# Why $\log\mathrm{det}(A (A^{-1}B)^{1/2}) = \log\mathrm{det}(A^{1/2}B^{1/2})$

$$A,B$$ are positive definite matrices. Show $$\log\mathrm{det}\left( A^{1/2} (A^{−1/2} B A^{−1/2})^{1/2} A^{1/2} \right) = \log\mathrm{det}(A^{1/2}B^{1/2})$$

I have known : $$A^{1/2} (A^{−1/2} B A^{−1/2})^{1/2} A^{1/2} = A (A^{-1}B)^{1/2}$$ But I can't get $$\log\mathrm{det}(A (A^{-1}B)^{1/2}) = \log\mathrm{det}(A^{1/2}B^{1/2})$$ I hope someone will help me. Thank you.

• Doesn't it follow straight from $A(A^{-1}B)^{1/2}=AA^{-1/2}B^{1/2}=A^{1/2}B^{1/2}$ ? – J.F May 3 at 13:40
• @J.F, If $A$ and $B$ do not commute, then $(A^{-1}B)^{1/2}$ need not coincide with $A^{-1/2}B^{1/2}$. But their determinants should do. – Sangchul Lee May 3 at 13:53

Note that $$\det(AB) = \det(A)\det(B)$$ and $$\det(A^{1/2}) = \sqrt{\det(A)}$$. Use these to pull apart the determinants into products and powers of real numbers, then check if they're equal.