Positive Operator: Why Is \sqrt{A^*A} Positive?

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SUMMARY

The discussion centers on the positivity of the operator \(\sqrt{A^*A}\), where \(A^*\) denotes the adjoint of a linear operator \(A\). It is established that \(A*A\) is a self-adjoint operator with a spectral decomposition, leading to the conclusion that \(\sqrt{A^*A}\) is positive. The discussion references key concepts such as positive semi-definite products and the unique positive square root of positive operators in complex Hilbert spaces. The use of singular value decomposition and Cholesky decomposition is suggested for further exploration of this topic.

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Dragonfall
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Given a linear operator A, why is \sqrt{A^*A} positive? Where A* is the adjoint.
 
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I'll give a matrix argument. First a simple question, take a column vector a, what is the rank ? What is the rank of a*a? Same question for a row vector b?

I would suggest writing the dimensions and checking when the condition is true then carry on with singular value decomposition or Cholesky if you like. you might end up with positive SEMI-definite product! but if you modify the definitions of course, you can recover the same result with some caution.
 
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The way I'd approach it is that A*A is self-adjoint, and hence has a spectral decomposition, and hence the sqrt of A*A is \sum_a\sqrt{a}\left| a\right>\left< a\right|. Somehow, this is positive.
 
This follows from the two following facts/theorems:
1) Every positive operator (on a complex Hilbert space) has a unique positive square root, and
2) A*A is a positive operator.
 

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