Derivative of matrix square root

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To compute the derivative of the square root of a positive definite matrix function A(x) with respect to x, implicit differentiation can be applied to the equation S(x)*S(x) = A(x). This leads to the conclusion that the derivative S'(x) is the unique solution to a Sylvester equation. The discussion emphasizes the importance of defining the principal square root of a matrix, which retains positive definiteness. The Chain Rule is also suggested as a useful approach in this context. Understanding these mathematical principles is crucial for accurately deriving the desired result.
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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