260
IX – Multivariate Differential and Integral Calculus
The first integral, extended to R and in fact to a compact subset, is a C
∞
function of t like ϕ, and the third one can be made to approach X(0) = 1
by using a Dirac sequence ϕ n (Chap. V, § 8, n
◦ 27). So X(t) is indeed C
∞ .
This remark has already been made in the classical case [Chap. VII, § 1, n
◦ 2,
(iii)].
Finally note the useful formula
det [exp(X)] = exp [Tr(X)] ,
(15.24)
where, for any square matrixX, Tr(X) denotes the sum of the diagonal elements of X. Setting D(t) = det[exp(tX)] indeed defines an obviously continuous homomorphism from R to R, and so D(t) = exp(ct) for some c = D
(0)
to be computed. As the map det : M n (R) −→ R is much more than differentiable, the multivariate chain rule shows that D
(0) = det
(1) exp
(0)X, where
det
(1) is the derivative at X = 1 of the map X → det X and exp
(0) = 1
that of the exp map at the origin. Hence c = det
(1)X = T (X) with a real
valued linear function of X. By (23),
T (UXU
−1 ) = T (X) for all U ∈ GL n (R) ,
(15.25)
and so T (UX) = T (XU ). As it is not hard to check that every Y ∈ M n (R)
is the sum of invertible matrices (Y + λ1 is invertible provided −λ is not an
eigenvalue of Y ), it can be deduced that
T (XY − Y X) = 0
(15.26)
for all matrices X and Y . Setting T (X) = a
j
i X
i
j , where the X
i
j are the
coefficients of X, and by writing (26) explicitly for a matrix Y all of whose
entries, apart from one, are non-zero, it immediately follows that a
j
i = 0 for
i = j, so that T (X) is a linear combination of the diagonal entries of X. Its
value remains invariant if its entries undergo an arbitrary permutation, for
this amounts to replacing X by UXU
−1 , where the matrix U permutes the
vectors of the canonical basis for R
n . Hence T (X) is proportional to the trace
of X. It remains to check that T (X) = Tr(X) for a matrix X with non-zero
trace; we leave it to the reader to choose this matrix in such a way as to
minimize calculations.
This proof of (24) is slightly longer than the classical proof, but it teaches
the reader, if he does not already know it, that, up to a constant factor,
relation (26) characterizes the function X → Tr(X).
Exercise 1. Show that, for every linear functional X → f (X) on M n (R),
where R denotes an arbitrary commutative field, there is a unique matrix A
such that f (X) = Tr(AX).
Exercise 2. Let E be an n-dimensional vector space over R, M (E) the
set of linear maps E −→ E, (a i ) a basis for E and f (x 1 , . . . , x n ) the unique
n-linear alternating form equals 1 on the basis vectors, i.e. is the determinant
of the x i with respect to this basis. Hence if u ∈ M (E), then
IX – Multivariate Differential and Integral Calculus
The first integral, extended to R and in fact to a compact subset, is a C
∞
function of t like ϕ, and the third one can be made to approach X(0) = 1
by using a Dirac sequence ϕ n (Chap. V, § 8, n
◦ 27). So X(t) is indeed C
∞ .
This remark has already been made in the classical case [Chap. VII, § 1, n
◦ 2,
(iii)].
Finally note the useful formula
det [exp(X)] = exp [Tr(X)] ,
(15.24)
where, for any square matrixX, Tr(X) denotes the sum of the diagonal elements of X. Setting D(t) = det[exp(tX)] indeed defines an obviously continuous homomorphism from R to R, and so D(t) = exp(ct) for some c = D
(0)
to be computed. As the map det : M n (R) −→ R is much more than differentiable, the multivariate chain rule shows that D
(0) = det
(1) exp
(0)X, where
det
(1) is the derivative at X = 1 of the map X → det X and exp
(0) = 1
that of the exp map at the origin. Hence c = det
(1)X = T (X) with a real
valued linear function of X. By (23),
T (UXU
−1 ) = T (X) for all U ∈ GL n (R) ,
(15.25)
and so T (UX) = T (XU ). As it is not hard to check that every Y ∈ M n (R)
is the sum of invertible matrices (Y + λ1 is invertible provided −λ is not an
eigenvalue of Y ), it can be deduced that
T (XY − Y X) = 0
(15.26)
for all matrices X and Y . Setting T (X) = a
j
i X
i
j , where the X
i
j are the
coefficients of X, and by writing (26) explicitly for a matrix Y all of whose
entries, apart from one, are non-zero, it immediately follows that a
j
i = 0 for
i = j, so that T (X) is a linear combination of the diagonal entries of X. Its
value remains invariant if its entries undergo an arbitrary permutation, for
this amounts to replacing X by UXU
−1 , where the matrix U permutes the
vectors of the canonical basis for R
n . Hence T (X) is proportional to the trace
of X. It remains to check that T (X) = Tr(X) for a matrix X with non-zero
trace; we leave it to the reader to choose this matrix in such a way as to
minimize calculations.
This proof of (24) is slightly longer than the classical proof, but it teaches
the reader, if he does not already know it, that, up to a constant factor,
relation (26) characterizes the function X → Tr(X).
Exercise 1. Show that, for every linear functional X → f (X) on M n (R),
where R denotes an arbitrary commutative field, there is a unique matrix A
such that f (X) = Tr(AX).
Exercise 2. Let E be an n-dimensional vector space over R, M (E) the
set of linear maps E −→ E, (a i ) a basis for E and f (x 1 , . . . , x n ) the unique
n-linear alternating form equals 1 on the basis vectors, i.e. is the determinant
of the x i with respect to this basis. Hence if u ∈ M (E), then