Inverse of an operator does not exist, can't see why

In summary, we discussed the non-existence of an inverse for the operator g_{\mu \nu} \Box - \partial_{\mu}\partial_{\nu} when converted to momentum space. We explored the possible form of an inverse and found that it would result in a contradiction. Additionally, we looked at the non-trivial projection operator and showed that it has no inverse through a simple equation.
  • #1
Elwin.Martin
207
0
I feel kind of lame, but here's my situation:
We start with the operator [itex]g_{\mu \nu} \Box - \partial_{\mu}\partial_{\nu}[/itex] and convert to momentum space to get [itex]-g_{\mu \nu} k^{2} - k_{\mu}k_{\nu}[/itex].

Apparently it's easy to see that this has no inverse?

I'm told that if it *did* it would be of the form
[itex]Ag^{\nu \lambda} +B k^{\nu}k^{\lambda}[/itex]
but I don't see why it would be in this form, to start with. Are these A's and B's just constant coefficients...?

I understand that given the form above, we can simply multiply our inverse and original operator and we get that [itex]-Ak^{2}\delta_{\mu}^{\lambda} +A k_{\mu}k^{\lambda}=\delta_{\mu}^{\lambda}[/itex]
. . . but I don't see why this is an issue, or rather, I can't see why this equation has no solution.

Thanks for any and all help,
E_Martin
 
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  • #2
Mentz114 said:
I don't think these equations have a solution

[tex]
-Ak^{2}\delta_{\mu}^{\lambda} +A k_{\mu}k^{\lambda}=\delta_{\mu}^{\lambda}
[/tex]

but I can't prove it right now.

:P then you're in agreement with Ryder...I don't see it though. I have the feeling it doesn't, but I cannot show it.
 
  • #3
Elwin.Martin said:
:P then you're in agreement with Ryder...I don't see it though. I have the feeling it doesn't, but I cannot show it.

Sorry, I deleted my post because I'm looking for a solution and it might be on.

Have you found a solution ?
[Edit] Any solution leads to a contradiction, so there isn't one. Try writing out all 16 equations.
 
Last edited:
  • #4
Mentz114 said:
Sorry, I deleted my post because I'm looking for a solution and it might be on.

Have you found a solution ?
[Edit] Any solution leads to a contradiction, so there isn't one. Try writing out all 16 equations.

OH, WOW...thanks!

Just writing out the first component works, haha...fail.

Thanks again!
 
  • #5
Non-trivial projection operator has no inverse; [itex]P^{2}=P[/itex]. Consider
[tex]P_{ab} = g_{ab} - \frac{k_{a}k_{b}}{k^{2}},[/tex]
[tex]P^{ac}= g^{ac} - \frac{k^{a}k^{c}}{k^{2}}.[/tex]
then
[tex]P_{ab}P^{ac} = \delta^{c}_{b} - \frac{k_{b}k^{c}}{k^{2}} \equiv P^{c}_{b}[/tex]
 

Related to Inverse of an operator does not exist, can't see why

1. What is an inverse of an operator?

An inverse of an operator is a mathematical operation that "undoes" the original operation. For example, the inverse of addition is subtraction, and the inverse of multiplication is division.

2. Why might an inverse of an operator not exist?

An inverse of an operator may not exist if the original operation is not reversible or if there are certain restrictions on the values that can be used. For example, the inverse of taking the square root of a number only exists for non-negative numbers.

3. How can I determine if the inverse of an operator exists?

This can be determined by examining the properties of the operation. For an inverse to exist, the operation must be both one-to-one and onto. This means that each input has a unique output and that every possible output has a corresponding input.

4. Can an operator have more than one inverse?

Yes, it is possible for an operator to have multiple inverses. For example, both addition and multiplication have multiple inverses (subtraction and division for addition, and division and multiplication for multiplication). However, an operator can only have one left inverse and one right inverse.

5. How is the concept of an inverse operator used in science?

The concept of an inverse operator is used in many scientific fields, such as physics, engineering, and computer science. It allows for the simplification of complex equations and helps in solving problems that involve multiple operations. It is also used in data analysis and signal processing to "reverse" the effects of certain operations and retrieve original data.

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