dannan in7has showed the following inequality by using the well known arithmetic-geometric inequality, forn-square positive definite matricesAandB:
ndetAdetBm/n≤trAmBm, 2.6
wheremis a positive integer. If we takeAX,BA−1, andm1 in2.6, then we get
ndetXdetA−11/n≤trXA−1. 2.7
That is,
n
detX detA
1/n
≤trXA−1. 2.8
In particular, if we takeXIin2.5and2.8, then we arrive at
trAtrA−1
n −
n
2
≤κFA,
n
1 detA
1/n
≤trA−1.
2.9
Also, from the well-known Cauchy-Schwarz inequality, sincen2≤trAtrA−1, one can obtain
0< n≤2trAtrA
−1
n −n≤κFA. 2.10
Furthermore, from arithmetic-geometric means inequality, we know that
ndetA1/n≤trA. 2.11
Sincen≤trA/detA1/n, we write 0< n≤2trA/detA1/n−n. Thus by combining2.9
and2.11we arrive at
2 trA
Lemma 2.2. LetAbe a positive definite matrix. Then
trA3/2trA−1/2
trA −
n
2 ≥0. 2.13
Proof. Letλibe positive real numbers fori1,2, . . . , n. We will show that
k
i1 λ3/2
i
k
i1 λ−i1/2
≥ k 2 k i1 λi 2.14
for allk1,2, . . . , n. The proof is by induction onk. Ifk1,
λ3/2 1 ·λ
−1/2 1 λ1≥
1
2λ1. 2.15
Assume that inequality2.14holds for somek. that is,
k
i1 λ3/2
i
k
i1 λ−i1/2
≥ k 2 k i1 λi
. 2.16
Then
k1
i1 λ3/2
i
k1
i1 λ−i1/2
k
i1 λ3/2
i λ3k/12
k
i1
λ−i1/2λ−k11/2
k
i1 λ3/2
i
k
i1 λ−i1/2
k
i1
λ3/2
i λ−k11/2λi−1/2λ3k1/2
λk1
≥ k 2 k i1 λi k i1
λiλk1 λk1
≥ k
2
k
i1 λi12
k
i1
λiλk1 λk21
k1
2
k1
i1 λi
.
2.17
The first inequality follows from induction assumption and the inequality
a2b2 ab ≥
ab
2 ≥
ab 2.18
Theorem 2.3. LetAbe positive definite real matrix. Then
0≤2n trA
3/2
trAdetA1/2n −n≤κFA, 2.19
whereκFAis the Frobenius condition number.
Proof. LetX >0 andA >0. Then from inequality1.9we can write
A, XFX, A−1F
X2 F
−
A, A−1 F
2
≤ AFA−1F
2 2.20
where A, BF trATBand · denotes the Frobenius norm. Then we get
trAXtrXA−1
X2 F
−n
2
≤ κFA
2 . 2.21
SetXA1/2. Then
trA3/2trA−1/2
trA −
n
2
≤ κFA
2 . 2.22
SincetrA3/2trA−1/2/trA−n/2≥0 byLemma 2.2andndetA−1/21/n≤
trA−1/2,
trA3/2
trA n
detA−1/21/n− n 2 ≤
trA3/2trA−1/2
trA −
n
2 ≤
κFA
2 . 2.23
Hence
2n trA
3/2
trAdetA1/2n −n≤κFA. 2.24
Let λi be positive real numbers fori 1,2, . . . , n. Now we will show that the left side of
inequality2.19is positive, that is,
2
n
i1 λ3/2
i ≥ n i1 λi n i1 λ1/2n
i
. 2.25
By the arithmetic-geometric mean inequality, we obtain the inequality
1 n n i1 λi n i1 λ1/2
i ≥ n i1 λi n i1 λ1/2n
i
So, it is enough to show that
2
n
i1 λ3/2
i ≥ 1 n n i1 λi n i1 λ1/2
i
. 2.27
Equivalently, 2n n i1 λ3 i ≥ n i1 λ2 i n i1 λi
. 2.28
We will prove by induction. Ifk1, then
2λ31≥λ21·λ1λ31. 2.29
Assume that the inequality2.28holds for somek. Then
2k1
k1
i1 λ3 i 2k k i1 λ3
i 2 k
i1 λ3
i 2kλ3k12λ3k1
≥ k i1 λ2 i k i1 λi 2 k i1 λ3
i λ3k1
2λ3k1
≥ k i1 λ2 i k i1 λi 2 k i1 λ2
iλk1λiλ2k1
2λ3k1
≥ k i1 λ2 i k i1 λi k i1 λ2
iλk1 k
i1
λiλ2k1λ3k1
k1 i1 λ2 i k1 i1 λi . 2.30
The first inequality follows from induction assumption and the second inequality follows from the inequality
a3b3≥a2bab2 2.31
Theorem 2.4. LetAandBbe positive definite real matrices. Then
cosA, IcosB, I≤ 1
2cosA, B 1. 2.32
In particular,
cosA, A−1≤cosA, IcosA−1, I≤ 1 2
cosA, A−11≤1. 2.33
Proof. We consider the right side of inequality1.10:
a, x x, b ≤ 1
2 a, babx
2. 2.34
We can extend this inequality to matrices as follows:
A, XF X, BF ≤ 1
2 A, BF AFBFX
2
F 2.35
whereA, X, B∈Rn×n. Since A, XFtrATX, it follows that
trATXtrXTB≤ 1 2
trATBAFBFX2F, 2.36
LetXbe identity matrix andAandBpositive definite real matrices. According to inequality
2.36, it follows that
trAtrB≤ 1
2trABAFBFn, trAtrB
√
nAF√nBF ≤
1 2
trAB
AFBF 1
.
