﻿ A Beautiful Theorem Concerning Determinants
San José State University

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Thayer Watkins
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 A Beautiful Theorem Concerning Determinants

Let A be a square matrix with complex elements. There is a theorem for determinants that says that the determinant of the exponential function of a matrix is equal to the exponential function of the trace of the matrix; in symbols

#### det(exp(A)) = exp(tr(A))

First consider the special simple case when A is a diagonal matrix, say Λ=diag(λ1, λ2, …, λn). In this case exp(Λ) is also a diagonal matrix, namely diag(exp(λ1), exp(λ2), …, exp(λn)).

The determinant of exp(Λ) is just the product

#### det(exp(Λ) = exp(λ1)·exp(λ2)···exp(λn) which reduces to det(exp(Λ) = exp(λ1+λ2+ … + λn) which is the same as det(exp(Λ) = exp(tr(Λ))

Next consider the special case of matrices which are diagonalizable. For such a matrix A there exists a matrix P such that A is equal to P-1ΛP where Λ is a diagonal matrix. The exponential function of a square matrix is defined in terms of the same sort of infinite series that defines the exponential function of a single real number; i.e.,

#### exp(A) = I + A + (1/2!)A² + (1/3!)A³ + …

where I is the appropriate identity matrix.

When P-1ΛP is substituted into A² the result is

#### A² = (P-1ΛP)(P-1ΛP) = P-1Λ(PP-1)ΛP = P-1Λ(PP-1)ΛP P-1Λ(I)ΛP = P-1ΛΛP = P-1Λ²P

Likewise all powers of A are similarly reduced. Thus

#### exp(A) = P-1[I + Λ = (1/2!)Λ² + (1/3!)Λ³ + …]P

Thus it is also true that exp(A) is equal to P-1exp(Λ)P. Furthermore the determinant of a product of square matrices is the product of their determinants. Thus

#### det(exp(A)) = det(P-1exp(Λ)P) = det(P-1)det(exp(Λ)det(P) = det(P-1)det(P)det(exp(Λ) = det(P-1P)det(exp(Λ) = det(I)det(exp(Λ) = exp(tr(Λ))

Another theorem concerning determinants is that the trace of a matrix is equal to the sum of its eigenvalues (taking into account their multiplicities). Thus

#### tr(Λ) = tr(A) and hence det(exp(A)) = exp(tr(A))

Now for the general case: For any n×n matrix A there exists an n×n matrix Q such that

#### QAQ-1 = J and hence A = Q-1JQ

where J is of the canonical Jordan form; i.e., it has the eigenvalues of A on the principal diagonal, the elements next to the principal diagonal on the right are 0 or 1 and all other elements are zero.

From the construction for the previous case it follows that

Therefore

#### det(exp(A)) = det(Q-1exp(J)Q) = det(exp(J))

But all of the elements below the principal diagonal are zero for exp(J) as well as for J. Thus det(exp(J)) is equal to the product of all elements on its principal diagonal. But the elements on the principal diagonal of exp(J) are the exponential function of the elements on the principal diagonal of J. (See Powers of Jordan canonical form matrices

Thus

#### det(exp(J)) = Πexp(Jj,j) = exp(ΣJj,j) = exp(tr(J))

But the trace of J is equal to the trace of A so

#### det(exp(A)) = exp(tr(A))

(To be continued.)