Let $A$ be a positive definite matrix and $B$ be a positive semi-definite matrix. Under what conditions does $\det(A+B)=\det(A)+\det(B)$ hold?

We all know that for any two positive semi-definite matrices $A$ and $B$, $\det(A+B) \geq \det(A)+\det(B)$ and strict inequality holds when $A$ and $B$ are both positive definite matrices having order $\geq 2$. My question is that under what conditions on $A$ (positive definite) and $B$ (positive semi-definite) the reverse also holds true, that is, the equality holds? Please help.

Best regards,

Prasenjit Ghosh.

  • $\begingroup$ Sorry, what are these abbreviations? I guess, 'p.d.' stands for 'positive definite', but 'n.n.d.'? $\endgroup$ – Berci May 10 '13 at 7:59
  • $\begingroup$ I am extremely sorry for not using the complete terminology. Yes,by "p.d." I meant positive definite and by "n.n.d." I meant non-negative definite (positive semi-definite). $\endgroup$ – Prasenjit May 10 '13 at 8:21
  • $\begingroup$ The 0 matrix and identity works. Maybe a few more $\endgroup$ – Hawk May 10 '13 at 8:28

When $A\succ0$ and $B\succeq0$, $\det(A+B)=\det(A)+\det(B)$ if and only if $B=0$ or $n=1$. Hints:

  1. $\det(A+B)=\det(A)+\det(B)$ is equivalent to $\det(I+A^{-1/2}BA^{-1/2})=1+\det(A^{-1/2}BA^{-1/2})$.
  2. So you are essentially asking when does $\det(I+X)=1+\det(X)$ hold for a positive semidefinite $X$. WLOG you may assume that $X$ is diagonal.
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    $\begingroup$ I thought at first you were wrong about what comes after #2, but I was mistaken about that. ☺ +1. $\endgroup$ – Harald Hanche-Olsen May 10 '13 at 8:45
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    $\begingroup$ @HaraldHanche-Olsen Uhh, sorry for the confusion. There is actually no need to reduce $X$ to a diagonal matrix, as we may think directly in terms of eigenvalues, but I am too used to transform everything to a simpler form first, despite doing so may be redundant. $\endgroup$ – user1551 May 10 '13 at 8:50
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    $\begingroup$ I know, I do suffer the simplification bug myself. I think my students are mystified sometimes when I rephrase a question just so I don't have to deal with a random constant through a lengthy calculation, but I keep doing it anyway. $\endgroup$ – Harald Hanche-Olsen May 10 '13 at 9:48

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