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I ran into this problem recently.

If $A$ is a $n\times n$ matrix with $1$s in the diagonal and all off diagonal entries have absolute value less than $1$, is $A$ invertible?

It it definitely true for a $2\times 2$ matrix and seems to be true for larger matrices, just by trying to compute the determinant, however, my expression seems to get messy, and I was wondering if there was a easier method to solve this. If not, direction on how to argue from definition of determinant would be appreciated. Although I can see that an argument might come from there, I am having problems with that too.

Thank you for any help, and any help is much appreciated.

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3 Answers 3

up vote 3 down vote accepted

As Will points out, this isn't true. But you can salvage this idea using the weaker notion of diagonal dominance: if $A$ has 1s along the diagonal and $$\sum_{j\neq i} |a_{ij}| < 1,$$ then $A$ is invertible (the Levy–Desplanques theorem). Notice that for 2x2 matrices, this condition is just that the off-diagonal entries have magnitude less than 1.

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well, no. With $n$ by $n,$ and $n \geq 3,$ let all the off-diagonal entries be $$ \frac{-1}{n-1}. $$ So the vector with all entries 1 is an eigenvector with eigenvalue 0.

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The 2x2 case works because it is diagonally dominant. So to explain what happened here, the off-diagonal entries EACH having magnitude less than one isn't enough. They need to be small enough. So if the sum of the absolute values of the off-diagonal elements in a row is strictly less than one and the diagonal-entry has magnitude one, then the matrix will be invertible. Gershgorin proved it so.

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