Could one provide an example of a non-diagonalizable $2n\times 2n$ matrix with strictly positive entries, or prove that such matrix doesn't exist?
Every $2\times 2$ matrix with strictly positive entries is diagonalizable. One can prove it starting with one double eigenvalue, from where one off-diagonal entry necessarily equals $0$, contradiction.
My computations (without exact proof for the moment) show that e.g. a $(2n+1)\times (2n+1)$ matrix, where the rows $r_i=r_j$ for $i,j\neq n+1$ is non-diagonalizable.
I have no answer for the matrices of order $2n\times 2n, n>1.$
Motivation
My question is inspired by this one, and a research I did with special square matrices with positive entries.
Here you are: $$ A = \begin{pmatrix} 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&4&2 \end{pmatrix} $$ It has rank 2, but the only eigenvalue different from zero is 5, that is simple since it is the spectral radius (Perron-Frobenius), so it lacks a zero eigenvector.