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1

$begingroup$

Suppose we have a function $xmapsto A(x)$ where $A(x)$ is an $ntimes n$ matrix with differentiable matrix elements $[A(x)]_ij=a_ij(x)$. Then it is natural to take
beginalign*
fracdAdx := beginbmatrix
fracd a_11dx & cdots & fracd a_1ndx
\
vdots & ddots & vdots
\
fracda_n1dx & cdots & fracda_nndx
endbmatrix
endalign*

My question is as follows; suppose we have some function $f$ that is sufficiently "nice" (differentiable, etc.). Then, is there a "chain-rule like" expression for
beginalign*
fracddx f(A(x)).
endalign*
For example, is there a general criterion for when we can write
beginalign*
fracddx f(A(x)) = fracd f(A)dAfracdAdx ?
endalign*

This question came to me in a more concrete example. Suppose $A(x)$ is differential, positive definite matrix for all $xinmathbbR$. Is there a known formula for
$dS/dx$ where $S(x) := A^1/2(x)$ is the unique positive definite square root of $A(x)$? One result that follows by differentiating the expression $S^2(x) =A(x)$ is
beginalign*
S(x)fracdSdx + fracdSdxS(x) = fracdAdx
endalign*
which can be viewed as a Sylvester equation for the matrix $dS/dx$. If $S$ and $dS/dx$ commute, then this can be solved as
$$fracdSdx = frac12A^-1/2fracdAdx$$
which is exactly what we would get if we naively applied the chain rule. Does this Sylvester equation have a closed form solution for the general case?

Thanks!

calculus linear-algebra matrix-calculus

share|cite|improve this question

edited Mar 29 at 16:37

UglyMousanova19

asked Mar 29 at 15:30

UglyMousanova19UglyMousanova19

464

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  DId you mean to say if $S$ and $fracdSdx$ commute? becasue they usually do not.
  $endgroup$
  – Mark Fischler
  Mar 29 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's what I meant. Fixed the typo, thanks!
  $endgroup$
  – UglyMousanova19
  Mar 29 at 16:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The chain rule is quite general, and in its finest form just says $d(fcirc g) = df circ dg$. Of course, there is a fair bit of intepretation that must be done to come up with what that means in your case, which I don't have time to trace down. But rest assured that what you've written can be given a precise meaning where it (or a formula very similar) will be true.
  $endgroup$
  – Paul Sinclair
  Mar 30 at 1:13
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose we have a function $xmapsto A(x)$ where $A(x)$ is an $ntimes n$ matrix with differentiable matrix elements $[A(x)]_ij=a_ij(x)$. Then it is natural to take
beginalign*
fracdAdx := beginbmatrix
fracd a_11dx & cdots & fracd a_1ndx
\
vdots & ddots & vdots
\
fracda_n1dx & cdots & fracda_nndx
endbmatrix
endalign*

My question is as follows; suppose we have some function $f$ that is sufficiently "nice" (differentiable, etc.). Then, is there a "chain-rule like" expression for
beginalign*
fracddx f(A(x)).
endalign*
For example, is there a general criterion for when we can write
beginalign*
fracddx f(A(x)) = fracd f(A)dAfracdAdx ?
endalign*

This question came to me in a more concrete example. Suppose $A(x)$ is differential, positive definite matrix for all $xinmathbbR$. Is there a known formula for
$dS/dx$ where $S(x) := A^1/2(x)$ is the unique positive definite square root of $A(x)$? One result that follows by differentiating the expression $S^2(x) =A(x)$ is
beginalign*
S(x)fracdSdx + fracdSdxS(x) = fracdAdx
endalign*
which can be viewed as a Sylvester equation for the matrix $dS/dx$. If $S$ and $dS/dx$ commute, then this can be solved as
$$fracdSdx = frac12A^-1/2fracdAdx$$
which is exactly what we would get if we naively applied the chain rule. Does this Sylvester equation have a closed form solution for the general case?

Thanks!

calculus linear-algebra matrix-calculus

share|cite|improve this question

edited Mar 29 at 16:37

UglyMousanova19

asked Mar 29 at 15:30

UglyMousanova19UglyMousanova19

464

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  DId you mean to say if $S$ and $fracdSdx$ commute? becasue they usually do not.
  $endgroup$
  – Mark Fischler
  Mar 29 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's what I meant. Fixed the typo, thanks!
  $endgroup$
  – UglyMousanova19
  Mar 29 at 16:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The chain rule is quite general, and in its finest form just says $d(fcirc g) = df circ dg$. Of course, there is a fair bit of intepretation that must be done to come up with what that means in your case, which I don't have time to trace down. But rest assured that what you've written can be given a precise meaning where it (or a formula very similar) will be true.
  $endgroup$
  – Paul Sinclair
  Mar 30 at 1:13
  

  
  
  
  

add a comment | 

$begingroup$

Suppose we have a function $xmapsto A(x)$ where $A(x)$ is an $ntimes n$ matrix with differentiable matrix elements $[A(x)]_ij=a_ij(x)$. Then it is natural to take
beginalign*
fracdAdx := beginbmatrix
fracd a_11dx & cdots & fracd a_1ndx
\
vdots & ddots & vdots
\
fracda_n1dx & cdots & fracda_nndx
endbmatrix
endalign*

My question is as follows; suppose we have some function $f$ that is sufficiently "nice" (differentiable, etc.). Then, is there a "chain-rule like" expression for
beginalign*
fracddx f(A(x)).
endalign*
For example, is there a general criterion for when we can write
beginalign*
fracddx f(A(x)) = fracd f(A)dAfracdAdx ?
endalign*

This question came to me in a more concrete example. Suppose $A(x)$ is differential, positive definite matrix for all $xinmathbbR$. Is there a known formula for
$dS/dx$ where $S(x) := A^1/2(x)$ is the unique positive definite square root of $A(x)$? One result that follows by differentiating the expression $S^2(x) =A(x)$ is
beginalign*
S(x)fracdSdx + fracdSdxS(x) = fracdAdx
endalign*
which can be viewed as a Sylvester equation for the matrix $dS/dx$. If $S$ and $dS/dx$ commute, then this can be solved as
$$fracdSdx = frac12A^-1/2fracdAdx$$
which is exactly what we would get if we naively applied the chain rule. Does this Sylvester equation have a closed form solution for the general case?

Thanks!

calculus linear-algebra matrix-calculus

share|cite|improve this question

edited Mar 29 at 16:37

UglyMousanova19

asked Mar 29 at 15:30

UglyMousanova19UglyMousanova19

464

$endgroup$

share|cite|improve this question

edited Mar 29 at 16:37

UglyMousanova19

asked Mar 29 at 15:30

UglyMousanova19UglyMousanova19

464

share|cite|improve this question

edited Mar 29 at 16:37

UglyMousanova19

asked Mar 29 at 15:30

UglyMousanova19UglyMousanova19

464

asked Mar 29 at 15:30

UglyMousanova19UglyMousanova19

464

asked Mar 29 at 15:30

UglyMousanova19UglyMousanova19

464