2.37
From the definition of the cosine of the angle between two given realn×nmatrices, we get
cosA, IcosB, I≤ 1
2cosA, B 1. 2.38
In particular, forBA−1we obtain that
cosA, IcosA−1, I≤ 1 2
Also, Chehab and Raydan in8have proved the following inequality for positive definite real matrixAby using the well-known Cauchy-Schwarz inequality:
cosA, A−1≤cosA, IcosA−1, I. 2.40
By combining inequalities2.39and2.40, we arrive at
cosA, A−1≤cosA, IcosA−1, I≤ 1 2
cosA, A−11 2.41
and since1/2cosA, A−1 1 n/2A
FA−1F 1/2andn ≤ κFA, we arrive at
1/2cosA, A−1 1≤1. Therefore, proof is completed.
Theorem 2.5. LetAbe a positive definite real matrix. Then
n√nAF
trA ≤κFA. 2.42
Proof. According to the well-known Cauchy-Schwarz inequality, we write
n
i1 λiA
2
≤
n
i1 λ2
iA
n, 2.43
whereλiAare eigenvalues ofA. That is,
trA2≤ntrA2. 2.44
Also, from definition of the Frobenius norm, we get
trA≤√nAF. 2.45
Then, we obtain that
cosA, I √trA
nAF ≤1. 2.46
Likewise,
When inequalities2.40and2.47are combined, they produce the following inequality:
cosA, A−1≤cosA, I,
n κFA ≤
trA
√
nAF.
2.48
Therefore, finally we get
n√nAF
trA ≤κFA. 2.49
Note that Tarazaga in9has given that ifAis symmetric matrix, a necessary condition to be positive semidefinite matrix is that trA≥ AF.
Wolkowicz and Styan in10have established an inequality for the spectral condition numbers of symetric and positive definite matrices:
κ2A≥1 2s
m−s/p, 2.50
wherep√n−1,mtrA/n, ands A2F/n−m21/2.
Also, Chehab and Raydan in8have given the following practical lower bound for the Frobenius condition numberκFA:
κFA≥max
n,
√
n
cos2A, I,1
2s
m−s/p
. 2.51
Now let us compare the bound in2.49and the lower bound obtained by the authors in8
for the Frobenius condition number of positive definite matrixA. Since 0≤ AF/trA≤1,A2F/trA2≤ AF/trA. Thus, we get
n√nA2 F
trA2 ≤
n√nAF
trA ≤κFA. 2.52
All these bounds can be combined with the results which are previously obtained to produce practical bounds forκFA. In particular, combining the results given by Theorems
2.1,2.3, and2.5and other results, we present the following practical new bound:
κFA≥max
2 trA
detA1/n−n,2n
trA3/2
trAdetA1/2n −n,
n√nAF
trA ,1 2s
m−s/p
Example 2.6.
A
⎡ ⎢ ⎢ ⎣
4 1 0 2 1 5 1 2 0 1 6 3 2 2 3 8
⎤ ⎥ ⎥
⎦. 2.54
Here trA 23, AF √179, detA 581, and have n 4. Then, we obtain that 2trA/detA1/n−n5.369444, 2ntrA3/2/trAdetA1/2n−n5.741241,n√nA
F/trA
4.653596, and 1 2s/m−s/p 2.810649. SinceκFA 6.882583, in this example, the best
lower bound is the second lower bound given byTheorem 2.3.
Acknowledgments
The authors thank very much the associate editors and reviewers for their insightful comments and kind suggestions that led to improving the presentation. This study was supported by the Coordinatorship of Selc¸uk University’s Scientific Research Projects.
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