Sum of squared logarithms - an inequality relating positive definite matrices and their matrix logarithm

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1 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 R E S E A R C H Open Access Sum of squared logarithms - an inequality relating positive definite matrices and their matrix logarithm Mircea Bîrsan 1,,PatrizioNeff 1* and Johannes Lankeit 1 * Correspondence: patrizio.neff@uni-due.de 1 Lehrstuhl für Nichtlineare Analysis und Modellierung, Fakultät für Mathematik, Universität Duisburg-Essen, Essen, Germany Full list of author information is available at the end of the article Abstract Let y 1, y, y, a 1, a, a 0, )besuchthaty 1 y y = a 1 a a and y 1 + y + y a 1 + a + a, y 1 y + y y + y 1 y a 1 a + a a + a 1 a. log y 1 ) +logy ) +logy ) log a 1 ) +loga ) +loga ). This can also be stated in terms of real positive definite -matrices P 1, P : If their determinants are equal, det P 1 =detp,then tr P 1 tr P and tr Cof P 1 tr Cof P log P 1 F log P F, where log is the principal matrix logarithm and P F = i,j=1 P ij denotes the Frobenius matrix norm. Applications in matrix analysis and nonlinear elasticity are indicated. MSC: 6D05; 6D07 Keywords: matrix logarithm; elementary symmetric polynomials; inequality; characteristic polynomial; positive definite matrices; means 1 Introduction Convexity is a powerful source for obtaining new inequalities; see, e.g., [1, ]. In applications coming from nonlinear elasticity, we are faced, however, with variants of the squared logarithm function; see the last section. The function logx)) is neither convex nor concave. Nevertheless, the sum of squared logarithms inequality holds. We will proceed as follows: In the first section, we will give several equivalent formulations of the inequality, for example, in terms of the coefficients of the characteristic polynomial Theorem 1), in terms of elementary symmetric polynomials Theorem ), in terms of means Theorem 5) or in terms of the Frobenius matrix norm Theorem 7). A proof of the inequality will be given in Section, and some counterexamples for slightly changed variants of the inequality are discussed in Section. In the last section, an application of the sum of squared loga- 01 Bîrsan et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page of 16 rithms inequality in matrix analysis and in the mathematical theory of nonlinear elasticity is indicated. Formulations of the problem All theorems in this section are equivalent. Theorem 1 For n =or n =let P 1, P R n n be positive definite real matrices. Let the coefficients of the characteristic polynomials of P 1 and P satisfy tr P 1 tr P and tr Cof P 1 tr Cof P and det P 1 = det P. log P 1 F log P F. For n =, we will now give equivalent formulations of this statement. The case n =can be treated analogously. For its proof, see Remark 15. By orthogonal diagonalization of P 1 and P, the inequalities can be rewritten in terms of the eigenvalues y 1, y, y and a 1, a, a,respectively. Theorem Let the real numbers a 1, a, a >0and y 1, y, y >0be such that y 1 + y + y a 1 + a + a, y 1 y + y y + y 1 y a 1 a + a a + a 1 a, 1) y 1 y y = a 1 a a. log y 1 ) +log y ) +log y ) log a 1 ) +log a ) +log a ). ) The elementary symmetric polynomials, see, e.g.,[,p.178] e 0 y 1, y, y )=1, e 1 y 1, y, y )=y 1 + y + y, e y 1, y, y )=y 1 y + y 1 y + y y, e y 1, y, y )=y 1 y y are known to have the Schur-concavity property i.e., e k is Schur-convex) [1, 4]; see 16). It is possible to express the problem in terms of these elementary symmetric polynomials as follows. Theorem Let a 1, a, a >0and y 1, y, y >0satisfy e 1 y 1, y, y ) e 1 a 1, a, a ), e y 1, y, y ) e a 1, a, a ), e y 1, y, y )=e a 1, a, a ).

