Taylor expansion for matrix logarithm

[Backpacker]

A paper I'm reading states the that: for positive hermitian matrices A and B, the Taylor expansion of [itex]\log(A+tB)[/itex] at t=0 is

[itex]\log(A+tB)=\log(A) + t\int_0^\infty \frac{1}{B+zI}A \frac{1}{B+zI} dz + \mathcal{O}(t^2).[/itex]

However, there is no source or proof given, and I cannot seem to find a derivation of this identity anywhere! Any help would be appreciated. Thanks.

 

[I like Serena]

A paper I'm reading states the that: for positive hermitian matrices A and B, the Taylor expansion of [itex]\log(A+tB)[/itex] at t=0 is

[itex]\log(A+tB)=\log(A) + t\int_0^\infty \frac{1}{B+zI}A \frac{1}{B+zI} dz + \mathcal{O}(t^2).[/itex]

However, there is no source or proof given, and I cannot seem to find a derivation of this identity anywhere! Any help would be appreciated. Thanks.

Welcome to PF, Backpacker!

I don't recognize your formula, but:

$$\log(A+tB)=\log(A(I+tA^{-1}B)= \log A + \log(I+tA^{-1}B) = \log A + tA^{-1}B + \mathcal{O}(t^2)$$

 

[Stephen L Adler]

For a derivation see: https://vpn.ias.edu/ckfinder/userfiles/files/,DanaInfo=www.sns.ias.edu,SSL+matrixlog_tex.pdf

 

[wabbit]

$$\log(A(I+tA^{-1}B)= \log A + \log(I+tA^{-1}B) $$

This doesn't seem quite right, unless ## A ## and ## B ## commute.