How can I find a solution valid for all cases?

  • #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!
 

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
7,583
1,472


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


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
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413


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...
 

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