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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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I'm just now learning about Sylvester's identity, but doesn't the 2nd equality hold from letting $A = (A^*B^{-1})$ and $B = A$ in Sylvester's identity? And similarly for the last. – breeden Feb 4 '14 at 1:57
Do you mean the last expression $\det(I+A^{*}B^{-1}A)$? – triomphe Feb 4 '14 at 2:01
I updated my answer to show you what I mean. – breeden Feb 4 '14 at 2:08
Yes that gives the first two and the fourth. Third still seems incorrect. – triomphe Feb 4 '14 at 2:10
It shows $\det(I + A^*B^{-1}A) = \det(I + B^{-1}AA^*) = \det(I + AB^{-1}A^*)$ from the equality above. – breeden Feb 4 '14 at 2:13
up vote 3 down vote accepted

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 $$ \det(I+AB^{-1}A^*)=3, \ \ \det(I+A^*B^{-1}A)=2. $$

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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 4 '14 at 1:45
@breeden: I think that $\det(I+B^{-1}AA^*)=2$. – Martin Argerami Feb 4 '14 at 1:48
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 4 '14 at 1:53
@MartinArgerami The problem arose by the different mutual information defined by authors in the following two papers. So they both can't be right?… eq (5) and eq (1) page 20. – triomphe Feb 4 '14 at 1:55
@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 4 '14 at 2:04

$\begin{align*} \det(I + B^{-1}AA^*) &= \det(B)\det(I + B^{-1}AA^*)\det(B^{-1}) \\ &= \det(B(I + B^{-1}AA^*)B^{-1}) \\ &= \det(BB^{-1} + BB^{-1}AA^*B^{-1}) \\ &= \det(I + AA^*B^{-1}) \end{align*}$

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

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

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Bah, I give up. Linear algebra is hard. :P – breeden Feb 4 '14 at 2:15

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