CONCENTRATION OF EIGENVALUE SUMS AND GENERALIZED LIEB’S CONCAVITY THEOREM
5.6 Supporting Materials .1 Mixed Discriminant.1Mixed Discriminant
5.6.2 Exterior Algebra
Here we give a brief review of exterior algebras on the vector spaceCn. For more details, one may refer to [14, 100]. For the convenience of our use, the notations in our paper might be different from those in other materials. For any 1 ≤ k ≤ n, let
∧k(Cn)denote the vector space of thekt h exterior algebra ofCn, equipped with the inner product
h·,·i∧k : ∧k(Cn)× ∧k(Cn) −→ C
hu1∧ · · · ∧uk,v1∧ · · · ∧vki∧k =det
hu1,v1i hu1,v2i · · · hu1,vki hu2,v1i hu2,v2i · · · hu2,vki
... ... . . . ...
huk,v1i huk,v2i · · · huk,vki
,
wherehu,vi=u∗vis the standardl2inner product onCn.
LetL(∧k(Cn)) denote the space of all linear operators from∧k(Cn)to itself. For any matrices A(1),A(2),· · · ,A(k) ∈Cn×n, we can define an element inL(∧k(Cn)):
M(k)(A(1),A(2),· · ·,A(k)):
∧k(Cn) −→ ∧k(Cn) v1∧v2∧ · · · ∧vk 7−→ X
σ∈Sk
A(σ(1))v1∧A(σ(2))v2∧ · · · ∧A(σ(k))vk, (5.45) whereSk is the symmetric group of orderk. Apparently, the map
(A(1),A(2),· · ·,A(k)) 7−→ M(k)(A(1),A(2),· · ·,A(k))
is symmetric in A(1),A(2),· · ·,A(k) and is linear in each single A(i). For simplicity, we will use the following notations for any matrices A,B,C ∈Cn×n:
M0(k)(A)= 1
k!M(k)(A,· · ·, A), (5.46a) M1(k)(A;B)= 1
(k −1)!M(k)(A,B,· · · ,B), (5.46b) M2(k)(A,B;C)= 1
(k −2)!M(k)(A,B,C,· · ·,C). (5.46c) To avoid confusion, we define M1(1)(A;B) = M01(A), M2(1)(A,B;C) = 0, and M2(2)(A,B;C) = M1(2)(A;B). Obviously the identity operator in L(∧k(Cn)) is M0(In). We will be using the following properties:
• Invertibility: if A∈Cn×nis invertible, then (M0(k)(A))−1=M0(k)(A−1).
• Adjoint: for anyA∈Cn×n,(M0(k)(A))∗ = M0(k)(A∗), with respect to the inner producth·,·i∧k.
• Positiveness: If A ∈ Hn, then M0 (A) is Hermitian; if A ∈ H+n, then M0(k)(A) 0; if A∈H++n , thenM0(k)(A) 0.
• Product properties: for any A,B,C,D ∈Cn×n, we have
M0(k)(AB)= M0(k)(A)M0(k)(B), (5.47a) M1(k)(A;B)M0(k)(C)= M1(k)(AC;BC), (5.47b) M0(k)(C)M1(k)(A;B)= M1(k)(C A;C B), (5.47c) M1(k)(A;C)M1(k)(B;D)= M2(k)(AD,C B;C D)+M1(k)(AB;C D). (5.47d)
• Derivative properties: for any differentiable functions A(t),B(t) : R −→
Cn×n, we have
∂
∂tM0(k)(A(t))= M1(k)(A0(t);A(t)) (5.48a)
∂
∂tM1(k)(A(t);B(t))= M1(k)(A0(t);B(t))+M2(k)(A(t),B0(t);B(t)).
