Derivative of matrix square root

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Discussion Overview

The discussion revolves around computing the derivative of the square root of a matrix-valued function A(x) with respect to a scalar x. The context includes considerations of matrix properties, particularly focusing on positive definite matrices and the implications for the square root operation.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant inquires about the method to compute the derivative of the square root of a matrix-valued function A(x).
  • Another participant questions how the square root of a matrix is defined, citing examples that illustrate ambiguity in the definition.
  • A participant clarifies that A is positive definite, indicating that the principal square root should be considered.
  • There is a suggestion to apply the Chain Rule, noting that A(x) represents a one-parameter family of positive definite matrices.
  • One participant proposes using implicit differentiation on the equation S(x)*S(x)=A(x) to derive S'(x), suggesting it leads to a Sylvester equation.

Areas of Agreement / Disagreement

Participants express varying viewpoints on the definition of the matrix square root and the implications for differentiation. The discussion does not reach a consensus on the method or definition.

Contextual Notes

The discussion highlights potential ambiguities in defining the square root of a matrix and the assumptions regarding the properties of A(x). The application of the Chain Rule and implicit differentiation introduces additional mathematical considerations that remain unresolved.

cpp6f
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If I have a matrix valued function A(x) of some scalar x, how do I compute the derivative of the square root of A with respect to x? It seems like it should be simple, but I can't find it anywhere on the internet. Thanks!
 
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hi cpp6f! :smile:

how would you define the square root of a matrix? :confused:

eg the square root of the 2x2 identity matrix could be itself, or [1,0;0,-1] :redface:
 
I forgot to mention that A is positive definitive. So the principle square root of A (the square root that is also positive definitive)
 
cpp6f said:
I forgot to mention that A is positive definitive. So the principle square root of A (the square root that is also positive definitive)

See what you can do with the Chain Rule is what you are saying is A(X) is a one parameter family of positive definite square matrices.
 
To save you a bit of footwork - implicit differentiation on the equation S(x)*S(x)=A(x) shows that S'(x) is the unique solution to a Sylvester equation.
 

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