Calculating the Inverse of Operator L in R^2

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The discussion centers on finding the inverse of the operator L defined as L = \widehat{1} + |u><v| in R^2. The correct inverse, under the condition that <u|v>=0 (orthogonality), is identified as L^{-1} = 1 - |u><v|. A suggested method for deriving the inverse involves assuming a form L^{-1} = a\widehat{1} + b|u><v| and solving for coefficients a and b through the equation LL^{-1} = L^{-1}L = \widehat{1}. One participant acknowledges having previously derived a similar inverse without the orthogonality assumption. The conversation highlights strategies for algebraic manipulations in operator theory.
Bimmel
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hello,

given is the Operator L=\widehat{1}+|u><v|, where \widehat{1} means the unity-tensor.

Whats the inverse of L?

I calculated the inverse of L in R^2 but I don't get it back to the bra-ket-notation. Can somebody help?


BTW: Sorry for my bad english!
 
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The inverse of this operator, assuming <u|v>=0, that is orthogonality, is given by 1-|u><v|.

When in doubt about this matter put

L^{-1}=a\widehat{1}+b|u&gt;&lt;v|

and using LL^{-1}=L^{-1}L=\widehat{1} fix coeficients a and b. This is a general strategy for this kind of algebraic manipulations.


Jon
 
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Thx for your fast answer!

I think a had already that kind of inverse operator, but I didn't assume that <u|v>=0. :shy:

Tobi
 

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