Effect of adding a matrix to both numerator and denominator of a ratio between determinants of two matrices

Question

Assume matrix $A$ is symmetric and positive definite, and matrices $B$ and $C$ are symmetric and positive semi-definite. Originally I have ratio between determinants

$$\frac{\det(A+B)}{\det(A)}$$

which is obviously greater than or equal to 1.

How would this ratio change (increase or decrease) when I add another matrix $C$ inside the determinant on both numerator and determinator, as follows?

$$\frac{\det(A+B+C)}{\det(A+C)}$$

My intuition is that

$$\frac{\det(A+B+C)}{\det(A+C)} \leq \frac{\det(A+B)}{\det(A)}$$

but I haven't been able to prove this. Any insight on this is appreciated!

Answer 1

It's true. Let $S=A+C$. Then \begin{aligned} \frac{\det(A+B)}{\det(A)} &=\det(I+A^{-1/2}BA^{-1/2})\\ &=\det(I+B^{1/2}A^{-1}B^{1/2})\quad(\text{because} \det(I+XY)=\det(I+YX))\\ &\ge\det(I+B^{1/2}S^{-1}B^{1/2})\quad(\text{because} B^{1/2}A^{-1}B^{1/2}\succeq B^{1/2}S^{-1}B^{1/2}\succeq0)\\ &=\det(I+S^{-1/2}BP^{-1/2})\\ &=\frac{\det(S+B)}{\det(S)}\\ &=\frac{\det(A+B+C)}{\det(A+C)}. \end{aligned}

Answer 2

The below is missing a step and possibly incorrect.

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Your intuition is correct. Throughout my proof below, positive (semi)-definite matrices are necessarily symmetric.

Denote $B' = A^{-1/2}BA^{-1/2}$ and $C' = A^{-1/2}CA^{-1/2}$. The statement that you are trying to prove can be rewritten as $$ \frac{\det(I + B' + C')}{\det(I + C')} \leq \frac{\det(I + B')}{\det(I)} \implies \det(I + B' + C') \leq \det(I + B')\det(I + C'). $$ We use $\leq$ to denote the Loewner ordering. That is, $A \leq B$ iff $B - A$ is positive semidefinite. Now, we note that $$ (I + C')^{1/2}(I + B')(I + C')^{1/2} = (I + C') + (I + C')^{1/2}(B')(I + C')^{1/2} \geq I + C' + B'. $$ It follows that $$ \det(I + B')\det(I + C') = \\ \det[(I + C')^{1/2}(I + B)(I + C')^{1/2}] \geq \\ \det[I + C' + B'], $$ which is what we wanted.