Is this some kind of Perron-Frobenius theorem? If one has a matrix $A$ whose entries are all $\geq 0$ (or $>0$) then there apparently exists a diagonal matrix $D > 0$ (i.e all the diagonal entries are positive) and a constant $\alpha >0$ such that $D^{-1} A D (id) = \alpha (id)$ where $(id)$ is a column vector of all $1$s. 


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*Is this above some restatement of the standard Perron-Frobenius theorem? Can someone kindly explain the equivalence? 



It would help if you could reference to sections in the book by Horn and Johnson - because thats the book that I have with me. 

The statement is incorrect if $A$ is merely nonnegative. The Perron-Frobenius theorem states (among many, many things) that if $A\ge0$, it has a nonnegative eigenvector corresponding to its spectral radius $\rho(A)$. It does not guarantee that this eigenvector is positive. Nor does it guarantee that $\rho(A)$ is positive. For example, consider
$$
A=\pmatrix{0&1\\ 0&0}.
$$
If $D^{-1}AD\mathbf1=\alpha\mathbf1$ for some positive diagonal matrix $D$ and some $\alpha>0$, then $A(D\mathbf1)=\alpha D\mathbf1$. In other words, $Av=\alpha v\ne0$ for some positive vector $v$. But this is clearly impossibly because by construction, the last entry of $Av$ must be zero. In fact, all eigenvalues of $A$ are zero, so $A$ does not have any positive eigenvalue $\alpha$, not to say any eigenvector (positive or not) for such $\alpha$.
The statement in your question is correct, however, when $A>0$. The Perron-Frobenius theorem states (again, among many, many things) that if $A>0$, $A$ has a positive eigenvector $v$ corresponding to its spectral radius $\rho(A)$. Since both $A$ and $v$ are positive, it follows that $\rho(A)>0$. So, if we set $\alpha=\rho(A)$ and $D=\operatorname{diag}(v)$, then $D$ is invertible and $D^{-1}AD\mathbf1=\alpha\mathbf1$.
