The second largest in the absolute value eigenvalue I have a left stochastic matrix $A$ with size $N\times N$. I also have a matrix $A_d=dA+\dfrac{1-d}{N}S$, here $S$ a matrix with all elements equal to $1$ and $d$  is a small parameter in [0,1].
I want to prove the fact that the second largest in the absolute value eigenvalue of the obtained matrix is equal to d.
My attempt: I know that $A$ is a left stochastic matrix, then obviously $A_d$ is also a left stochastic.
I proved the lemma, that if $A$ is a matrix such that $a_{i,j} \ge 0$ and $\sum_{j}a_{i,j} = 1$ (left stochastic) then $A$ has an eigenvalue $\lambda=1$ and that any eigenvalue $\lambda_i$: $|\lambda_i| \le 1$.
Here I stopped and have no idea how to go further. Thus, what is the next step of solution? Please, help me.

The matrix $S$ (usually called $J$) has rank one, and its one non-zero eigenvalue equals $N,$ with the eigenvector being (proportional to) the vector $\mathbb{1}$ of all ones. Notice that $\mathbb{1}$ is also the Perron-Frobenius (left) eigenvector of $A,$ so, your matrix definitely does have eigenvalues $1-d, d$ (with the same eigenvector $\mathbb{1}.$ If $A$ is self-adjoint, you are done. If $A$ is not self-adjoint, it is far from clear that your statement is correct, and all you can say with any degree of certainty is that the second eigenvalue is at least $d$ in absolute value. Unless $A$ is a multiple of $S, in which case your statement is also false.