3 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page of 16 e 1 log y1 ),log y ),log y ) ) e 1 log a1 ),log a ),log a ) ). Because y 1 y y = a 1 a a >0,wehave y 1 y + y y + y 1 y a 1 a + a a + a 1 a 1 y y + 1 y 1 a a + 1 a. Thus, we obtain the following theorem. Theorem 4 Let the real numbers a 1, a, a >0and y 1, y, y >0be such that y 1 + y + y a 1 + a + a, , ) y 1 y y a 1 a a y 1 y y = a 1 a a. log y 1 ) +log y ) +log y ) log a 1 ) +log a ) +log a ). 4) The conditions ) are also simple expressions in terms of arithmetic, harmonic and geometric and quadratic mean Ay 1, y, y )= y 1 + y + y, Hy 1, y, y )= Gy 1, y, y )= y 1 y y, Qy 1, y, y )= 1 y y + 1, y 1 y 1 + y + ) y. Theorem 5 Let a 1, a, a >0and y 1, y, y >0. Ay 1, y, y ) Aa 1, a, a ), Ha 1, a, a ) Hy 1, y, y ) reverse! ) and Gy 1, y, y )=Ga 1, a, a ) imply Qlog y 1, log y, log y ) Qlog a 1, log a, log a ). We denote by a i =: d i, y i =: x i and arrive at Theorem 6 Let the real numbers d i and x i be such that d 1, d, d >0,x 1, x, x >0and x 1 + x + x d 1 + d + d, x 1 x + x x + x 1 x d 1 d + d d + d 1 d, 5) x 1 x x = d 1 d d.

4 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 4 of 16 log x 1 ) +log x ) +log x ) log d 1 ) +log d ) +log d ). 6) If we again view x i and d i as eigenvalues of positive definite matrices, an equivalent formulation of the problem can be given in terms of their Frobenius matrix norms: Theorem 7 For n {, }, let P 1, P R n n be positive definite real matrices. Let P 1 F P F and P 1 1 F P 1 F and det P 1 = det P. log P 1 F log P F. Let us reconsider the formulation from Theorem 5.If we denote c i := log a i, z i := log y i, from Ha 1, a, a ) Hy 1, y, y ), we obtain e z 1 + e z + e z e c 1 + e c + e c. Theorem 8 Let the real numbers c 1, c, c and z 1, z, z be such that e z 1 + e z + e z e c 1 + e c + e c, e z 1 + e z + e z e c 1 + e c + e c, 7) z 1 + z + z = c 1 + c + c. z 1 + z + z c 1 + c + c. 8) InordertoproveTheorem8, one can assume without loss of generality that z 1 + z + z = c 1 + c + c =0. 9) Thus, we have the equivalent formulation Theorem 9 Let the real numbers c 1, c, c and z 1, z, z be such that e z 1 + e z + e z e c 1 + e c + e c, e z 1 + e z + e z e c 1 + e c + e c, 10) z 1 + z + z = c 1 + c + c =0.

5 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 5 of 16 z 1 + z + z c 1 + c + c. 11) Let us prove that Theorem 8 can be reformulated as Theorem 9. Indeed, let us assume that Theorem 9 is valid and show that the statement of Theorem 8 also holds true. We denote by s the sum s = z 1 + z + z = c 1 + c + c and we designate z i = z i s, c i = c i s i =1,,). the real numbers z i and c i satisfy the hypotheses of Theorem 9 and we obtain z 1 + z + z c 1 + c + c. This inequality is equivalent to z i ) s c i ) s, which, by virtue of the condition 7),reducesto z 1 + z + z c 1 + c + c. Thus, Theorem 8 is also valid. By virtue of the logical equivalence A B C) C A B) for any statements A, B, C, wecanformulatetheinequality11) i.e., Theorem9) inthe following equivalent manner. Theorem 10 Let the real numbers c 1, c, c and z 1, z, z be such that z 1 + z + z = c 1 + c + c =0 and z 1 + z + z < c 1 + c + c. 1) one of the following inequalities holds: e z 1 + e z + e z < e c 1 + e c + e c or e z 1 + e z + e z < e c 1 + e c + e c. 1) We use the statement of Theorem 10 for the proof. Before continuing, let us show that our new inequality is not a consequence of majorization and Karamata s inequality [5]. Consider z =z 1,...,z n ) R n + and c =c 1,...,c n ) R n + arranged already in decreasing order z 1 z z n and c 1 c c n.if k z i k c i 1 k n 1), n z i = n c i, 14)

