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Suppose A is a $n\times n$ square matrix$$A= \begin{bmatrix} 0 &1 &0 &\cdots &0\\ 0 &0 &1&\cdots &0\\ \vdots &\vdots&\vdots & &\vdots\\ 0&0&0&\cdots&1\\ 1 &0&0&\cdots&0 \end{bmatrix} $$ How to prove A is a irreducible a matrix.

A is irreducible if and if only $(I+A)^{n-1}$ is a postive matrix which means that all the entris is positive. Note that $(I+A)^{n-1}=\sum_{k=0}^{n-1} \binom {n-1} {k}A^k$ and A is nonnegative, thus $A$ is irreducible if and only if $\sum_{k=0}^{n-1}A^k$ is a positive matrix.

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  • $\begingroup$ Something is wrong with the way you have written the matrix. Does the principal diagonal have all zeros? If so, why is there a $1$ in the bottom right hand corner? This doesn't match with the diagonal above the principal diagonal which has all ones, presumably. Please clarify what happens to the principal diagonal, and the diagonal above that. $\endgroup$ Commented Aug 28, 2019 at 3:57
  • $\begingroup$ @астонвіллаолофмэллбэрг Sorry, already corrected. $\endgroup$ Commented Aug 28, 2019 at 4:14
  • $\begingroup$ The simplest observation to be made is that $A$ is a cyclic permutation matrix, i.e. $Ae_k=e_{k-1}$ (where $e_0\equiv e_n$). This in particular means that $A^n=I$. (It is also an example of a circulant matrix, which may be useful as well.) $\endgroup$ Commented Aug 28, 2019 at 4:16
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    $\begingroup$ @Semiclassical Thanks for your comment! I have solved this problem with your observation. $\endgroup$ Commented Aug 28, 2019 at 4:28

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Suppose $e_1 = (1,0,\cdots,0),e_2 = (0,1,0,\cdots,0),\cdots,e_n=(0,0,\cdots,0,1)$. Let $e_0= e_n$.

We have $$Ae_k=e_{k-1} ,k=1,2,\cdots,n$$

We have $(\sum_{k=0}^{n-1}A^k)e_i = (e_1,e_2,\cdots,e_n)(1,1,\cdots,1)^T$ for $i=1,2,\cdots,n$.

Then we can conclude that$$\sum_{k=0}^{n-1}A^k=(\sum_{k=0}^{n-1}A^k)I=(\sum_{k=0}^{n-1}A^k)(e_1,e_2,\cdots,e_n) = \begin{bmatrix} 1&1&\cdots &1\\ 1&1&\cdots &1\\ \vdots &\vdots& & 1\\ 1& 1&\cdots &1 \end{bmatrix} $$

which is a positive matrix.

Thus the matrix $A$ is a irreducible matrix.

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