The $ K $-Th Derivatives of the Immanant and the $\Chi $-Symmetric Power of an Operator

$The $ K $-Th Derivatives of the Immanant and the $\Chi $-Symmetric Power of an Operator$

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2 Immanant

We will write Mn(C) to represent the vector space of the square matrices of order n with complex entries. Let A Mn(C) and χ be an irreducible character of C. We deﬁne the immanant of∈ A as:

n d (A)= χ(σ) a . χ X Y iσ(i) σ∈Sn i=1

In other words, dχ : Mn(C) C is a map taking an n n matrix to its immanant. This is a differentiable−→ map. For X M (C×), we denote by ∈ n DdχA(X) the directional derivative of dχ at A in the direction of X. We denote by A(i j) the n n square matrix that is obtained from A by replacing the i-th row| and j-th× column with zero entries, except entry (i, j) which we set to 1. We define the immanantal adjoint adj (A) as the n n χ × matrix in which the entry (i, j) is dχ(A(i j)). This agrees with the definition of permanental adjoint in [13, p. 237], but| not with the usual adjugate matrix (we would need to consider the transpose in that case). This is a matter of convention. We obtain the following result similar to the Jacobi formula for the determinant.

Theorem 2.1. For each X M (C), ∈ n T Ddχ(A)(X) = tr(adjχ(A) X).

Proof. For each 1 j n, let A(j; X) be the matrix obtained from A by ≤ ≤ replacing the j-th column of A by the j-th column of X and keeping the rest of the columns unchanged. Then the given equality can be restated as

n Dd (A)(X)= d (A(j; X)). (1) χ X χ j=1

2 On the other hand we note that Ddχ(A)(X) is the coeﬃcient of t in the polynomial dχ(A + tX). Using the fact that the immanant is a multilinear function of the columns we obtain the desired result. Again using the fact that the immanant is a multilinear function, we notice that for any 1 i n, we have ≤ ≤ n d (A)= a d (A(i j)). χ X ij χ | i=1

Using this and (1), we get that

n n Dd (A)(X)= x d (A(i j)). (2) χ X X ij χ | i=1 j=1

We will generalize the expressions (1) and (2) for the derivatives of all orders of the immanant. We now turn to derivatives. Let V1,...,Vn be n vector spaces over C, 1 k and let φ : V1 ... Vn C be a multilinear form. For A, X ,...X × × −→ 1 k ∈ V1,...,Vn, the k-th derivative of φ at A in the direction of (X ,...,X ) is given by the expression

k k 1 k ∂ 1 k D φ(A)(X ,...,X ) := φ(A + t1X + ... + tkX ). ∂t1 ...∂tk t1=...=tk=0

k C This is a multilinear function of Mn ( ). In particular, if we consider φ = dχ we have the deﬁnition of the k-th derivative of the immanant.

3 First Expression

We start by introducing some notation. Given a matrix A Mn(C), we will represent by A the i-th column of A, i = 1,...,n . ∈ [i] { } Let k be a natural number, 1 k n. Take A, X1,...,Xk M (C), ≤ ≤ ∈ n and t1,...,tk k unknowns. Let Qk,n be the set of strictly increasing maps 1,...,k 1,...,n and G the set of increasing maps. { } −→{ } k,n We will denote by A(α; X1,...,Xk) the matrix of order n obtained from A replacing the α(j) column of A by the α(j) column of Xj.

3 Theorem 3.1 (First expression). For every 1 k n, ≤ ≤ Dkd (A)(X1,...,Xk)= d A(α; Xσ(1),...,Xσ(k)). χ X X χ σ∈Sk α∈Qk,n

In particular,

Dkd (A)(X,...,X)= k! d A(α; X,...,X). χ X χ α∈Qk,n

k 1 k Proof. Just like in the case of the ﬁrst derivative, we have that D dχ(A)(X ,...,X ) 1 is the coeﬃcient of t1 ...tk in the expansion of the polynomial dχ(A + t1X + k ... + tkX ). Using the linearity of the immanant function in each column, we obtain the desired equality.