6 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 6 of 16 we say that z majorizes c, denotedbyz c. The following result is well known [6, p.89], [4, 5]. If f : R R is convex, then z c n f z i ) n f c i ). 15) Afunctiong : R n R which satisfies z c gz 1,...,z n ) gc 1,...,c n ) 16) is called Schur-convex. In Theorem 8, the convex function to be considered would be f t) =t. Do conditions 7) upon rearrangement of z, c R + if necessary) yield already majorization z c? This is not the case, as we explain now. Let the real numbers z 1 z z and c 1 c c be such that e z 1 + e z + e z e c 1 + e c + e c, e z 1 + e z + e z e c 1 + e c + e c, 17) z 1 + z + z = c 1 + c + c. These conditions do not imply the majorization z c, z 1 c 1, z 1 + z c 1 + c, z 1 + z + z = c 1 + c + c. 18) Therefore, our inequality i.e., z1 + z + z c 1 + c + c ) does not follow from majorization in disguise. Indeed, let and z 1 = , z = , z = c 1 = 1 + 1, c = 1 + 1, c = 1. we have z 1 > z > z and c 1 > c > c,togetherwith e z 1 + e z + e z = > =e c 1 + e c + e c, e z 1 + e z + e z = > =e c 1 + e c + e c, z 1 + z + z = c 1 + c + c =0, but the majorization inequalities 18) are not satisfied, since z 1 < c 1. Proof of the inequality Of course, we may assume without loss of generality that c 1 c c and z 1 z z and the same for a i, d i, x i, y i ).

7 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 7 of 16 The proof begins with the crucial lemma. Lemma 11 Let the real numbers a b candx y zbesuchthat a + b + c = x + y + z =0, a + b + c = x + y + z. 19) the inequality e a + e b + e c e x + e y + e z 0) is satisfied if and only if the relation a x 1) holds, or equivalently, if and only if c z ) holds. Proof Let us denote by r := a + b + c )>0.,from19), it follows b + c = a, b + c = r a, y + z = x, y + z = r x, and we find b = 1 a + r a )), c = 1 a r a )), y = 1 x + r x )), z = 1 x r x )). ) In view of 19)anda b c, x y z, one can show that [ ] [ r a, x, r, b, y r, r ] [, c, z r, r ]. 4) Indeed, let us verify the relations 4). We have r a r 6 1 a + b + c ) a a + b + c ) b + c 5a and a b + c ) b +a + b) 5a and b + c) b + c ) 4a ab b 0 and b + c bc a b)a + b) 0 and b c) 0,

8 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 8 of 16 which hold true since a b and a + b a + b + c = 0. Similarly, we have r b r b r 4 4b a + b + c ) 5b a + c 5b a +a + b) a +ab 4b 0 a b)a +b) 0, which holds true since a b and a +b a + b + c =0.Also,wehave r c r r c r 4 a + b ) c and 5c a + b a + b + c ) c 1 6 a + b + c ) a + b ) a + b) and 5a + b) a + b a b) 0 and 4a +10ab +4b 0 a b) 0 and a +b)a + b) 0, which hold true since a +b a + b + c =0anda + b a + b + c =0.Onecanshowin the same way that x [ r, r], y [ r, r ], z [ r, r ], so that 4)hasbeenverified. We prove now that the inequality 1) holds if and only if )holds.indeed,using),4 and 4)weget c z a r a ) x r x ) a r + 1 ) a ) x r r + 1 ) x ) r a x, since the function t t + 1 t )isdecreasingfort [ 1,1]. Let us prove next that the inequalities 0)and1) are equivalent. To accomplish this, we introduce the function f :[ r, r] R by ) f x)=e x x+ r + e x ) )/ x r + e x ) /. 5) Taking into account )and4) 1,theinequality0) can be written equivalently as f a) f x), 6) which is equivalent to a x, since the function f defined by 5) is monotone increasing on [ r, r], as we show next. To this aim, we denote by cos ϕ := x [ ) [ 1 x ], r,1 i.e. ϕ := arccos 0, π ]. r

