Can we prove that the given expression $I-L$ is invertible? I would like to prove whether the matrix $I-L$ has an inverse or not, based on the following expressions.
Let $I$ be an identity matrix of size $n\times n$, and let $L = (l_{ij})_{n\times n}$ be a non-negative square matrix. Moreover,
$$T=\begin{bmatrix}
    1 & t_{12} & t_{13} & \dots  & t_{1n} \\
    t_{21} & 1 & t_{23} & \dots  & t_{2n} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    t_{n1} & t_{n2} & t_{n3} & \dots  & 1
\end{bmatrix}$$
where

*

*$t_{ij}$ is non-negative $\forall i,j$ and there is at least one $t_{ij}$ which must be zero;


*lower triangular of $T$  is written as $t_{ji}=1/t_{ij}$ if $t_{ij}\neq 0$, and $t_{ji}=0$ if $t_{ij}=0$; and


*$L=k.T$, where $k=\frac{1}{\max \{R, S\}}$ with
$R=\max_i \sum_{j=1}^{n}t_{ij}$, and $S=\max_{j} \sum_{i=1}^{n}t_{ij}$; and


*$T$ is an irreducible matrix i.e., it digraph is strongly connected.
Now I want to prove whether $I-L$ invertible using determinant $\det(I-L)\neq 0$, where $I$ is an identity matrix of size $n\times n$.  I have tried to check using numerical simulations (by generating many random matrices of type $T$) and it seems that the determinant is different from zero. i.e., $I-L$  has an inverse. But I wanted to be certain using mathematical proof. I realize that $max(R,S)$ is the largest row or column sum of $T$ and I tried to show using $max(R,S)I−T$  is invertible (instead of $I-L$). But I got stuck. Please I need help? Thanks in advance!
 A: This isn't true. Suppose $T$ is a symmetric $\{0,1\}$-matrix that has exactly $R$ ones on each row, such as
$$
T=\pmatrix{1&1&0&1\\ 1&1&1&0\\ 0&1&1&1\\ 1&0&1&1}\text{ (with $R=3$)}.
$$
Since $T$ is a symmetric $\{0,1\}$-matrix, we have $R=S$. Now $I-L=\frac{1}{R}(RI-T)$ is singular because $(RI-T)e=0$ (where $e$ is the vector of ones).
More generally, suppose $n=2(m+1)\ge4$ is even. Let $a_1,a_2,\ldots,a_m$ be any $m$ positive numbers. Let $T$ be the circulant matrix whose first row is $(1,a_1,\ldots,a_m,0,a_m^{-1},\ldots,a_1^{-1})$. E.g.
$$
T=\pmatrix{1&2&0&\frac12\\ \frac12&1&2&0\\ 0&\frac12&1&2\\ 2&0&\frac12&1}.
$$
By construction, all requirements about $T$ are satisfied. Since $T$ is circulant, we again have $R=S$ and $Te=Re$. Therefore  $I-L=\frac{1}{R}(RI-T)$ is singular.
A: To simplify, we can assume WLOG that
$\text{max row sum of }T\geq \text{max column sum of }T$
(if not, run the argument on $T':= T^T$)
There are two cases to consider
(i) $\text{min row sum of }T= \text{max row sum of }T = k^{-1}$
this implies all rows have the same sum so the ones vector, $\mathbf 1$, is an eigenvector
$L\mathbf 1 = k\cdot T\mathbf 1 = k \cdot k^{-1}\mathbf 1=\mathbf 1\implies \mathbf 1 \in \ker \big(\mathbf 1 - \mathbf L\big)$, ie. $\big(\mathbf 1 - \mathbf L\big)$ is singular.
(ii) $\text{min row sum of }T\lt \text{max row sum of }T = k^{-1}$
then $\big(I-L\big)$ is weakly diagonally dominant, and the dominance is strict in at least one row, and the underlying graph has one communicating class (irreducible). Taussky's refinement of Gerschgorin Discs applies and we conclude $\det\big(I-L\big)\neq 0$
I gave a proof of Taussky's refinement under "Optional Second" here:
Prove that this block matrix is positive definite
