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.
Best Answer
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.