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?  
 A: Since my comment provided sufficient clarification:
When there's a stationary state, your system will evolve towards that state. In your case, the two left eigenvectors are $(−1,1)$ and $(3,10)$ with corresponding eigenvalues $−0.3$ and $1$. Every other state of the system can be decomposed into those two states. The first state exihibits oscillating behaviour, but it dies out as $0.3<1$. The other state is stationary. So whatever your initial state, you'll evolve towards that stationary state.
A: Call the states $a$ and $b$. The chain is irreducible since $P(a,b)=1\ne0$ and $P(b,a)=0.3\ne0$. One can go from $b$ to $b$ with positive probability in one step (probability $0.7$) hence the period of $b$ is $1$.  The chain is irreducible hence the period of every state is $1$ (this can also be proved directly). That is, the chain is aperiodic.
A: Intutive answer;
the markov chain is aperiodic since it is reached from any node to itself in not periodic interval e.g 2, 4, 6,......
here the P11 makes it as a aperiodic since the chain re goes to node after 2,3,5 e.g no periodicity
A: 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
$$
\left(\frac3{13}x+\frac{10}{13}y\right)\begin{bmatrix}1\\1\end{bmatrix}
$$
