Powers of a Positive Matrix in the Limit

According to PF theorem, the characteristic polynomial of $A$ is in the form $\chi_A(x)=(x-\lambda)f(x)$ where $\lambda>0$ and the roots of $f$ have modulus $<\lambda$. Moreover, there is a (unique up to a factor) vector $v>0$ s.t. $Av=\lambda v$. Since $A^T$ is primitive, there is a (unique up to a factor) vector $u>0$ s.t. $u^TA=\lambda u^T$. Since $u^Tv>0$, we can choose the above factors s.t. $u^Tv=1$.

There is a basis in the form $v,\cdots$ s.t., for this change of basis of matrix $P\in M_n(\mathbb{R})$, $A=Pdiag(\lambda,B_{n-1})P^{-1}$ where $\chi_B(x)=f(x)$; then $B$ has a spectral radius $\rho(B)<\lambda$, that is, $\rho(\lambda^{-1}B)=\mu<1$.

Thus $(\lambda^{-1}A)^k=Pdiag(1,(\lambda^{-1}B)^k)P^{-1}$ tends, when $k\rightarrow\infty$, to the rank $1$ projector $M=Pdiag(1,0_{n-1})P^{-1}$; moreover, for every $\epsilon>0$, $||(\lambda^{-1}A)^k-M||=O((\dfrac{\mu+\epsilon}{\lambda})^k)$.

A projector is uniquely defined by $im(M)$ and $\ker(M)=(im(M^T))^{\perp}$, that is, by $im(M), im(M^T)$.

Notice that $Mv=\lim_k (\lambda^{-1}A)^kv=v,u^TM=\lim_k u^T(\lambda^{-1}A)^k=u^T$. Then the rank $1$ projector $M$ is uniquely defined by $im(M)=span(v),im(M^T)=span(u)$.

Now $R=vu^T$ is also a projector with image $span(v)$ and $im(R^T)=span(u)$. Then $M=vu^T$.