Define first the diagonal matrix element $$d_i=\frac{2}{h^2}+V_i,$$ and the non-diagonal matrix element $$e_i=-\frac{1}{h^2}.$$ In this case the non-diagonal matrix elements are given by a mere constant. All non-diagonal matrix elements are equal.

With these definitions the Schroedinger equation takes the following form $$d_iu_i+e_{i-1}u_{i-1}+e_{i+1}u_{i+1} = \lambda u_i,$$ where $u_i$ is unknown. We can write the latter equation as a matrix eigenvalue problem $$\begin{equation} \left( \begin{array}{ccccccc} d_1 & e_1 & 0 & 0 & \dots &0 & 0 \\ e_1 & d_2 & e_2 & 0 & \dots &0 &0 \\ 0 & e_2 & d_3 & e_3 &0 &\dots & 0\\ \dots & \dots & \dots & \dots &\dots &\dots & \dots\\ 0 & \dots & \dots & \dots &\dots &d_{n_{\mathrm{step}}-2} & e_{n_{\mathrm{step}}-1}\\ 0 & \dots & \dots & \dots &\dots &e_{n_{\mathrm{step}}-1} & d_{n_{\mathrm{step}}-1} \end{array} \right) \left( \begin{array}{c} u_{1} \\ u_{2} \\ \dots\\ \dots\\ \dots\\ u_{n_{\mathrm{step}}-1} \end{array} \right)=\lambda \left( \begin{array}{c} u_{1} \\ u_{2} \\ \dots\\ \dots\\ \dots\\ u_{n_{\mathrm{step}}-1} \end{array} \right) \tag{4} \end{equation}$$ or if we wish to be more detailed, we can write the tridiagonal matrix as $$\begin{equation} \left( \begin{array}{ccccccc} \frac{2}{h^2}+V_1 & -\frac{1}{h^2} & 0 & 0 & \dots &0 & 0 \\ -\frac{1}{h^2} & \frac{2}{h^2}+V_2 & -\frac{1}{h^2} & 0 & \dots &0 &0 \\ 0 & -\frac{1}{h^2} & \frac{2}{h^2}+V_3 & -\frac{1}{h^2} &0 &\dots & 0\\ \dots & \dots & \dots & \dots &\dots &\dots & \dots\\ 0 & \dots & \dots & \dots &\dots &\frac{2}{h^2}+V_{n_{\mathrm{step}}-2} & -\frac{1}{h^2}\\ 0 & \dots & \dots & \dots &\dots &-\frac{1}{h^2} & \frac{2}{h^2}+V_{n_{\mathrm{step}}-1} \end{array} \right) \tag{5} \end{equation}$$ Recall that the solutions are known via the boundary conditions at $i=n_{\mathrm{step}}$ and at the other end point, that is for $\rho_0$. The solution is zero in both cases.