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For complex valued matrices $A,B$ where $B$ is invertible, does $$\det(I+B^{-1}AA^*)=\det(I+AA^*B^{-1})=\det(I+AB^{-1}A^*)=\det(I+A^*B^{-1}A)?$$ Here $A^*$ is the conjugate transform. I guess $\det(I+B^{-1}AA^*)=\det(I+AA^*B^{-1})$ holds by Sylvester's identity.

Correction $$\det(I+B^{-1}AA^*)=\det(I+AA^*B^{-1})=\det(I+A^*B^{-1}A)?$$

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triomphe
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2 Answers2

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Indeed, $\det(I+B^{-1}AA^*)=\det(I+AA^*B^{-1})$ holds by Sylvester's Identity.

The other equalities don't hold. Let $$ A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ B=\begin{bmatrix}1&0\\0&1/2\end{bmatrix}. $$ Then $$ \det(I+AB^{-1}A^*)=3, \ \ \det(I+A^*B^{-1}A)=2. $$

Martin Argerami
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  • It's probably worth mentioning that $\det(I + B^{-1}AA^*) = 1$ in this case and hence doesn't necessarily equal either of the other two quantities as well. – breeden Feb 04 '14 at 01:45
  • @breeden: I think that $\det(I+B^{-1}AA^*)=2$. – Martin Argerami Feb 04 '14 at 01:48
  • I had to remind myself. $A^$ means replace every entry by its conjugate. In this case (or any real case) $A^ = A$, right? – breeden Feb 04 '14 at 01:53
  • @MartinArgerami The problem arose by the different mutual information defined by authors in the following two papers. So they both can't be right? http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.217.4812&rep=rep1&type=pdf eq (5) and http://userver.ftw.at/~zemen/Cooperative/Lecture_12_handout.pdf eq (1) page 20. – triomphe Feb 04 '14 at 01:55
  • @MartinArgerami Doesn't $\det(I+AA^B^{-1})=\det(I+A^{}B^{-1}A)$? Meaning first is equal to fourth? By Sylvester's – triomphe Feb 04 '14 at 02:04
  • @breeden: $A^*$ means the transpose conjugate. – Martin Argerami Feb 04 '14 at 02:09
  • @triomphe: yes. Either that way, or calculating directly, the other two determinants equal to $2$. – Martin Argerami Feb 04 '14 at 02:10
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$\begin{align*} \det(I + B^{-1}AA^*) &= \det(B)\det(I + B^{-1}AA^*)\det(B^{-1}) \\ &= \det(B(I + B^{-1}AA^*)B^{-1}) \\ &= \det(BB^{-1} + BB^{-1}AA^*B^{-1}) \\ &= \det(I + AA^*B^{-1}) \end{align*}$

That shows the first equality. I don't see why the other two should be correct. I only mention this because the identity is much easier than Sylvester's identity.

Sylvester's identity says

$$\det(I + CD) = \det(I + DC)$$

Letting $C = A^*B^{-1}$ and $D = A$ we get

$$\det(I + A^*B^{-1}A) = \det(I + AA^*B^{-1}).$$

Similarly, if we let $C = B^{-1}A$ and $D = A^*$ we get

$$\det(I + B^{-1}AA^*) = \det(I + A^*B^{-1}A).$$

breeden
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