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Is there a notion of an ``average'' of a set of linear operators?

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$begingroup$

Say we have a set of (invertible) linear operators
$ A, B, C $. Is there any meaningful sense in which we could conceive of an ``average'' of these linear operators?

It would seem like the easy first answer would be some version of geometric mean, but for the fact that operators do not commute.

Another idea I thought about would be some (invertible) $D$ s.t. for $A' := A D^-1$, $B' := B D^-1$, and $C' := C D^-1$ we have $A' B' C' = I$, but again I can't see how this works given that operators don't commute.

I believe this question was asked previously here:

Do matrices have average and fluctuations?

But I don't think the response really answered the OP's question.

linear-algebra matrices definition lie-groups

share|cite|improve this question

asked Mar 30 at 4:07

seeker_after_truthseeker_after_truth

535

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Why geometric? What is the average of the three image vectors $Ax$, $Bx$, $Cx$? Well, $frac 1 3 (Ax + Bx + Cx)$ is one. But that's the image of $frac 1 3 (A + B + C)$. So is that sensible average of these three operators? Or do you have any particular reason for preferring a "multiplicative" one?
  $endgroup$
  – M. Vinay
  Mar 30 at 5:36
  
  

  
  
  
  


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  $begingroup$
  Operator spaces are Banach spaces. You can put probability measures on Banach spaces, e.g. Gaussian ones, and average over them.
  $endgroup$
  – Conifold
  Mar 30 at 6:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for these good points!
  $endgroup$
  – seeker_after_truth
  Mar 30 at 17:30
  

  
  
  
  

add a comment | 

0

$begingroup$

Say we have a set of (invertible) linear operators
$ A, B, C $. Is there any meaningful sense in which we could conceive of an ``average'' of these linear operators?

It would seem like the easy first answer would be some version of geometric mean, but for the fact that operators do not commute.

Another idea I thought about would be some (invertible) $D$ s.t. for $A' := A D^-1$, $B' := B D^-1$, and $C' := C D^-1$ we have $A' B' C' = I$, but again I can't see how this works given that operators don't commute.

I believe this question was asked previously here:

Do matrices have average and fluctuations?

But I don't think the response really answered the OP's question.

linear-algebra matrices definition lie-groups

share|cite|improve this question

asked Mar 30 at 4:07

seeker_after_truthseeker_after_truth

535

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Why geometric? What is the average of the three image vectors $Ax$, $Bx$, $Cx$? Well, $frac 1 3 (Ax + Bx + Cx)$ is one. But that's the image of $frac 1 3 (A + B + C)$. So is that sensible average of these three operators? Or do you have any particular reason for preferring a "multiplicative" one?
  $endgroup$
  – M. Vinay
  Mar 30 at 5:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Operator spaces are Banach spaces. You can put probability measures on Banach spaces, e.g. Gaussian ones, and average over them.
  $endgroup$
  – Conifold
  Mar 30 at 6:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for these good points!
  $endgroup$
  – seeker_after_truth
  Mar 30 at 17:30
  

  
  
  
  

add a comment | 

$begingroup$

Say we have a set of (invertible) linear operators
$ A, B, C $. Is there any meaningful sense in which we could conceive of an ``average'' of these linear operators?

It would seem like the easy first answer would be some version of geometric mean, but for the fact that operators do not commute.

Another idea I thought about would be some (invertible) $D$ s.t. for $A' := A D^-1$, $B' := B D^-1$, and $C' := C D^-1$ we have $A' B' C' = I$, but again I can't see how this works given that operators don't commute.

I believe this question was asked previously here:

Do matrices have average and fluctuations?

But I don't think the response really answered the OP's question.

linear-algebra matrices definition lie-groups

share|cite|improve this question

asked Mar 30 at 4:07

seeker_after_truthseeker_after_truth

535

$endgroup$

share|cite|improve this question

asked Mar 30 at 4:07

seeker_after_truthseeker_after_truth

535

share|cite|improve this question

asked Mar 30 at 4:07

seeker_after_truthseeker_after_truth

535

asked Mar 30 at 4:07

seeker_after_truthseeker_after_truth

535

asked Mar 30 at 4:07

seeker_after_truthseeker_after_truth

535