Taylor expansion for matrix logarithm

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SUMMARY

The Taylor expansion for the logarithm of positive Hermitian matrices A and B is given by the formula: log(A+tB) = log(A) + t∫(0 to ∞) (1/(B+zI))A(1/(B+zI)) dz + O(t^2). A user sought a derivation for this identity, which lacks a source in the referenced paper. Another participant provided an alternative expression: log(A+tB) = log(A) + log(I+tA^(-1)B) = log(A) + tA^(-1)B + O(t^2), noting that this holds true only if matrices A and B commute. A derivation can be found in the linked document.

PREREQUISITES
  • Understanding of positive Hermitian matrices
  • Familiarity with Taylor series expansions
  • Knowledge of matrix logarithms
  • Basic linear algebra concepts
NEXT STEPS
  • Study the derivation of matrix logarithms in the context of Hermitian matrices
  • Explore the properties of commutative matrices in linear algebra
  • Review Taylor series applications in matrix analysis
  • Examine the linked document for detailed derivations and proofs
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Mathematicians, researchers in linear algebra, and anyone studying matrix analysis or seeking to understand the properties of Hermitian matrices.

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

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

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

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

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:

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