Proof of transpose property of matrix exponential

Question

Using the fact that the matrix transpose distributes over infinite sums to show that $e^{(A^T)} = (e^A)^T$. I feel like this is really trivial, but I don't know quite how to prove this. How would I go about proving it?

Answer 1

The essential facts:

Transposition commutes with sums and powers: $$(A+B)^T=A^T+B^T,\qquad\qquad (A^n)^T=(A^T)^n.$$ The transposition is continuous: $$\lim_{n\to\infty}(A_n^T)=(\lim_{n\to\infty}A_n)^T.$$ And infinite sums are limits of partial sums.

Answer 2

Well, if you can use that distribution fact, then

$$e^{A^t}=\sum_{n=0}^\infty\frac1{n!}\left(A^{t}\right)^n=\sum_{n=0}^\infty\frac1{n!}\left(A^n\right)^t=\left(\sum_{n=0}^\infty\frac1{n!} A^n\right)^t=\left(e^A\right)^t$$

Answer 3

First, write out $$ e^{(A^T)} = I + \sum_1^\infty \frac{(A^T)^n}{n!} $$which does not require distributuivity of transpose, it is just exponentiation of an arbitrary matrix.

Then look at $$ e^A = I + \sum_1^\infty \frac{A^n}{n!} $$ And use distributivity of transpose and $(A^n)^T = (A^T)^N$to write $$ (e^A)^T = I + \sum_1^\infty \frac{(A^T)^n}{n!} $$