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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.2011-08-28
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    If $\bf{B}$ is a zero diagonal indefinite matrix then?2011-08-28
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    With relation, I mean any inequality between them. Ofcourse they are equal if off-diagonal entries of $\bf{B}$ are zero too.2011-08-28
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    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.2011-08-28
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    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 $det(\bf{A})$??2011-08-28
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    I hope I have clarified.2011-08-28
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    Since $\bf{B}$ increases only the off-diagonal elements of $\bf{A}$ can we deduce from Hadamard determinant inequality that $det(A+B)\leq\det(B)$?2011-08-28
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    shakera, you said more than one hour ago that you hoped you had *clarified* [the question] but, since then, you added the conditions that the entries of A are positive and the diagonal of B is zero and its off diagonal entries are nonnegative (or maybe positive?). As @Gerry explained, this becomes a wholly new question. As I already explained, you should stop doing that, this only alienates you potential answerers and prevents you to get useful answers.2011-08-28
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    I appreciate this. First I was looking a general answer.@Gerry explained that some conditions are needed for one way inequality to be hold. For this reason, I have added conditions on the entries of elements. Thanks all for your kind considerations.2011-08-28

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$A=\pmatrix{2&-1\cr-1&2\cr}$ 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 $N$ ?2011-08-28
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    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.2011-08-28
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    Thanks for your answer. Your example really helped me.2011-08-28
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    All entries of $\bf{A}$ are positive and entries of $\bf{B}$ are non-negative.2011-08-28
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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.