(5.48b) Next we consider the natural basis of∧k(Cn),
{ei1 ∧ei2∧ · · · ∧eik}1≤i1<i2<···<ik≤n,
which is orthogonal under the inner product h·,·i∧k. Then the trace function on L(∧k(Cn))is defined as
Tr : L(∧k(Cn)) −→ C
TrF = X
1≤i1<i2<···<ik≤n
hei1 ∧ei2∧ · · · ∧eik,F(ei1 ∧ei2 ∧ · · · ∧eik)i∧k. (5.49) It is not hard to check that this trace function is also invariant under cyclic per- mutation, i.e. TrF G = TrGF for any F,G ∈ L(∧k(Cn)). Then for any A(1),· · ·,A(k) ∈Cn×n, the trace Tr[M(k)(A(1),· · ·, A(k))] coincides with the defini- tion of the mixed discriminant, as one can check that
TrM(k)(A(1),· · ·,A(k))
= X
σ∈Sk
X
1≤i1<···<ik≤n
hei1 ∧ · · · ∧eik, A(σ(1))ei1∧ · · · ∧ A(σ(k))eiki∧k
= n!
(n−k)!D(A(1),· · ·,A(k),In,· · ·,In
| {z }
n−k
).
(5.50)
From this observation, we can now express thek-trace of a matrix A∈Cn×nas Trk[A]= TrM0(k)(A). (5.51) For those who are familiar with exterior algebra, it is clear that the spectrum of M0(k) is just {λi1λi2· · ·λik}1≤i1<i2<···<ik≤n, where λ1, λ2,· · ·, λn are the eigen- values of A. So in this way it is more convenient to see that TrM0(k)(A) = sum(spectrum ofM0(k)(A)) = P
1≤i1<···<ik≤nλi1λi2· · ·λik = Trk[A]. Our proof of Theorem 5.3.1 will base on the expression (5.51).
In fact, our proof the main theorem can be done without introducing the exterior algebra. We can instead go through the whole proof only using notations of mixed discriminant. The advantage of using exterior algebra is that it interprets thek-trace as the normal trace of operators in a space of higher dimension, so our k-trace functions have a nicer form that imitates the trace function in the original Lieb’s theorem. Also for the same reason, we are able to construct our proof by following the arguments of Lieb’s original proof in [67].
We next introduce some notations to simplify the expressions in what follows. For anynreal numbersλ1, λ2,· · ·, λn ∈R, we define the three symmetric forms
p(n,k) = X
1≤i1<i2<···<ik≤n
λi1λi2· · ·λik, 1≤ k ≤ n, (5.52a)
di(n,k) = X
1≤j1<j2<···<jk−1≤n i<{j1,j2,···,jk−1}
λj1λj2· · ·λjk−1, 2≤ k ≤ n, 1 ≤i ≤ n, (5.52b)
gi j(n,k) = X
1≤l1<l2<···<lk−2≤n i,j<{l1,l2,···,lk−2}
λl1λl2· · ·λlk−2, 3≤ k ≤ n, 1≤ i,j ≤ n, i , j. (5.52c)
For consistency, we definedi(n,k) = 1 ifk =1;gi j(n,k) =1 ifk =2 andi , j;gi j(n,k) =0 if k = 1 ori = j. Also we define p(n,k) = di(n,k) = gi j(n,k) = 0 ifk > n. Throughout this paper, whenever we are given some real numbers λ1, λ2,· · ·, λn, the quantities p(n,k),di(n,k),gi j(n,k)are always defined correspondingly with respect to{λi}1≤i≤n. The following relations are easy to verify with the definitions above, and will be useful in our proofs of lemmas and theorems. For anyn,k, and any 1≤ i,j ≤ nsuch that i, j, we have the expansion relations
p(n,k) = λidi(n,k) +di(n,k+1), di(n,k) = λjgi j(n,k)+gi j(n,k+1). (5.53) With the notations defined above, we give the following lemma. The proof is straightforward by definition, so we omit it here.
