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

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Discussion Overview

The discussion centers around the question of why the operator \(\sqrt{A^*A}\) is considered positive, where \(A^*\) denotes the adjoint of a linear operator \(A\). The scope includes theoretical aspects of linear operators and their properties in the context of functional analysis and operator theory.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests using a matrix argument and poses questions about the rank of column and row vectors, indicating that the conditions for positivity may depend on dimensions and definitions.
  • Another participant proposes that since \(A^*A\) is self-adjoint, it has a spectral decomposition, leading to the conclusion that \(\sqrt{A^*A}\) can be expressed in terms of its eigenvalues and eigenvectors, which implies positivity.
  • A third participant states two facts: every positive operator on a complex Hilbert space has a unique positive square root, and that \(A^*A\) is a positive operator, which they believe supports the claim.

Areas of Agreement / Disagreement

Participants present different approaches to the question, and while some points are supported by theorems, there is no consensus on a singular explanation or method for demonstrating the positivity of \(\sqrt{A^*A}\).

Contextual Notes

The discussion includes various assumptions about the properties of operators and their definitions, which may not be universally agreed upon. The implications of dimensionality and definitions in the context of positivity are also noted but remain unresolved.

Dragonfall
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Given a linear operator A, why is [tex]\sqrt{A^*A}[/tex] 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.
 
Last edited:
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 [tex]\sum_a\sqrt{a}\left| a\right>\left< a\right|[/tex]. 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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