Tassus Posted February 18, 2012 Share Posted February 18, 2012 I hope someone could explain how to calculate the following derivatives, Assume A, B are kxk full rank matrices I is identity matrix of the same order and τ is a nonnegative scalar variable and λ real number then define the following two functions φ(τ) = ln(det(τΑ+λΙ)) ψ(τ) = tr((τΑ+λΙ)-1)B which are the first and second derivatives of φ(.) and ψ(.) w.r.t. the argument τ? A proof and/or references should be usefull. Thank you Link to comment Share on other sites More sharing options...
DrRocket Posted February 18, 2012 Share Posted February 18, 2012 I hope someone could explain how to calculate the following derivatives, Assume A, B are kxk full rank matrices I is identity matrix of the same order and τ is a nonnegative scalar variable and λ real number then define the following two functions φ(τ) = ln(det(τΑ+λΙ)) ψ(τ) = tr((τΑ+λΙ)-1)B which are the first and second derivatives of φ(.) and ψ(.) w.r.t. the argument τ? A proof and/or references should be usefull. Thank you I see no reason to expect a simple or elegant expression for the derivatives that you seek. Take a look at simple case, say 2x2 diagonal matrices, and see how messy the expression is. It gets uglier from there. Link to comment Share on other sites More sharing options...
Tassus Posted February 18, 2012 Author Share Posted February 18, 2012 I see no reason to expect a simple or elegant expression for the derivatives that you seek. Take a look at simple case, say 2x2 diagonal matrices, and see how messy the expression is. It gets uglier from there. I think that there must be some expressions in terms of traces, determinants and things like that, notice that in the special case where λ=0 the function φ(τ) = ln(det(τΑ))=ln(τkdet(A))=klnτ+lndet(A) so the first derivative for example is just k/τ. So for the general problem I think that it can be written in a compact convenient form, for example in case of 2x2 matrices (k=2) φ(τ)=ln(τ2det(A)+λτtr(A)+λ2) which is an expression where the derivatives can easily be derived. Link to comment Share on other sites More sharing options...
Tassus Posted March 1, 2012 Author Share Posted March 1, 2012 Ι think that the derivatives are the following, Is there anyone to confirm this? Assume A= A' and B=B' then φ(τ) = ln(det(τΑ+λΙ)) φ'(τ) = tr(τΑ+λΙ)-1A ψ(τ) = tr(τΑ+λΙ)-1B ψ'(τ) = tr(τΑ+λΙ)-1B(τΑ+λΙ)-1)A ψ''(τ) = -2tr(τΑ+λΙ)-1B(τΑ+λΙ)-1)A(τΑ+λΙ)-1)A Link to comment Share on other sites More sharing options...
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