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This type of matrices $L$ is called Leslie type matrices in Population dynamics: $$L = \begin{pmatrix} f_{11} & f_{12} & f_{13} & \dotsm & f_{1,i-1} & f_{1,i}& \dotsm & f_{1,m-1}& f_{1m}\\ t_{21} & 0 & 0 & \dotsm & 0 & 0& \dotsm & 0 & 0 \\ 0 & t_{32} & 0 & \dotsm & 0& 0& \dotsm & 0 & 0 \\ \vdots \\0 & 0 & 0 & \dotsm & t_{i,i-1} & 0 & \dotsm &0&0 \\ \vdots \\0&0&0& \dotsm &0& 0 & \dotsm &t_{m,m-1}&0 \end{pmatrix}$$ where $0 < t_{ij} < 1,$ for $ 1 \leqslant i,j \leqslant m,$ and $ 0 < f_{1i}$ for $ i = 1,2, \dots,m.$

I would like to show that this matrix is irreducible. I got a theorem says that a matrix is irreducible if and only if the associated graph of the matrix is strongly connected.

Here is my the definition of a graph of a matrix: Let $L = [l_{ij}]_{1 \leq i,j \leq m}.$ A graph of a matrix $L$ denoted by $G(L),$ is defined to be a directed graph of $m$ nodes $ \lbrace N_1, N_2, \dots, N_m \rbrace$ in which there is a directed edge leading from $N_i$ to $N_j$ if and only if $l_{ij} \neq 0.$

Can somebody show me how to do this? Thank you.

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up vote 2 down vote accepted

The matrix is irreducible because the associated graph $G(L)$ contains a directed Hamiltonian cycle (a cycle that contains each vertex): There is an edge $(N_{i},N_{i-1})$ for $m\ge i\ge 2$ because $t_{i+1,i}>0$ and furthermore, because $f_{1m}>0$, there is also an edge $(N_1,N_m)$. Hence, there is a directed path from any two vertices to one another, which is the very definition of being strongly connected.

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