# Derivative of a scalar variable into matrix expressions

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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

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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.

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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.

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• 2 weeks later...

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

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