9 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 9 of 16 the function 5)can be written as [ f x)=hr, ϕ), where h :0, ) 0, π ] R, hr, ϕ)=e r cos ϕ + e r cosϕ+π/) + e r cosϕ π/). 7) We have to show that hr, ϕ) is decreasing with respect to ϕ [0, π ]. We compute the first derivative hr, ϕ) ϕ [ = r e r cos ϕ sin ϕ + e r cosϕ+π/) sin ϕ + π ) + e r cosϕ π/) sin ϕ π )]. 8) The function 8) has the same sign as the function Fr, ϕ):= 1 cos ϕ e r hr, ϕ), 9) r ϕ i.e., the function F :0, ) [0, π ] R given by Fr, ϕ)= sin ϕ e r sinϕ+π/) sin ϕ + π ) e r sinϕ π/) sin ϕ π ). 0) In order to show that Fr, ϕ) 0 for all r, ϕ) 0, ) [0, π ], we remark that lim r 0 Fr, ϕ)=0forfixedϕ [0, π ]andwecompute [ r Fr, ϕ)= e r sinϕ+π/) sin ϕ + π ) sin ϕ + π )] e r sinϕ π/) sin ϕ π ) sin ϕ π = [ e r sinϕ+π/) 1 cosϕ + π)+cos π ) )] = e r sinϕ π/) 1 cosϕ π)+cos π cos ϕ + 1 ) [e r sinϕ+π/) e r sinϕ π/) ] 0, since ϕ [0, π ]impliescos ϕ 1 and sinϕ + π ) sinϕ π ). Consequently, the function Fr, ϕ) is decreasing with respect to r and for any r, ϕ) 0, ) [0, π ]wehavethat ) Fr, ϕ) lim r 0 Fr, ϕ)=0. 1) From 9) and1), it follows that hr, ϕ) is decreasing with respect to ϕ [0, π ]. This means that f x) is increasing as a function of x [ r, r], i.e., therelation6) is indeed equivalent to a x and the proof is complete.

10 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 10 of 16 Consequence 1 Let the real numbers a b candx y zbesuchthat a + b + c = x + y + z =0, a + b + c = x + y + z. one of the following inequalities holds: e a + e b + e c e x + e y + e z, ) or e a + e b + e c e x + e y + e z. ) The inequalities ) and ) are satisfied simultaneously if and only if a = x, b = y and c = z. Proof According to Lemma 11,the inequality)is equivalent to a x, 4) while the inequality )is equivalent to a x. 5) Since one of the relations 4)and5) must hold, we have proved that one of the inequalities )and) issatisfied. They aresimultaneouslysatisfiedifandonlyifboth 4)and 5)holdtrue,i.e., a = x and consequently b = y, c = z). Consequence 1 Let the real numbers a b candx y zbesuchthat a + b + c = x + y + z =0, a + b + c = x + y + z and e a + e b + e c = e x + e y + e z. we have a = x, b = yandc= z. Proof Since by hypothesis e a +e b +e c e x +e y +e z holds, we can apply Lemma 11 to deduce a x and c z. On the other hand, by virtue of the inverse inequality e x + e y + e z e a + e b + e c and Lemma 11,weobtainx a and z c. In conclusion, we get a = x, c = z and b = y. Proof of Theorem 10 In order to prove 1), we define the real numbers t i = kz i i =1,,), wherek = we have c 1 + c + c z1 + z + >1. 6) z t 1 + t + t = c 1 + c + c =0 and t 1 + t + t = c 1 + c + c. 7)

