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Let $A$ be a symmetric and invertible matrix. I know that if $A$ has a constant diagonal and a constant off-diagonal ($A=\alpha I + \beta\tilde1$, where $I$ is an identity matrix, $\tilde{1}$ is a matrix of ones, and $\alpha,\beta$ are some scalars), then $A^{-1}$ has a constant diagonal.

What are other non-obvious possible structures of $A$ that guarantee that $A^{-1}$ has a constant diagonal vector?

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$A^{-1}$ has constant diagonal if and only if there exists a constant $C$ such that $\det(A_{ii})=C$ for all $i$ where $A_{ii}$ denotes the matrix with $i$th row and $i$th column deleted. –  Your Ad Here Oct 31 '12 at 20:16
    
Remark on that comment: That follows from Cramer's Rule. (@IHaveAStupidQuestion) –  k.stm Oct 31 '12 at 20:19
    
@K.Stm. I know, that was my point. –  Your Ad Here Oct 31 '12 at 20:22
    
@IHaveAStupidQuestion Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Aug 22 '13 at 7:25

1 Answer 1

$A^{−1}$ has constant diagonal if and only if there exists a constant $C$ such that $\det(A_{ii})=C$ for all $i$ where $A_{ii}$ denotes the matrix with $i$th row and $i$th column deleted.

As K. Stm. has pointed out, this follows from Cramer's rule.

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