Let $A=(a_{ij})$ be a non-negative square matrix of dimension $N$ and $n$ a natural number so that $a_{ij}^{(n)}>0$ for all $i,j \in \{1,...,N \}$. Here $a_{ij}^{(n)}$ denotes an entry of $A^n$.
Then the Perron-Frobenius theorem states, that $A$ has a positive eigenvalue $r > 0$ with geometric and algebraic multiplicity equal to $1$ and whose absolut value is strictly bigger than the absolute value of any other eigenvalue of $A$. Further more, there are two vectors $v$ and $w$ with strictly positive entries such that $v^{T}A = rv^{T}$ and $Aw=rw$. So $A$ has strictly positive left and right eigenvectors belonging to $r$.
Now I want to show that
$$ \lim_{n \rightarrow \infty} A^n/r^n = wv^{T}/v^{T}w. $$
First I want to looks at the case where $A$ is diagonalizable. I chose a matrix $U$ consisting of a basis of eigenvectors and replaced $A$ by $UDU^{-1}$, where $D$ is the diagonal matrix with all the eigenvalues on $A$ on the diagonal. W.l.o.g. I assumed that the first column of $U$ is the vector $w$. If I then use the fact that the absolute value of $r$ is bigger than the absolute value of any other eigenvalue, I get the following result:
$$ \lim_{n \rightarrow \infty} A^n/r^n = wv^{T} .$$
What am I doing wrong? Why is the $v^{T}w$ missing? Thanks for any advice!
It is not easy to spot mistakes in your derivation of the limit because it is not very explicit.
If $D$ is the diagonalised form of $A$ with the largest eigenvalue $r$ in the upper left corner then $D^n/r^n$ converges to a matrix with zeroes everywhere except 1 in the upper left corner.
Furthermore, if $A$ is diagonal(isable) then $v$ and $w$ are multiples of each other, so dividing by $v^Tw$ is just normalisation. Perhaps in your derivation of the limit you assumed that $v$ and $w$ were unit vectors? (in which case $wv^T$ would be a matrix with zeroes everywhere and $\pm1$ in the upper left corner)