Consider a matrix where the diagonal entries are anything but the off-diagonal entries are all one. I was able to find a formula for the determinant of this matrix, but what are other known properties? Does this matrix have a name? In particular is there a formula for its inverse?
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9$\begingroup$ It's the sum of a diagonal matrix and a rank one matrix, so the Sherman-Morrison formua would give an explicit formula for the inverse. en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula $\endgroup$– Terry TaoCommented Dec 23, 2019 at 4:12
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$\begingroup$ Never seen this before, so thank you! $\endgroup$– cgmilCommented Dec 23, 2019 at 4:30
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1$\begingroup$ @TerryTao: This matrix may have no inverses. $\endgroup$– user6976Commented Dec 23, 2019 at 4:44
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$\begingroup$ @MarkSapir That's certainly true; from the original formulation one can choose the diagonal entries to be one and then the matrix is clearly singular. If we're going to study the inverse we need to make more assumptions, but if there is an inverse it will have a certain form. (I used the Morrison formula to find that inverse.) Bonus points for other quantities such as eigenvalues/eigenvectors, though. $\endgroup$– cgmilCommented Dec 23, 2019 at 4:47
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$\begingroup$ @cgmil: The matrix can be invertible and the formula mat not give the inverse. For example matrix(2,1;1,1). $\endgroup$– user6976Commented Dec 23, 2019 at 6:17
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1 Answer
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Such a matriix has the form $J +D,$ where $D$ is a diagonal matrix, and $J$ is a square matrix with all entries $1$. One small remark is that if $D$ has two of its diagonal entries equal to $\lambda$, then $\lambda$ is also an eigenvalue of $J+D$. This is because the $\lambda$-eigenspace of $D$ is at least two-dimensional and the $0$-eigenspace of $J$ has codimension one.
answered Dec 23, 2019 at 12:49