Where the matrix A is replaced with $$G=\alpha A+(1-\alpha)\frac{1}{n}ee^T$$
Is it a sufficient condition a matrix is stochastic if the largest eigenvalue is 1? Or that in this case since A is stochastic and the largest eigenvalue of G is 1, that G is also stochastic, and thereby the columns sum to 1?
first the largest eigenvalue of a stochastic matrix is indeed $1$ but this is not a sufficient to prove that a matrix is stochastic (consider for instance add to zero. Take $ M=\begin{bmatrix}1& 0 & 1\\0&1& 0\\0&0&1\end{bmatrix} $ which has maximal eigenvalue 1 but isn't stochastic (see also e.g. Proof that the largest eigenvalue of a stochastic matrix is $1$)).
Now to show that $G$ is stochastic (assuming $A$ is stochastic), just apply the definitions: $$ \begin{align*} \sum_j G_{i,j} & = \sum_j (\alpha A_{i,j} + {(1-\alpha)\over n} ee^{T}_{i,j})\\ &= \alpha \sum_j A_{i,j} + {(1-\alpha)\over n} \sum_j ee^{T}_{i,j}\\ &= \alpha \times 1 + {(1-\alpha)\over n} \sum_j ee^{T}_{i,j} \\ &\text{(since $A$ is stochastic)}\\ &= \alpha + {(1-\alpha)\over n} \sum_j 1 \\ &= \alpha + (1-\alpha) = 1 \end{align*} $$
Does this answer your question?