1 k Proof. One simply has to observe that each summand in ∆χ(A; X ,...,X ) appears (n k)! times: once we ﬁx a permutation of the matrices X1,...,Xk, these summands− correspond to the possible permutations of the n k matrices equal to A. − As an immediate consequence of this result, we can obtain another formula for the derivative of order k of the immanant map. This generalizes formula (26) in [7]. Proposition 3.2 (First expression, rewritten). n! Dkd (A)(X1,...,Xk) = ∆ (A; X1,...,Xk). (3) χ (n k)! χ − 4 Laplace Expansion for Immanants

We start by generalizing the Laplace expansion, known for the determinant and the permanent, to all immanants. This expansion was proved ﬁrst for the determinant and the same arguments used can be used to prove the correspondent expansion for the permanent. These can be found in [11] and in [14]. The similarity of proofs is due to the fact that the determinant and the permanent of a direct sum of matrices is just the product of the determinants, or the permanents, of the summand blocks. However, if χ is any other irreducible character, there is no clear general relation between the immanant

4 of A and the immanant of any submatrix of A. So the Laplace expansion formula for any immanant is a little more complicated. Let α Q , and ∈ k,n S = σ S : σ( 1,...,k ) = Im α . kα { ∈ n { } } Fact 1. If α = β then S S = . 6 kα ∩ kβ ∅

Suppose σ Skα Skβ . Then σ( 1,...,k ) = Im α = Im β, and thus Im α = Im β. Since∈ α,∩ β Q , it follows{ that }α = β. ∈ k,n Fact 2. S = S n [ kα α∈Qk,n

If π S then π( 1,...,k )= j ,...,j . Suppose that j <...

Using the previous facts, we can conclude that for every σ Skβ , the value ∈ n χ(σ) a X Y tσ(t) σ S t=1 ∈ kβ does not depend on the values of the following entries of the matrix A:

I. The ﬁrst k rows of A and the n k columns of A in Im β. − II. Rows k +1,...,n of A and the k columns of A with index in Im β.

5 ∗ We now denote by A id β = (aij) the matrix of order n obtained by replacing every entry in I{ and| } II by zeros. We then have

n χ(σ) a = d (A id β ). X Y tσ(t) χ { | } σ S t=1 ∈ kβ

On the other hand,

n d (A id β ) = χ(σ) a∗ χ { | } X Y tσ(t) σ∈Sn t=1 n = χ(σ) a∗ . X Y tσ(t) σ∈S Sk t=1 α∈Qk,n α

n Now take α = β and σ S . Then a∗ = 0, because at least one of 6 ∈ kα Y tσ(t) t=1 the factors is zero. So

n χ(σ) a∗ =0, X Y tσ(t) σ∈Skα t=1 for every α Qk,n β . ∈ \{ } ∗ Moreover, for A id β , if σ Skβ then atσ(t) = atσ(t). So dχ(A id β ) = n { | } ∈ { | }

σ∈S χ(σ) atσ(t). Finally, kβ Y P t=1

n d (A) = χ(σ) a χ X Y tσ(t) σ∈Sn t=1 n = χ(σ) a X X Y tσ(t) α∈Qk,n σ∈Skα t=1 = d (A id α ). (4) X χ { | } α∈Qk,n

Deﬁnition 4.1. Let X be a square matrix of order n, k a natural number, 1 k n e α, β Q . We denote X[α β] as the square matrix of ≤ ≤ ∈ k,n |

6 order k obtained from X by picking the rows α(1),...,α(k) and the columns β(1),...,β(k). We denote X(α β) as the square matrix of order n k obtained from X by deleting the rows| α(1),...,α(k) and the columns β(1)− ,...,β(k).