Lemma 5.6.4. For any A,B ∈ C , and any diagonal matrix Λ ∈ C with diagonal entriesλ1, λ2,· · ·, λn, we have the following identities
TrM0(k)(Λ) = p(n,k), (5.54a)
TrM1(k)(A;Λ) =
n
X
i=1
Aiidi(n,k), (5.54b)
TrM2(k)(A,B;Λ) = X
1≤i,j≤n
(AiiBj j− AjiBi j)gi j(n,k), (5.54c)
for all1 ≤ k ≤ n, wherep(n,k),di(n,k),gi j(n,k)are defined with respect toλ1, λ2,· · ·, λn. We here provide an alternative of Lemma 5.4.4 using the following lemma.
Lemma 5.6.5. For any A∈Hn, we have
M0(k) exp(A) =exp M1(k)(A;In). Proof. We need to show that for anyv1∧v2∧ · · · ∧vk ∈ ∧k(Cn),
M0(k) exp(A)
(v1∧v2∧ · · · ∧vk) =exp M1(k)(A;In)
(v1∧v2∧ · · · ∧vk). (5.55) We use Taylor expansion ofexto expand
M0(k) exp(A) =M0(k) +
∞
X
j=0
1 j!Aj
, exp M1(k)(A;In) = +
∞
X
j=0
1
j! M1(k)(A;In)j. Then for any integers j1,j2, . . . ,jk ≥ 0, the coefficient of the term Aj1v1∧Aj2v2∧
· · · ∧ Ajkvk in the left hand side of (5.55) is 1 j1!j2!· · ·jk!,
and the coefficient of the same term in the right hand side of (5.55) is also 1
J! J j1
! J− j1 j2
!
· · · J− j1− j2− · · · − jk−1
jk
!
= 1
j1!j2!· · · jk!,
where(J = j1+ j2+· · ·+ jk).
An alternative proof of Lemma 5.4.4. Using Lemma 5.6.5 and the original GT in- equality for normal trace, we have
Trk[exp(A+B)]=TrM0(k) exp(A+B)
=Trexp M1(k)(A+B;In)
=Trexp M1(k)(A;In)+M1(k)(B;In)
≤ Trexp M1(k)(A;In)exp M1(k)(B;In)
=TrM0(k) exp(A)M0(k) exp(B)
=Trkexp(A)exp(B),
where we have used thatM1(k)(X;In)is linear in X. As shown by Petz [93], in the original GT inequality, the equality Tr[exp(A+ B)] =Tr[exp(A)exp(B)] holds for A,B ∈ Hnif and only if AB = B A. Therefore, according to our calculation above, the equality Trk[exp(A+B)]=Trk[exp(A)exp(B)] holds if and only if
M1(k)(A;In)M1(k)(B;In) =M1(k)(B;In)M1(k)(A;In). (5.56) However, one can check by definition that (5.56) is true if and only ifAB = B A. 5.6.3 Derivatives of Some Matrix Functions
Let us remind ourselves that a basic but important way to prove concavity of a differentiable function f(t) is by showing that f00(t) ≤ 0. Similarly, one way to prove concavity of a differentiable multivariate function f(x)is by showing that the second directional derivative ∂t∂22f(x+ty)|t=0 ≤ 0 for all allowed direction y. We will use this idea to prove the concavity of the k-trace functions (5.18) and (5.19).
For this purpose, we would need the following matrix derivative formulas.
• Consider a function A(t) : (a,b) −→ Hn, such that A(t) is differentiable on (a,b), then we have[132]
∂
∂t exp A(t) =Z 1
0 exp s A(t)
A0(t)exp (1−s)A(t)ds. (5.57) A0(t)denotes the derivative of A(t)with respect tot.
• Consider a function A(t): (a,b) −→ H++n , such that A(t)is differentiable on (a,b), then we have[67]
∂
∂t A(t)−1 =− A(t)−1A0(t) A(t)−1, (5.58) and
∂
∂t log A(t) = Z ∞
0 A(t)+τIn−1A0(t) A(t)+τIn−1dτ. (5.59)