Zero elements of an inverse matrix

Question

Consider a strictly diagonal dominant $n \times n$ matrix $L$ (i.e. $|L_{ii}| > \sum_{j \neq i} |L_{ij}| $ ) with $L_{ij} \leq 0$ for $i \neq j$ and $L_{ii} > 0$ for all $i$. Also $L$ is a symmetric matrix ($L^{T} = L$).

I would like to know for what kind of $L$, we can conclude that $(L^{-1})_{ij} = 0$ for $ i \neq j$.

I know that if $L$ is block diagonalizable, then the off-diagonal block of $L^{-1}$ would be zero. But what if $L$ is not block diagonalizable? Can we say anything about the conditions when $(L^{-1})_{ij} = 0$ for $ i \neq j$ ?

Answer 1

Note that because $LL^{-1}=I$ if the inverse is a diagonal matrix $D=[a_1,...,a_n]$then $L=[1/a_1,...,1/a_n]$

ie, the inverse of a diagonal matrix is diagonal too