Let α, β Q , and take A a square matrix of order k and B a square ∈ k,n matrix of order n k. Denote byα ¯ be the unique element of Qn−k,n with Im α = Im α. − We now deﬁne A B =(x ), M ij α|β as a square matrix of order n such that

x = 0 if i Im α and j Im β; • ij ∈ 6∈ x = 0 if i Im α and j Im β; • ij 6∈ ∈ x = a −1 −1 if i Im α and j Im β; • ij α (i)β (j) ∈ ∈

x = b −1 −1 if i Im α and j Im β. • ij α (i)β (j) 6∈ 6∈ In a sense, we pick rows α and β of A and rowsα ¯ and β¯ of B. If α = β = (1,...,k), this is the usual direct sum of A[1,...,k 1,...,k] and B(1,...,k 1,...,k). | | So, according to our deﬁnitions, it is easy to verify that

X id β = X[id β] X(id β). { | } | M | id|β

Applying the previous equality and (4) it follows that

d (X)= d (X[id β] X(id β)) = d (X id β ). χ X χ | M | X χ { | } β∈Qk,n id|β β∈Qk,n

For a ﬁxed α Q , using similar arguments, we get the Laplace expan- ∈ k,n sion for immanants:

d (X)= d (X[α β] X(α β)) = d (X α β ). (5) χ X χ | M | X χ { | } β∈Qk,n α|β β∈Qk,n

7 Example 4.2. Let A be a matrix of order 4. Let k = 2 and α = id. Then, we have

a11 a12 0 0 a11 0 a13 0     a12 a22 0 0 a12 0 a23 0 dχ(A) = dχ + dχ  0 0 a33 a34  0 a32 0 a34      0 0 a43 a44  0 a42 0 a44 a 0 0 a 0 a a 0  11 14  12 13  a12 0 0 a24 0 a22 a23 0 + dχ + dχ  0 a32 a33 0  a31 0 0 a34      0 a42 a43 0  a41 0 0 a44

0 a12 0 a14 0 0 a13 a14  0 a 0 a   0 0 a a  + d 22 24 + d 23 24 χ a 0 a 0  χ a a 0 0   31 33   31 32  a41 0 a43 0  a41 a42 0 0  We list some properties of the matrix X α β . { | } Proposition 4.3. Let α,β,α′, β′ Q . Then we have ∈ k,n X α β [α β]= X[α β] and X α β (α β)= X(α β). { | } | | { | } | | If β = β′, then both matrices X α β [α β′] and X α β (α β′) have a zero 6 { | } | { | } | column. If α = α′, then both matrices X α β [α′ β] and X α β (α′ β) have a zero row. 6 { | } | { | } | Proof. These are consequences of the deﬁnitions. We can now check that this formula generalises the known Laplace formulas for the determinant and the permanent (see [11] and [14]). If χ = ε then dε = det. For α Qk,n, denote α = α(1)+ ...+ α(k). Fixing α Qk,n and applying the previous∈ properties| we| have: ∈ det X = det(X α β ) X { | } β∈Qk,n = ( 1)|α| ( 1)|γ| det(X α β [α γ]) det(X α β (α γ)) − X X − { | } | { | } | β∈Qk,n γ∈Qk,n = ( 1)|α| ( 1)|β| det(X α β [α β]) det(X α β (α β)) − X − { | } | { | } | β∈Qk,n = ( 1)|α| ( 1)|β| det(X[α β]) det(X(α β)). (6) − X − | | β∈Qk,n

8 This is exactly the expression of the the Laplace expansion for determinants. With similar arguments we can prove the same result for the Laplace expansion of the permanent.

5 Second Expression

We now wish to separate the entries of the matrices X1,...,Xk from the entries of A. It is easy to express the determinant of a direct sum in terms of determinants of direct summands, and the same happens with the permanent. With immanants, the best one can do is use formula (5), which is what we do in this second expression. 1 k Take X ,...,X complex matrices of order n, and, for σ Sk, β Qk,n. Denoting by 0 the zero matrix of order n, we deﬁne ∈ ∈

σ σ(1) σ(k) Xβ = 0(β; X ,...,X ),

σ(p) the matrix whose β(p)-th column is equal to X[β(p)] and the remaining columns are zero, for 1 p k. ≤ ≤ Theorem 5.1 (Second Formula).

