 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016-resit/10/c/solution

In a submultiplicative matrix norm, the exponential series is absolutely convergent. Its Cauchy product can therefore be grouped by the total degree:

e^{A} e^{B} = \sum_{n = 0}^{\infty} \sum_{j = 0}^{n} \frac{A ^{j} B ^{n - j}}{j ! ( n - j )!} = \sum_{n = 0}^{\infty} \frac{1}{n !} \sum_{j = 0}^{n} (j n) A^{j} B^{n - j} .

When

A B = B A

, the inner sum is

(A + B)^{n}

e^{A} e^{B} = e^{A + B}

. The commutation assumption is used precisely in that binomial expansion; absolute convergence alone does not supply it.

 New to topics? Read the docs here!