11 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 11 of 16 If we apply the Consequence 1 for the numbers c 1 c c and t 1 t t,thenwe obtain that e t 1 + e t + e t e c 1 + e c + e c or e t 1 + e t + e t e c 1 + e c + e c. 8) In what follows, let us show that e z 1 + e z + e z < e t 1 + e t + e t. 9) Using the notations ρ := z 1 + z + z )and cos ζ := z [ ] 1 1 ρ,1, i.e., ζ := arccos z1 ρ ) [ 0, π ], we have kρ := t 1 + t + t )andcos ζ = t 1. With the help of the function h defined in kρ 7), we can write the inequality 9) intheform e ρ cos ζ + e ρ cosζ +π/) + e ρ cosζ π/) < e kρ cos ζ + e kρ cosζ +π/) + e kρ cosζ π/), [ hρ, ζ )<hkρ, ζ ), ρ, ζ ) 0, ) 0, π ], k >1. or 40) The relation 40) asserts that the function h defined in 7) is increasing with respect to the first variable r 0, ). To show this, we compute the derivative r hr, ϕ)=er cos ϕ cos ϕ + e r cosϕ+π/) cos ϕ + π ) + e r cosϕ π/) cos ϕ π By virtue of the Chebyshev s sum inequality, we deduce from 41)that ). 41) hr, ϕ)>0. 4) r Indeed, the Chebyshev s sum inequality [6,.17]assertsthat:ifa 1 a a n and b 1 b b n then n n n ) n a k b k a k b k ). k=1 k=1 k=1 In our case, we derive the following result: for any real numbers x, y, z such that x+y+z =0, the inequality xe x + ye y + ze z 1 x + y + z) e x + e y + e z) =0, 4) holds true, with equality if and only if x = y = z =0. Applying the result 4) to the function 41), we deduce the relation 4). This means that hr, ϕ) is an increasing function of r, i.e. the inequality 40) holds, and hence, we have proved 9).

12 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 1 of 16 One can show analogously that the inequality e z 1 + e z + e z < e t 1 + e t + e t 44) is also valid. From 8), 9) and44), it follows that the assertion 1) holdstrue. Thus, the proof of Theorem 10 is complete. Since the statements of the Theorems 8 and 10 are equivalent, we have proved also the inequality 8). Remark 14 The inequality 8) becomes an equality if and only if z i = c i, i =1,,. Proof Indeed, assume that z1 +z +z = c 1 +c +c. we can apply the Consequence 1 and we deduce that e z 1 + e z + e z e c 1 + e c + e c or e z 1 + e z + e z e c 1 + e c + e c. 45) Taking into account 7) 1, in conjunction with 45), we find e z 1 + e z + e z = e c 1 + e c + e c or e z 1 + e z + e z = e c 1 + e c + e c. 46) By virtue of 46), we can apply the Consequence 1 to derive z 1 = c 1, and consequently z = c, z = c. Let us prove the following version of the inequality 6) for two pairs of numbers d 1, d and x 1, x : Remark 15 If the real numbers d 1 d >0andx 1 x >0aresuchthat x 1 + x d 1 + d and x 1 x = d 1 d =1, 47) then the inequality log x 1 ) +log x ) log d 1 ) +log d ) 48) holds true. Note that the additional condition 1 x x 1 d1 + 1 d is automatically fulfilled. Proof Since x 1 x = d 1 d =1andd 1 d >0,x 1 x >0,wehavex 1 1, d 1 1and log x 1 = log x 0, log d 1 = log d 0, so that the inequality 48)isequivalenttolog x 1 log d 1, i.e.,wehavetoshowthatx 1 d 1.