Dkd A(X1,...,Xk)= d (Xσ[α β] A(α β)), χ X X χ β | M | σ∈Sk α,β∈Qk,n α|β in particular

Dkd A(X,...,X)= k! d (X[α β] A(α β)) χ X χ | M | α,β∈Qk,n α|β Proof. We have proved that

Dkd A(X1,...,Xk)= d A(β; Xσ(1),...,Xσ(k)). χ X X χ σ∈Sk β∈Qk,n By the Laplace expansion for immanants, for every β Q , we have that ∈ k,n σ(1) σ(k) dχA(β; X ,...,X )= = d (A(β; Xσ(1),...,Xσ(k)) α β ) X χ { | } α∈Qk,n = d (A(β; Xσ(1),...,Xσ(k))[α β] A(β; Xσ(1),...,Xσ(k))(α β)). X χ | M | α∈Qk,n α|β

9 Now we just notice that

A(β; Xσ(1),...,Xσ(k))[α β]= Xσ[α β] | β | and A(β; Xσ(1),...,Xσ(k))(α β)= A(α β). | | This concludes the proof of the formula.

6 Formulas for the k-th derivative for the m- th χ-symmetric tensor power

In this section we wish to establish a formula for the k-th derivative of the χ-symmetric tensor power of a matrix. Before we can do this, we need quite a bit of deﬁnitions, including the very deﬁnition of this matrix. We start with some classical results that can be found in [13, chapter 6]. Let χ be an irreducible character of Sm and χ(id) K = χ(σ)P (σ), χ m! X σ∈Sm where id stands for the identity element of Sm. Kχ is a linear operator on mV , and it is also an orthoprojector. Is is called a symmetriser map. ⊗ The range of Kχ is called the symmetry class of tensors associated with the irreducible character χ and it is represented by V = K ( mV ). χ χ ⊗ It is well known that the alternating character χ(σ) = εσ (sign of the permutation σ) leads to the symmetry class mV ; on the other hand the principal character χ(σ) 1 leads to the symmetry∧ class mV . ≡ ∨ Given a symmetriser map Kχ, we denote v v ... v = K (v v ... v ). 1 ∗ 2 ∗ ∗ m χ 1 ⊗ 2 ⊗ ⊗ m

These vectors belong to Vχ and are called decomposable symmetrised tensors. Let Γ be the set of all maps from the set 1,...,m into the set m,n { } 1,...,n . This set can also be identiﬁed with the collection of multiindices {(i ,...,i} ): i n . If α Γ , this correspondence associates to α the { 1 m j ≤ } ∈ m,n m-tuple (α(1),...,α(m)). In the set Γm,n we will consider the lexicographic order. The set ασ : σ S Γ { ∈ m} ⊆ m,n 10 −1 is the orbit of α. The group Sm acts on Γm,n by the action (σ, α) ασ where σ S and α Γ . The stabiliser of α is the subgroup−→ of S ∈ m ∈ m,n m deﬁned as G = σ S : ασ = α . α { ∈ m } Let e ,...,e be an orthonormal basis of the vector space V . Then { 1 n} e⊗ = e e ... e : α Γ { α α(1) ⊗ α(2) ⊗ ⊗ α(m) ∈ m,n} is a basis of the m-th tensor power of V . So the set of all decomposable symmetrised tensors spans Vχ. However, this set need not to be a basis of Vχ, because its elements might not be linearly independent, some of them may even be zero. Let

Ω = Ω = α Γ : χ(σ) =0 . (7) χ { ∈ m,n X 6 } σ∈Gα With simple calculations, we can conclude that χ(id) e∗ 2 = χ(σ). (8) k αk m! X σ∈Gα

So the nonzero decomposable symmetrised tensors are e∗ : α Ω . Now, { α ∈ } let ∆ be the system of distinct representatives for the quocient set Γm,n/Sm, constructed by choosing the ﬁrst element in each orbit, for the lexicographic order of indices. It is easy to check that ∆ G , where G is the set of ⊆ m,n m,n all increasing sequences of Γm,n. Let

