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How $\det(\bf{A}+\bf{B})$ is related with $\det(\bf{A})$ where $\bf{A}$ is either semi-definite or positive definite matrix but $\bf{B}$ is a zero diagonal indefinite matrix. $\det(\cdot)$ denotes the determinant and all matrices are square. Off-diagonal elements of $\bf{B}$ are all positive. All elements of $\bf{A}$ are positive.

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If there is no restriction on B, choose any C and consider B=C-A. Then you are trying to relate det(C) with det(A) for a unspecified C, hence there is no possibility of any relation whatsoever. – Did Aug 28 '11 at 9:45
If $\bf{B}$ is a zero diagonal indefinite matrix then? – shakera Aug 28 '11 at 9:47
With relation, I mean any inequality between them. Ofcourse they are equal if off-diagonal entries of $\bf{B}$ are zero too. – shakera Aug 28 '11 at 9:50
shakera, honestly I think you should think deeply to determine the exact situation you are interested in and then rewrite precisely your question to describe that. Otherwise, as with some of your other questions, the same ballet of people saying (with reason) they do not understand the question will occur again, leading to the frustration of everybody involved. – Did Aug 28 '11 at 9:54
If $\bf{B}$ is semi-definite then $det(\bf{A}+\bf{B})$ $\geq$ $det(\bf{A})$...I have a situation in which $\bf{B}$ is zero-diagonal indefinite. What will be then the relation between $det(\bf{A}+\bf{B})$ and $det(\bf{A})$?? – shakera Aug 28 '11 at 9:59
up vote 3 down vote accepted

$A=\pmatrix{2&-1\cr-1&2\cr}$ is positive definite with determinant 3. $B=\pmatrix{0&b\cr b&0\cr}$ has zero diagonal and is indefinite. The determinant of $A+B$ is $4-(b-1)^2$ which can be greater than, less than, or equal to 3. This suggests that in general you can't say very much.

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If elements of $A$ are positive then determinant is again 3. But determinant of $A+B$ becomes $4-(b+1)^2$ hence $det(A+B)\leq det(A)$ for all positive value of $b$. Is it correct for general $N$ ? – shakera Aug 28 '11 at 10:17
So now you want the entries (not elements) of $A$ to be positive, and you want the entries of $B$ to be non-negative. That's a whole new question. Maybe you should post it as a new question, but maybe first you should think about whether it's really the question you want to ask, whether you can find any simple examples one way or the other, and so on. – Gerry Myerson Aug 28 '11 at 10:33
Thanks for your answer. Your example really helped me. – shakera Aug 28 '11 at 10:40
All entries of $\bf{A}$ are positive and entries of $\bf{B}$ are non-negative. – shakera Aug 28 '11 at 10:42

If $A \in \mathbb R^{n \times n}$ is positive definite, then it is invertible. Hence,

$$\det (A+B) = \det (A (I_n + A^{-1} B)) = \det (A) \cdot \det (I_n + A^{-1} B)$$

If $n=2$, then

$$\det (I_n + A^{-1} B) = 1 + \det (A^{-1} B) + \mbox{tr} (A^{-1} B) = 1 + (\det (A))^{-1} \cdot \det (B) + \mbox{tr} (A^{-1} B)$$

and, thus,

$$\begin{array}{rl} \det (A+B) &= \det (A) \cdot \det (I_n + A^{-1} B)\\\\ &= \det (A) \cdot (1 + (\det (A))^{-1} \cdot \det (B) + \mbox{tr} (A^{-1} B))\\\\ &= \det (A) + \det (B) + \det (A) \cdot \mbox{tr} (A^{-1} B)\end{array}$$

If $n > 2$, it gets a bit messy.

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