13 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 1 of 16 Indeed, if we insert x = 1 x 1 and d = 1 d 1 into the inequality 47) 1 then we find x x 1 d1 + 1 d1, which means that x 1 d 1 since the function t t + 1 t is increasing for t [1, ). This completes the proof. Alternative proof of Remark 15 Let x = d =1.47)impliesx 1 +x +x d 1 +d +d and x 1 x x = d 1 d d = 1 as well as x 1 x + x x + x 1 x =1+x + x 1 1+d + d 1 = d 1 d + d d + d 1 d, 49) because x 1 x =1=d 1 d,andtheorem6 provides the assertion. 4 Some counterexamples for weakened assumptions Example 16 UnlikeintheDcaseinRemark15, for two triples of numbers the second condition 18) of Theorem, namelyy 1 y + y y + y 1 y a 1 a + a a + a 1 a, cannot be removed. Let y 1 = e 6, y =1, y = e 6, a 1 = e 4, a = e 4, a = e 8. y 1 y y = a 1 a a =1and y 1 + y + y > e 6 e e 4 >e 4 > a 1 + a + a, but log y 1 ) +log y ) +log y ) =6+0+6 < =log a 1 ) +log a ) +log a ). Example 17 The condition y 1 y y = a 1 a a cannot be weakened to y 1 y y a 1 a a.indeed, let y = y = a 1 = a =1,y 1 = e, a = e. y 1 + y + y = e e = a 1 + a + a, y 1 y + y 1 y + y y = e + e +1 1+e + e = a 1 a + a 1 a + a a, y 1 y y = e e = a 1 a a. But nevertheless log y 1 ) +log y ) +log y ) =1+0+0<0+0+4=log a 1 ) +log a ) +log a ). A counterexample for the two variable case can be constructed analogously.

14 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 14 of 16 Example 18 Even with an analogous condition, the inequality 4)doesnotholdforn =4 numbers without further assumptions). Indeed, let y 1 = e, y = y = e 7, y 4 = e 15, a 1 = a = e 6, a = e 7, a 4 = e 19. y 1 y y y 4 = a 1 a a a 4 =1.Also, y 1 + y + y + y 4 = e + e 7 + e 7 + e 15 >0+e 7 +e 6 + e 19 = a 1 + a + a + a 4. Furthermore, y 1 y + y 1 y + y 1 y 4 + y y + y y 4 + y y 4 = e 8 + e 8 + e 14 + e 14 + e 8 + e 8 and a 1 a + a 1 a + a 1 a 4 + a a + a a 4 + a a 4 = e 1 + e 1 + e 1 + e 1 + e 1 + e 1. Since e >e +1,wehavee 14 > e 1 + e 1 + e 1 and, therefore, y 1 y + y 1 y + y 1 y 4 + y y + y y 4 + y y 4 a 1 a + a 1 a + a 1 a 4 + a a + a a 4 + a a 4. Nevertheless, for the sum of squared logarithms, the reverse inequality log y 1 ) +log y ) +log y ) +log y 4 ) = =4 < 48 = =log a 1 ) +log a ) +log a ) +log a 4 ) holds true. Example 19 The inequality 4) does not remain true either, if the function logy) isre- placed by its linearization y 1). Indeed, let y 1 =9,y =5,y = 1 45, a 1 =10,a =1,a = y 1 + y + y >14>11.1=a 1 + a + a and y 1 y + y 1 y + y y > = a 1 a + a 1 a + a a. But y 1 1) +y 1) +y 1) = ) 44 ) 9 <81< =a 1 1) +a 1) +a 1).

