Can Kronecker Exponentiation be Simplified for Matrices in Different Forms?

[prospero]

Homework Statement

I am trying to derive the Kronecker exponentiation relation:

e^A \otimes e^B = e^{A \oplus B}

with A,B as n-by-n and m-by-m matrices.

Homework Equations

Kronecker sum is defined as:

A \oplus B = A \otimes I_m + B \otimes I_n

The Attempt at a Solution

I first tackle the simpler case of A,B and I all being 2-by-2 matrices.

A \otimes I = \begin{bmatrix} A_{11} I & A_{12} I \\ A_{21} I & A_{22} I \end{bmatrix}

and

I \otimes B = \begin{bmatrix} B & 0 \\ 0 & B \end{bmatrix}

Then I can represent A \otimes I as the sum of one diagonal and two nilpotent matrices. This makes their matrix exponentiation easier. I \otimes B is a diagonal block matrix and exponentiating it, we get a block matrix with blocks of e^B along the diagonal, i.e.:

e^{I \otimes B} = \begin{bmatrix} e^B & 0 \\ 0 & e^B \end{bmatrix}

I tried multiplying the matrices together and it might have worked but it was taking too much time and it is clearly not feasible to prove the general case.

However I notice that e^{I \otimes B} = I \otimes e^B.

Now if we can write e^{A \otimes I} = e^A \otimes I the problem is over since ( e^A \otimes I ) ( I \otimes e^B ) = e^A \otimes e^B.

However I am not aware of a property of matrix exponentiation that justifies the last step.

 

[brmath]

Here is a way to tackle your last question. The statement is true if A is diagonal. Can you show it is true if A is diagonalizable? Can you show it is true if A is not diagonalizable but is in Jordan form? I believe if A is in Jordan form with blocks $J_1, J_2 ...$ etc then $e^A$ is just $e^{J_1}, e^{J_2},$ etc.

Since most of the properties of the underlying matrices are the same under the Kronecker product, I think this would most likely work out.