Matrix $A \ge 0$ is a semi primitive matrix, if for some $k$ matrix $A^{k}$ contains a positive column.
Prove, if characteristic polynomial of stochastic matrix $A$ equal to $t^{(n-1)}(t-1)$, then A is semi primitive and $\lim_{k \to \infty}A^{k} = A^{n-1}$.
Please give me some hints to prove it. And sorry for my bad English. I'm not an English native speaker.
By the Cayley Hamilton theorem, we have $A^{n} = A^{n-1}$. It follows that $A^{k} = A^{n-1}$ for all $k \geq n$, so it is clear that $\lim_{k \to \infty}A^k = A^{n-1}$.
As for proving that $A$ is primitive: first, show that $A^{n-1}$ must have a rank of $1$.
Now, note that $B = A^{n-1}$ is a non-negative, stochastic matrix with rank $1$. Because $B$ has rank $1$, we can write it in the form $$ B = uv^T $$ for column-vectors $u,v$. Because $B$ is non-negative, both $u$ and $v$ must be non-negative. Because $B$ is stochastic, we can take $u$ to be the vector $(1,\dots,1)^T$. That is, the $i$th column of $B$ is equal to $(v_i,\dots,v_i)^T$. Because $B \neq 0$, we must have $v \neq 0$. However, if the $i$th entry of $v$ is positive, then the $i$th column of $B$ must be positive.