Compare $\det(I+A(\sum_{j=1}^{N}B_{j}))$ and $\sum_{j=1}^{N} \det(I+A B_{j})$

Question

Matrix $A$ is symmetric positive definite, and matrices $B_{j}$, $j=1,\dots,N$ are symmetric positive semidefinite and have the same size as $A$.

I wish to compare which is larger:

$\det(I+A(\sum_{j=1}^{N}B_{j}))$ or $\sum_{j=1}^{N}\det(I+A B_{j})$?

Or by rewriting it a little bit:

$$\det(I+A(\sum_{j=1}^{N}B_{j}))=\det(A)\det(A^{-1}+\sum_{j=1}^{N}B_{j})$$ $$\sum_{j=1}^{N}\det(I+A B_{j})= \det(A)\sum_{j=1}^{N}\det(A^{-1}+B_{j})$$ We can equivalently compare $\det(A^{-1}+\sum_{j=1}^{N}B_{j})$ and $\sum_{j=1}^{N}\det(A^{-1}+B_{j})$

Where $A^{-1}$ is still a symmetric positive definite matrix.

I think the latter one is larger but I can't prove this.

Any input is appreciated!

I suspect that this can be nicely solved using the concavity of the log-det function and Jensen's inequality. — Ben Grossmann, May 21 '20 at 21:35
If we Jensen's inequality on log det function, isn't it that we will have summation of log det, instead of summation of det (which is what we want)? — Chuanhao Li, May 21 '20 at 23:08

score 2 · Accepted Answer · answered May 22 '20 at 06:18

2

Neither of them is always larger than the other. E.g. when $A=B_1=I_2$ and $B_2=tI_2$, $$ \det(I+AB_1+AB_2)-\det(I+AB_1)-\det(I+AB_2)=(t+2)^2-2^2-(t+1)^2=2t-1 $$ is positive when $t>\frac12$ and negative when $0\le t<\frac12$.

answered May 22 '20 at 06:18

user1551

139,064

Compare $\det(I+A(\sum_{j=1}^{N}B_{j}))$ and $\sum_{j=1}^{N} \det(I+A B_{j})$

1 Answers1