Characterization of two-step 2x2 stochastic matrices Show that:

A 2 x 2 stochastic matrix is two-step transition matrix of a Markov chain if and only if the sum of its principal diagonal terms is greater than or equal to $1$.


Encode the Markov matrix $\begin{pmatrix}1-a & a\\ b & 1-b\end{pmatrix}$ by the couple $(a,b)$ with $a$ and $b$ between $0$ and $1$. The square of this matrix is encoded by the couple
$$
(a(2-a-b),b(2-a-b)),
$$ 
hence the question is to determine for which $a$ and $b$ between $0$ and $1$, there exists $x$ and $y$ between $0$ and $1$ such that
$$
a=x(2-x-y),\qquad b=y(2-x-y).
$$
If $x$ and $y$ exist, their sum $s$ solves $s(2-s)=a+b$ hence $(s-1)^2=1-a-b$, in particular 
$$
a+b\le1.
$$ 
Assume this condition holds. Then $s=1\pm\sqrt{1-a-b}$, hence 
$$
x=a/(2-s)=a/(1\mp\sqrt{1-a-b}),
$$
and
$$y=b/(2-s)=b/(1\mp\sqrt{1-a-b}).
$$
Since $1+\sqrt{1-a-b}$ is between $1$ and $2$, at least the choice $+$ in $\mp$ yields solutions $x$ and $y$ between $0$ and $1$.
Finally the Markov matrix $\begin{pmatrix}1-a & a\\ b & 1-b\end{pmatrix}$ is the square of a Markov matrix if and only if 
$$
a+b\le1.
$$
