- #1
prospero
- 1
- 0
Homework Statement
I am trying to derive the Kronecker exponentiation relation:
[tex] e^A \otimes e^B = e^{A \oplus B} [/tex]
with A,B as n-by-n and m-by-m matrices.
Homework Equations
Kronecker sum is defined as:
[tex] A \oplus B = A \otimes I_m + B \otimes I_n [/tex]
The Attempt at a Solution
I first tackle the simpler case of A,B and I all being 2-by-2 matrices.
[tex] A \otimes I = \begin{bmatrix} A_{11} I & A_{12} I \\ A_{21} I & A_{22} I \end{bmatrix} [/tex]
and
[tex] I \otimes B = \begin{bmatrix} B & 0 \\ 0 & B \end{bmatrix} [/tex]
Then I can represent [itex] A \otimes I [/itex] as the sum of one diagonal and two nilpotent matrices. This makes their matrix exponentiation easier. [itex] I \otimes B [/itex] is a diagonal block matrix and exponentiating it, we get a block matrix with blocks of [itex] e^B [/itex] along the diagonal, i.e.:
[tex] e^{I \otimes B} = \begin{bmatrix} e^B & 0 \\ 0 & e^B \end{bmatrix} [/tex]
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 [itex] e^{I \otimes B} = I \otimes e^B [/itex].
Now if we can write [itex] e^{A \otimes I} = e^A \otimes I [/itex] the problem is over since [itex] ( e^A \otimes I ) ( I \otimes e^B ) = e^A \otimes e^B [/itex].
However I am not aware of a property of matrix exponentiation that justifies the last step.
Last edited: