<s>[INST] why is this Markov Chain aperiodic I have this Matrix: 
$$P=\begin{pmatrix}  0 & 1 \\ 0.3 & 0.7 \end{pmatrix}$$
this markov chain is said to be aperiodic, I dont understand how it comes to it. Period $\delta$ is the gcd of the set of all diagonal elements, right? if $\delta>1$, $P$ is periodic, if $\delta=1$, then aperiodic. 
but here it is not $\delta=1$, is it? or do i have to transit the matrix to some certain form?  
 [/INST] Since the Jordan Normal Form of the matrix is
$$
\begin{bmatrix}
0&1\\
0.3&0.7
\end{bmatrix}
=\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}
\begin{bmatrix}
-0.3&0\\0&1
\end{bmatrix}
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}^{-1}
$$
we have
$$
\begin{align}
\begin{bmatrix}
0&1\\
0.3&0.7
\end{bmatrix}^n
&=\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}
\begin{bmatrix}
-0.3&0\\0&1
\end{bmatrix}^n
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}^{-1}\\
&\to
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}
\begin{bmatrix}
0&0\\0&1
\end{bmatrix}
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}^{-1}\\
&=\frac1{13}\begin{bmatrix}
3&10\\3&10
\end{bmatrix}
\end{align}
$$
Thus, the initial state $\begin{bmatrix}x\\y\end{bmatrix}$ tends toward the fixed point state
$$
\left(\frac3{13}x+\frac{10}{13}y\right)\begin{bmatrix}1\\1\end{bmatrix}
$$
</s>