∆ = ∆ Ω. ∩ It can be proved that the set e∗ : α ∆ is linearly independent. We have { α ∈ } already seen that the set e∗ : α Ω , spans V , so there is a set ∆, such { α ∈ } χ that ∆ ∆ Ω and b ⊆ ⊆ ′ ∗ b := e : α ∆ , (9) E { α ∈ } b is a basis for Vχ. It is also known that this basis is orthogonal if χ is a linear character. In general, if χ does not have degree one, there are no known orthonormal bases of V (S ) formed by decomposable symmetrised tensors. Let =(v : χ m E α α ∆) be the orthonormal basis of the m-th χ-symmetric tensor power of the ∈ b 11 vector space V obtained by applying the Gram-Schmidt orthonormalization procedure to ′. Let B be the t t change of basis matrix, from to E × E ′ =(e∗ : α ∆). This means that for each α ∆, E α ∈ ∈ b b v = b e∗ . α X γα γ γ∈∆ b We note that this matrix B does not depend on the choice of the orthonormal basis of V , since the set ∆ is independent of the vectors, and has a natural order (the lexicographic order),b which the basis inherits. More- over, the Gram-Schmidt process only depends on the numbersE e∗ , e∗ , and, h α βi by [13, p. 163], these are given by formula

m χ(id) e∗ , e∗ = χ(σ) e , e . h α βi m! X Yh α(t) βσ(t)i σ∈Sm t=1

Hence, they only depend on the values of e , e = δ and are thus indepen- h i ji ij dent of the vectors themselves.

It is known that if T is a linear operator on V, then Vχ is invariant for its m-th fold tensor power mT . Thus, the χ-symmetric power T , denoted by ⊗ m Kχ(T ) is deﬁned as the restriction of T to Vχ. There is a close connection of this χ-symmetric tensor power of ⊗T and the immanant, as the following result already shows (this is a rephrasing of [13, p. 230]).

Theorem 6.1. Suppose χ is an irreducible character of the group Sm. Let E = e ,...,e be an orthonormal basis of the inner product space V . Let { 1 n} T L(V,V ) be the unique linear operator such that M(T, E)= A. ∈If α, β Γ , then ∈ m,n χ(id) K (T )(e∗ ), e∗ = d (AT [α β]). (10) h χ α βi m! χ |

T As we came across the immanant of a transpose, we note that dχ(A )=

12 dχ(A), where χ is also an irreducible character of Sm, deﬁned as χ(σ) := χ(σ).

m d (AT ) = χ(σ) (AT ) χ X Y iσ(i) σ∈Sm i=1 m = χ(σ) a (i = σ−1(j)) X Y σ(i)i σ∈Sm i=1 m −1 = χ(σ) a −1 (σ = τ ) X Y jσ (j) σ∈Sm j=1 m = χ(τ −1) a X Y jτ(j) τ∈Sm j=1 m = χ(τ) a X Y jτ(j) τ∈Sm j=1

= dχ(A)

*** χ tem o mesmo ∆ que o χ? [E’ so’ curiosidade] b Now we want to define Kχ(A), the m-th χ-symmetric tensor power of the matrix A. A natural way to do this is to fix an orthonormal basis E in V , and consider the linear endomorphism T such that A = M(T, E). Then the basis is orthonormal and one can define E K (A) := M(K (T ), ) χ χ E The matrix K (A) has order t = ∆ , with Q t. χ | | | m,n| ≤ It is important to notice that thisb matrix does not depend on the choice of the orthonormal basis E of V . This is an immediate consequence of the

13 formula (10): for α, β ∆, the (α, β) entry of K (A) is ∈ χ b ∗ ∗ Kχ(T )vβ, vα = bγβKχ(T )e , bδαe h i X γ δ γ,δ∈∆ b = b b K (T )e∗ , e∗ X γβ δα χ γ δ γ,δ∈∆ b χ(id) = b b d (AT [γ δ]) m! X γβ δα χ | γ,δ ∈∆b χ(id) = b b d (A[δ γ]T ) m! X γβ δα χ | γ,δ ∈∆b χ(id) = b b d (A[δ γ]). m! X γβ δα χ | γ,δ∈∆ b This deﬁnition admits, as special cases, the m-th compound and the m-th induced power of a matrix, as deﬁned in [13, p. 236]. The matrix Kχ(A) is called the induced matrix in [13, p. 235], in the case when the character has degree one.