15 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 15 of Conjecture for arbitrary n The structure of the inequality in dimensions n =andn =andextensivenumerical sampling strongly suggest that the inequality holds for all n N if the n corresponding conditions are satisfied. More precisely, in terms of the elementary symmetric polynomials, we expect the following: Conjecture 0 Let n N and y i, a i >0for i =1,...,n. If for all i =1,...,n 1we have e i y 1,...,y n ) e i a 1,...,a n ) and e n y 1,...,y n )=e n a 1,...,a n ), then n log y i ) n log a i ). 6 Applications The investigation in this paper has been motivated by some recent applications. The new sum of squared logarithms inequality is one of the fundamental tools in deducing a novel optimality result in matrix analysis and the conditions in the form ) had been deduced inthecourseofthatwork.optimalityinthematrixproblemsuggestedthesumofsquared logarithms inequality. Indeed, based on the present result in [7], it has been shown that for all invertible Z C and for any definition of the matrix logarithm as possibly multivalued solution X C of exp X = Z it holds min log Q Z F = log U p Z F = log H F, Q Q=I min sym log Q Z F = sym log U p Z 50) F = log H F, Q Q=I where sym X = 1 X + X )isthehermitianpartofx C and U p is the unitary factor in the polar decomposition of Z into unitary and Hermitian positive definite matrix H Z = U p H. 51) This result 50) generalizes the fact that for any complex logarithm and for all z C \{0} min ϑ π,π] [ log C e iϑ z ] = log R z, min ϑ π,π] [ Re log C e iϑ z ] = log R z. 5) The optimality result 50) can now also be viewed as another characterization of the unitary factor in the polar decomposition. In addition, in a forthcoming contribution [8],we use 50) tocalculatethegeodesicdistanceoftheisochoricpartofthedeformationgradient SL, R) toso, R) in the canonical left-invariant Riemannian metric on F det F 1 SL, R), to the effect that dist geod F det F 1 ), SO, R) = dev log F T F F, 5)

16 Bîrsan et al. Journal of Inequalities and Applications 01, 01:168 Page 16 of 16 where dev X = X 1 tr X)I is the orthogonal projection of X R to trace free matrices. Thereby, we provide a rigorous geometric justification for the preferred use of the Henckystrain measure log F T F F in nonlinear elasticity and plasticity theory [9]. Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed fully to all parts of the manuscript. Notably all ideas have emerged by continuous discussions among them. Author details 1 Lehrstuhl für Nichtlineare Analysis und Modellierung, Fakultät für Mathematik, Universität Duisburg-Essen, Essen, Germany. Department of Mathematics, University A.I. Cuza of Iaşi, Iaşi, Romania. Acknowledgements The first author MB) was supported by the German state grant: Programm des Bundes und der Länder für bessere Studienbedingungen und mehr Qualität in der Lehre. Received: 1 January 01 Accepted: 8 March 01 Published: 1 April 01 References 1. Guan, K: Schur-convexity of the complete elementary symmetric functions. J. Inequal. Appl. 006, Article ID ). doi: /jia/006/6764. Roventa, I: A note on Schur-concave functions. J. Inequal. Appl. 159, ). Steele, JM: The Cauchy-Schwarz Master Class: an Introduction to the Art of Mathematical Inequalities. Cambridge University Press, Cambridge 004) 4. Khan, AR, Latif, N, Pečarić, J: Exponential convexity for majorization. J. Inequal. Appl. 105, ) 5. Karamata, J: Sur une inégalité relative aux fonctions convexes. Publ. Math. Univ. Belgrad 1, ) 6. Hardy, GH, Littlewood, JE, Pólya, G: Inequalities. The University Press, Cambridge 194) 7. Neff, P, Nagatsukasa, Y, Fischle, A: The unitary polar factor Q = U p minimizes LogQ * Z) and sym * LogQ * Z) in the spectral norm in any dimension and the Frobenius matrix norm in three dimensions 01, submitted) 8. Neff, P, Eidel, B, Osterbrink, F, Martin, R: The isotropic Hencky strain energy log U measures the geodesic distance of the deformation gradient F GL + n)toson) in the unique left-invariant Riemannian metric on GL + n)whichis also right On)-invariant 01, in preparation) 9. Hencky, H: Über die Form des Elastizitätsgesetzes bei ideal elastischen Stoffen. Z. Techn. Physik 9, ) doi: /109-4x Cite this article as: Bîrsan et al.: Sum of squared logarithms - an inequality relating positive definite matrices and their matrix logarithm. Journal of Inequalities and Applications 01 01:168.