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Trace of a particular matrix product

  1. Apr 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Claim: If ##A \in \mathcal{M}_n (\mathbb{C})## is arbitrary, and ##D## is a matrix with ##\beta## in its ##(i-j)##-th entry, and ##\overline{\beta}## in its ##(j-i)##-th, where ##i \ne j##, and with zeros elsewhere, then ##Tr(AD) = a_{ij} \beta + a_{ji} \overline{\beta}##

    2. Relevant equations


    3. The attempt at a solution

    I am having difficulty with sum notation. By definition, the ##(l-m)##-th entry of the matrix product ##AD## is

    ##\displaystyle (AD)_{lm} = \sum_{k=1}^n a_{lk} D_{km}##

    And so the trace should be

    ##\displaystyle Tr(AD) = \sum_{q = 1}^n ((AD)_{lm})_q = \sum_{q=1}^n \sum_{k=1}^n a_{lk} D_{km}##

    Given the description of the matrix ##D##, it would seem that ##(D)_{lm} = 0## whenever ##l \ne i##, ##l \ne j##, ##m \ne i##, or ##m \ne j##. However, I am unsure about this and am having difficulty properly splitting up the sum. Could someone guide me along?
     
  2. jcsd
  3. Apr 10, 2015 #2

    Zondrina

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    There are only two elements of that sum which survive. The ##a_{ij}D_{ij}## term and the ##a_{ji}D_{ji}## term. Every other term will be zero.

    You and I both know who those survivors are ;).
     
  4. Apr 10, 2015 #3
    Intuitively it is obvious. But shouldn't we use the definition to formally show it is valid?
     
  5. Apr 10, 2015 #4

    Zondrina

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    Okay so we know:

    $$Tr(AD) = \sum_{i=1} \sum_{j=1} A_{ij} D_{ji}$$

    Maybe this notation will be more convenient for this problem.
     
  6. Apr 10, 2015 #5
    Perhaps, although I am having double the difficulty with two sums involved.
     
  7. Apr 10, 2015 #6

    Zondrina

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    It might be easier to think about it in more steps. What would the matrix product ##AD## produce on its own? Take the trace of the resulting matrix, which happens to be the sum of the two diagonal entries.

    How can this help you argue only two terms survive?
     
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