# On a determinant inequality of positive definite matrices

Assume that $B$ and $A$ are two positive definite matrices. Take $B^*$ a block diagonal matrix with block $B_{11}$ and $B_{22}$ of $B$. This means the following: $$B=\left[\begin{array}{ll} B_{11}& B_{12}\\ B_{21}&B_{22} \end{array}\right], B^*=\left[\begin{array}{ll} B_{11}& 0\\ 0 &B_{22} \end{array}\right]$$ It can be seen that $2B^*\succ B$ and therefore: $$\det(I+2AB^*)\geq \det(I+AB).$$ So I managed to prove that, however I have another conjecture that the following is true: $$2\det(I+AB^*)\geq \det(I+AB).$$ My question is about the proof of this conjecture.

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A quick counterexample to your conjecture is

\begin{equation*} A = \begin{pmatrix} 13 & 3 & -13 & -5\\ 3 & 4 & -3 & 4\\ -13 & -3 & 13 & 5\\ -5 & 4 & 5 & 10\\ \end{pmatrix},\quad B = \begin{pmatrix} 15 & 3 & -14 & -1\\ 3 & 18 & -6 & 3\\ -14 & -6 & 19 & -1\\ -1 & 3 & -1 & 1 \end{pmatrix} \end{equation*}

For this choice, $2\det(I+AB^*) \approx 6\times 10^4$, while $\det(I+AB)\approx 8\times 10^4$. The $A$ above is semidefinite; if that displeases you, then just add a trivial $\epsilon I$ to it, the counterexample will still hold.

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Many thanks for the answer. Unfortunately I forgot to write an important assumption about $B$. All diagonal elements are equal: $b_{11}=...=b_{nn}=K$. I am not sure if this is significant or not for the problem. As a matter of fact I was trying to find an upper bound like $C\det(I+AB^*)\geq \det(I+AB)$ for some $C$ independent of the dimension of $B$. I wonder if this is possible. My numerical evaluation of the current problem shows that with this additional assumption, we can have $C=2$. I might have mistaken somewhere. Should I post this as question, if it can be an independent problem? – Arash Sep 26 '13 at 10:45
Even with the all diagonal elements equal this does not hold; take the matrix $B$ that I have above, set its diagonal to 15 (it still remains posdef); the conjecture still fails. Perhaps for it to hold $C$ has to depend on $n$ (the size of the matrices)---but am not sure about that. You might want to pose it as a separate problem. – Suvrit Sep 26 '13 at 16:15

From the comments, a true version is $$2^n\det(I+AB^*)\geq \det(I+AB), \quad (1)$$ where $n$ is the dimension of $A, B$.

The proof is easy, as $2^n\det(I+AB^*)\geq\det(I+2AB^*)$. Can $2$ be replaced by a smaller constant in (1)?

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