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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

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

  2. 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

  3. $L=k.T$, 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

  4. $T$ 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!

user3840
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    Have you considered the case where all $t_{i,j}=1$? i.e. where $T = \mathbf {11}^T$? – user8675309 Jan 26 '22 at 19:11
  • Yes, it will have an inverse. Because all rows of $max(R,S)I−T$ are different by reducing the diagonal elements into one (deviding all elements by $n-1)$) as max(R,S)=n. – user3840 Jan 26 '22 at 22:14
  • I don't follow your response. $k^{-1} L\mathbf 1=\big(k^{-1}I -T\big)\mathbf 1 = \big(k^{-1}I-\mathbf {11}^T\big)\mathbf 1$ $= k^{-1}\mathbf 1 -\mathbf 1\big( \mathbf 1^T \mathbf 1\big)= n \mathbf 1 - n \mathbf 1 =\mathbf 0$ so $L$ is singular – user8675309 Jan 26 '22 at 23:57
  • Yes, you are right. I made a mistake in the original question. I had to say there is at least one $t_{ij}$ which must be zero. Because T must be an incomplete matrix from the decision problem I am working on (which is called incomplete pairwise comparison matrix). Now I have edited the question. Thanks a lot! Now I am still thinking about it. – user3840 Jan 27 '22 at 07:36

2 Answers2

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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.

user1551
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  • @user1551Using MATLAB, I got $det(I-L)=6.21\times 10^{-17}$ which is different from zero (actually its numerical approximation is zero). But I don't understand the implication of $(RI-T)e$. Thanks. – user3840 Jan 27 '22 at 10:02
  • @user3840 $(I-L)e=\frac1R(RI-T)e=0$. Therefore the homogeneous system of linear equations $(I-L)x=0$ has a non-trivial solution $x=e$. This means $I-L$ is singular. – user1551 Jan 27 '22 at 10:04
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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

user8675309
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