How can I find a solution valid for all cases?

  • Thread starter weetabixharry
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In summary, the equation tr\left\{\textbf{AB}\right\} = \sigma can be solved for all possible values of \textbf{B} that will allow it to hold, given the only constraint that tr\left\{\textbf{A}\right\} = 1. By considering the (i,j)th element of \boldsymbol{AB}, it can be determined that \boldsymbol{B} must be diagonal and that the b_{ii} terms can be eliminated from the summation. This solution may be simplified by making use of specific examples.
  • #1
weetabixharry
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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!
 
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  • #2


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?
 
  • #3


Stephen Tashi said:
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?

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...
 
  • #4


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:

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

[/itex]

Therefore the trace is:

[itex]
{tr}\left\{ \boldsymbol{AB}\right\}
=\sum_{i=1}^{N}\sum_{j=1}^{N}a_{ij}b_{ji}=\sigma
[/itex]

Our only constraint is:

[itex]

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

[/itex]

Therefore, we can write:

[itex]

\begin{eqnarray*}
{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}
\end{eqnarray*}
[/itex]

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:

[itex]

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

[/itex]

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]:

[itex]
\boldsymbol{B}=\sigma \boldsymbol{I}
[/itex]
 
  • #5


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?
 
  • #6


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...
 

1. What is the process for finding a solution valid for all cases?

The process for finding a solution valid for all cases involves identifying the problem, gathering data and information, analyzing the data, and then using logical reasoning and scientific principles to develop and test potential solutions. This process may also involve conducting experiments, using mathematical models, and consulting with other experts in the field.

2. How do I ensure that my solution will work for all cases?

To ensure that a solution will work for all cases, it is important to thoroughly test and validate the solution. This may involve conducting experiments, collecting data, and analyzing the results. It is also important to consider potential limitations and alternate scenarios that could affect the solution's effectiveness.

3. What factors should I consider when developing a solution for all cases?

When developing a solution for all cases, it is important to consider factors such as the problem's complexity, the available resources, and any potential ethical or societal implications. It is also important to consider the potential consequences of the solution and how it may impact different populations.

4. How can I ensure that my solution is scientifically sound?

To ensure that a solution is scientifically sound, it is important to use established scientific principles and methods. This may involve conducting experiments, using mathematical models, and consulting with other experts in the field. It is also important to thoroughly analyze and interpret the data and results to ensure they support the proposed solution.

5. Can a single solution be valid for all cases?

In many cases, a single solution may not be valid for all cases. This is because different problems and scenarios may require different approaches and solutions. However, by using scientific principles and methods, it is possible to develop a solution that is valid for a wide range of cases and can be adapted for specific scenarios.

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