How to calculate this $\frac{0}{0}$ limit? Let $R=\left[\begin{array}{ccc}
\frac{1}{2} & \frac{1}{2} & 0\\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3}\\
0 & 0 & 1
\end{array}\right]$, $P=\left[\begin{array}{ccc}
\frac{13}{36} & \frac{-17}{36} & \frac{4}{36}\\
\frac{-17}{36} & \frac{25}{36} & \frac{-8}{36}\\
\frac{4}{36} & \frac{-8}{36} & \frac{4}{36}
\end{array}\right]$ , $Q=\left[\begin{array}{ccc}
0 & 0 & 0\\
0 & \frac{1}{9} & \frac{-1}{9}\\
0 & \frac{-1}{9} & \frac{1}{9}
\end{array}\right]$,  $y(0)=\left[\begin{array}{c}
1 \\
2 \\
5
\end{array}\right]$. 
Define 
$$
r(k)=\frac{y^{T}(0)(R^{T})^{k-1}PR^{k-1}y(0)}{y^{T}(0)(R^{T})^{k-1}QR^{k-1}y(0)}
$$
The problem I'd like to look at is how to represent the value of $\lim_{k\rightarrow\infty}r(k)$ with the property of matrix $P$ and $Q$ (e.g. eigenvalue).
Note that we have $\lim_{k\rightarrow\infty}R^{k}=\left[\begin{array}{ccc}
0 & 0 & 1\\
0 & 0 & 1\\
0 & 0 & 1
\end{array}\right]$, the row and column sum of $P$ and $Q$ are $0$ and actually $\lim_{k\rightarrow\infty}r(k)$ exists. Because  $\lim_{k\rightarrow\infty} R^{k}y(0)= \left[\begin{array}{c}
5 \\
5 \\
5
\end{array}\right]$, thus both the numerator and denominator converge to $0$ which leads to a $\frac{0}{0}$ indefinite form. It is believed that the value of $\lim_{k\rightarrow\infty}r(k)$ is encoded in matrix pair $(P,Q)$, but I'm unable to come up with a formula or even the connection between $\lim_{k\rightarrow\infty}r(k)$ and $(P,Q)$, could anyone help me on that ?
 A: Actually, $r(k)=\frac{13}{16}$ for all $k$. So, the limit is also $\frac{13}{16}$.
We have $R=VDV^{-1}$, where $D=\mathrm{diag}(1,\frac56,0)$ and
$$
V = \pmatrix{1&3&1\\ 1&2&-1\\ 1&0&0},
\ V^{-1} = \pmatrix{0&0&1\\ \tfrac15&\tfrac15&\tfrac{-2}5\\ \tfrac25&\tfrac{-3}5&\tfrac15}.
$$
Therefore, for any $3\times3$ matrix $A$,
\begin{align*}
&y^{T}(0)(R^{T})^{k-1}AR^{k-1}y(0)\\
=& y(0)^T(VD^kV^{-1})^TA(VD^kV^{-1})y(0)\\
=& (V^{-1}y(0))^T D^k(V^TAV)D^k (V^{-1}y(0)).\tag{1}
\end{align*}
Now, straightforward calculations show that
$$
V^TPV = \frac1{36}\pmatrix{0&0&0\\ 0&13&6\\ 0&6&72},
\ V^TQV = \frac19\pmatrix{0&0&0\\ 0&4&-2\\ 0&-2&1}.
$$
Therefore
$$
D^k(V^TPV)D^k = \frac1{\color{red}{36}}\pmatrix{0&0&0\\ 0&\color{red}{13}\left(\tfrac56\right)^{2k}&0\\ 0&0&0},
\ D^k(V^TQV)D^k = \frac1{\color{red}{9}}\pmatrix{0&0&0\\ 0&\color{red}{4}\left(\tfrac56\right)^{2k}&0\\ 0&0&0}.\tag{2}
$$
Since the second entry of $V^{-1}y=(5,\frac{-7}{5},\frac15)^T$ is nonzero, it follows from $(1)$ and $(2)$ that
$$
\frac{y^{T}(0)(R^{T})^kPR^{k-1}y(0)}{y^{T}(0)(R^{T})^kQR^{k-1}y(0)}
=\frac{13/36}{4/9}=\frac{13}{16}=0.8125,
$$
which is independent of $k$.