Denote by immχ(A) the square matrix with rows and columns indexed by ∆, whose (γ, δ) entry is d (A[γ δ]) (one could call the elements of this χ | matrixb immanantal minors indexed by ∆, the usual minors are gotten by considering the alternating character, inb which case ∆ = Qm,n). With this deﬁnition, we can rewrite the previous equation as b χ(id) K (A)= B∗ imm (A)B. (11) χ m! χ

1 n Finally, denote by miximmχ(X ,...,X ) the square matrix with rows and columns indexed by ∆, whose (γ, δ) entry is ∆ (X1[γ δ],...,Xn[γ δ]), χ | | so that miximmχ(A,...,Ab) = immχ(A). We use the same shorthand as with the mixed immanant: for k n, ≤ 1 k 1 k miximmχ(A; X ,...,X ) := miximmχ(A,...,A,X ,...X )

Before our main formula, we recall a general result about derivatives, which follows from the deﬁnition.

14 Lemma 6.2. If f and g are two maps such that f g is well deﬁned, with g linear, then ◦

Dk(f g)(A)(X1,...,Xk)= Dkf(g(A))(g(X1),...,g(Xk)). ◦ Theorem 6.3. According to our previous notation, we have

χ(id) DkK (A)(X1,...,Xk)= B∗ miximm (A; X1,...,Xk)B χ (m k)! χ − and, using the notation we have already established, the (α, β) entry of this matrix is χ(id) b b d (X[δ γ]σ[ρ τ] A[δ γ](ρ τ)). m! X γβ δα X X χ | τ | M | | γ,δ σ∈Sk ρ,τ∈Qk,m ρ|τ ∈∆b Proof. Notice that the map A A[δ γ] is linear, so we can apply the previous 7→ | lemma in order to compute the derivatives of the entries of the matrix Kχ(A). The (α, β) entry of the k-th derivative of the m-th χ-symmetric tensor power k 1 k of A, i.e., the (α, β) entry of the matrix D Kχ(A)(X ,...,X ) is:

χ(id) b b Dkd (A[δ γ])(X1[δ γ],...,Xk[δ γ]). k! X γβ δα χ | | | γ,δ ∈∆b

To abbreviate notation, for ﬁxed γ, δ ∆, we will write C := A[δ γ], and ∈ | Zi := Xi[δ γ], i =1,...,k. Using formula (3),b we get | Dkd (A[δ γ])(X1[δ γ],...,Xk[δ γ]) = Dkd (C)(Z1,...,Zk) χ | | | χ m! = ∆ (C; Z1,...,Zk). (m k)! χ − k 1 k So the (α, β) entry of D Kχ(A)(X ,...,X ) is

χ(id) m! b b ∆ (C; Z1,...,Zk)= m! X γβ δα (m k)! χ γ,δ∈∆ b − χ(id) b b ∆ (A[δ γ]; X1[δ γ],...,Xk[δ γ]) (m k)! X γβ δα χ | | | γ,δ − ∈∆b

15 1 k According to the deﬁnition of miximmχ(A; X ,...,X ), we have

χ(id) DkK (A)(X1,...,Xk)= B∗ miximm (A; X1,...,Xk)B. χ (m k)! χ − This establishes the ﬁrst formula. For the entries of the matrix, we use the formula in theorem 5.1:

