I'm following a paper and trying to figure out a step that they made. Say that we have the following matrix:
\begin{bmatrix}((1-\alpha)(1-s_0)+\alpha(1-s_1))v_f&q_0(1-s_0)v_f&q_1(1-s_1)v_f \\((1-\alpha)s_0(1-\mu_0)+\alpha s_1)v_0&q_0s_0(1-\mu_0)v_{m0}&q_1s_1\mu_1v_{m0} \\((1-\alpha)s_0\mu_0+\alpha s_1(1-\mu_1))v_1 & q_0s_0\mu_0v_{m1} & q_1s_1(1-\mu_1)v_{m1} \end{bmatrix}
This is a biological model. I don't think the details are important because I think my problem is with the mathematics. So, let's say that $a_1$ defines the first column (reproductive output of females), $a_2$ defines the second column (reproductive output of males type 0), and $a_3$ defines the third column (reproductive output of males type 1). It is required that all females (relative frequency $y_f$) have the same reproductive output as all males ($y_0$ + $y_1$ ). Accordingly:
\begin{equation}a_1y_f = a_2y_0 + a_3y_1 \end{equation}
This is equation 1. Let \begin{equation} y = (y_{f},y_0,y_1) \end{equation}
be the dominant right eigenvector of the matrix. This eigenvector is given by \begin{equation} A y = \lambda y \end{equation}
They now say that substituting equation 1 into this equation results in:
\begin{equation} \lambda y = 2a_1y_f \end{equation}
I get this step. This is just following the rules of multiplication of matrices by a vector:
\begin{equation} \lambda y = a_1y_f + a_2y_0 + a_3y_1 \end{equation}
Using equation 1:
\begin{equation} \lambda y = a_1y_f + a_1y_f \end{equation} \begin{equation} \lambda y = 2a_1y_f \end{equation}
But then the authors go from here to:
\begin{equation} \lambda = 2a_{11} \end{equation}
Which I don't understand - $a_{11}$ is the first element of the matrix. I think the problem is my lack of mathematical knowledge to understand the logic, so I'm hoping that someone can help me figure out how this is possible.