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Homework Help: Kronecker exponentiation

  1. Jan 19, 2014 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    Kronecker sum is defined as:

    [tex] A \oplus B = A \otimes I_m + B \otimes I_n [/tex]

    3. 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]


    [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: Jan 19, 2014
  2. jcsd
  3. Jan 19, 2014 #2
    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.
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