Taylor expansion for matrix logarithm

In summary, the paper states that for positive hermitian matrices A and B, the Taylor expansion of log(A+tB) at t=0 is given by the formula \log(A+tB)=\log(A) + t\int_0^\infty \frac{1}{B+zI}A \frac{1}{B+zI} dz + \mathcal{O}(t^2). However, a source or proof is not provided and the derivation of this identity is not readily available.
  • #1
Backpacker
9
0
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.
 
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  • #2
Backpacker said:
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! :smile:

I don't recognize your formula, but:

$$\log(A+tB)=\log(A(I+tA^{-1}B)= \log A + \log(I+tA^{-1}B) = \log A + tA^{-1}B + \mathcal{O}(t^2)$$
 
  • #4
I like Serena said:
$$\log(A(I+tA^{-1}B)= \log A + \log(I+tA^{-1}B) $$
This doesn't seem quite right, unless ## A ## and ## B ## commute.
 

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