Dkd (C)(Z1,...,Zk)= d (Zσ[ρ τ] C(ρ τ)), χ X X χ τ | M | σ∈Sk ρ,τ∈Qk,m ρ|τ recalling that σ σ(1) σ(k) Zτ = 0(τ; Z ,...,Z ), where 0 denotes the zero matrix of order m. So, the (α, β) entry of the k-th derivative of Kχ(A) is:

1 k The formula obtained for the higher order derivatives of Kχ(A)(X ,...,X ) generalises the expressions obtained by Bhatia, Jain and Grover ([7], [8]). We will prove this for the derivative of the m-th compound, establishing that, from the formula in theorem 6.3, one can establish formula (2.5) in [10], from which the main formula for the derivative of the m-th compound of A is obtained. So we take χ = ε then

K (A)(X1,...,Xk)= m(A)(X1,...,Xk). χ ∧ In this case ∆ = Q and the basis e∧ : α Q is orthogonal and it is m,n { α ∈ m,n} easy to see (byb direct computation or using formula (8)) that every vector has norm 1/√m!. So the matrix B of order n is diagonal and its diagonal m entries are equal to √m!. We now notice two properties that we will use in our computations:

16 I. For any square matrices X Mk(C), Y Mn−k(C) and functions α, β Q , ∈ ∈ ∈ k,n det X Y =( 1)|α|+|β| det X det Y. M − α|β This is a consequence of formula (6). We again notice that if γ = β, the matrices 6 (X Y )[α γ] and (X Y )(α γ) M | M | α|β α|β have a zero column. Now, using the Laplace expansion for the determinant along α, det X Y = ( 1)|α| ( 1)|γ| det((X Y )[α γ]) det((X Y )(α γ)) M − X − M | M | α|β γ∈Qk,n α|β α|β = ( 1)|α|+|β| det((X Y )[α β]) det((X Y )(α β)) − M | M | α|β α|β = ( 1)|α|+|β| det X det Y. − II. For α, β Q and ρ, τ Q , we have ∈ m,n ∈ k,m det(X[α β]σ[ρ τ]) = k!∆(X1[α β][ρ τ],...,Xk[α β][ρ τ]) X | τ | | | | | σ∈Sk

To check this, consider the columns of the matrices involved. Remember that

X[α β]σ[ρ τ] = [Xσ(1)[α β][ρ τ] ... Xσ(k)[α β][ρ τ] ) | τ | | | [1] | | [k] and the matrices that appear in the ﬁrst sum are the same as the ones that appear in the mixed discriminant.

17 We are now ready to prove the result. If we replace in theorem 6.3 d = det, we have that that the (α, β) entry of Dk m (A)(X1,...,Xk) is χ ∧ 1 b b det(X[δ γ]σ[ρ τ] A[δ γ](ρ τ)) m! X γβ δα X X | τ | M | | γ,δ∈Qm,n σ∈Sk ρ,τ∈Qk,m ρ|τ 1 = m! det(X[α β]σ[ρ τ] A[α β](ρ τ)) m! X X | τ | M | | σ∈Sk ρ,τ∈Qk,m ρ|τ = ( 1)|ρ|+|τ| det(A[α β](ρ τ)) det(X[α β]σ[ρ τ]) X X − | | | τ | σ∈Sk ρ,τ∈Qk,m = k! ( 1)|ρ|+|τ| det(A[α β](ρ τ))∆(X1[α β][ρ τ],...,Xk[α β][ρ τ]) X − | | | | | | ρ,τ∈Qk,m

We denoted by ∆(B1, ..., Bn) the mixed discriminant. The formula we obtained is formula (2.5) in [10], if you take into account that in this paper the roles of the letters k and m are interchanged. Using similar arguments one can obtain the formula for the k-th derivative of m(A)(X1,...,Xk) in [8]. ∨ Acknowledgements. We would like to thank Prof. Rajendra Bhatia for presenting this problem to us and for all the suggestions and discussions we had on this subject. We would also like to thank Prof. Jos´eDias da Silva for sharing with us his understanding of these topics and for all the discussions we had.

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