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How can I find a solution valid for all cases?

  1. Nov 10, 2011 #1
    How can I find a solution valid for "all" cases?

    I have an equation:

    [itex]tr\left\{\textbf{AB}\right\} = \sigma[/itex]

    where [itex]tr\left\{\right\}[/itex] denotes the matrix trace. The square matrix [itex]\textbf{A}[/itex] is independent of both the square matrix [itex]\textbf{B}[/itex] and the real scalar [itex]\sigma[/itex].

    I want to determine all possible values of [itex]\textbf{B}[/itex] that will allow the above equation to hold for all [itex]\textbf{A}[/itex], given the only constraint:

    [itex]tr\left\{\textbf{A}\right\} = 1[/itex]

    For example, I can see that [itex]\textbf{B}=\sigma \textbf{I}[/itex] will always be valid (where [itex]\textbf{I}[/itex] is the identity matrix). But can I guarantee that there are no other possible values for [itex]\textbf{B}[/itex]?

    I have been pondering this problem for some time and cannot see a way of approaching it. Any advice would be greatly appreciated!
  2. jcsd
  3. Nov 11, 2011 #2

    Stephen Tashi

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    Re: How can I find a solution valid for "all" cases?

    Let [itex] A_{i,j} [/itex] denote a matrix whose [itex] (i,j) [/itex] entry is 1 and the rest of whose entries are zero. If the equation was true for all matrices it would work for each [itex] A_{i,j} [/itex]. Is that going to be possible?
  4. Nov 11, 2011 #3
    Re: How can I find a solution valid for "all" cases?

    I think we can only consider the [itex]A_{i,i}[/itex] matrices (i.e. a diagonal entry is 1), in order to satisfy the trace constraint on [itex]A[/itex]. However, I'll look into an approach like this. I have found a possible solution in a similar way - by considering the (i,j)th element. I'll check it over and post it here if I think it has any chance of being correct...
  5. Nov 11, 2011 #4
    Re: How can I find a solution valid for "all" cases?

    I hope someone will have the time to read through the following attempt and point out any errors in my reasoning:

    The [itex]\left( i,i\right) ^{th}[/itex] element of [itex]\boldsymbol{AB}[/itex] can be written as:

    \left[ \boldsymbol{AB}\right] _{i,i}=\sum_{j=1}^{N}a_{ij}b_{ji}


    Therefore the trace is:

    {tr}\left\{ \boldsymbol{AB}\right\}

    Our only constraint is:


    {tr}\left\{ \boldsymbol{A}\right\} =\sum_{i=1}^{N}a_{ii}=1


    Therefore, we can write:


    {tr}\left\{ \boldsymbol{AB}\right\} &=&\sigma {tr}\left\{
    \boldsymbol{A}\right\} \\
    \sum_{i=1}^{N}\sum_{j=1}^{N}a_{ij}b_{ji} &=&\sigma \sum_{i=1}^{N}a_{ii}

    Clearly, the right hand side is only a function of diagonal terms.
    Therefore, the [itex]a_{ij}[/itex] ([itex]i\neq j[/itex]) terms must be eliminated from the left
    hand side by assuming all [itex]b_{ji}[/itex] ([itex]j\neq i[/itex]) terms are zero.

    Therefore [itex]\boldsymbol{B}[/itex] must be diagonal and:


    \sum_{i=1}^{N}a_{ii}b_{ii}=\sigma \sum_{i=1}^{N}a_{ii}


    The only way (I think) [itex]\boldsymbol{B}[/itex] can satisfy this in general is if
    the [itex]b_{ii}[/itex] terms can be removed from the summation (i.e. they are a
    constant). Clearly, that constant is [itex]\sigma [/itex]:

    \boldsymbol{B}=\sigma \boldsymbol{I}
  6. Nov 12, 2011 #5


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    Re: How can I find a solution valid for "all" cases?

    I didn't quite follow your argument about diagonal terms, but I think you have at least started on the right track. I also think you're making it a bit more complicated than it needs to be. I would begin like this: Let A be an arbitrary matrix such that Tr A=1. Let B be an arbitrary matrix such that Tr(AB)=σ. Then Tr(AB)=σTr(A).

    Now start making specific choices of A, to see how they constrain B. For example, what does the equality Tr(AB)=σTr(A) say when A11=1 and all other components of A are =0?
  7. Nov 12, 2011 #6
    Re: How can I find a solution valid for "all" cases?

    Thanks for the advice. I think this idea of choosing various specific examples for [itex]\textbf{A}[/itex] is a very good one and could simplify matters significantly. I'll see what I can come up with...
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