Explicit Scheme, final stability analysis

The eigenvalues of $\mathbf{A}$ are ${\lambda }_i=1-\alpha {\mu }_i$, with ${\mu }_i$ being the eigenvalues of $\hat{B}$. To find ${\mu }_i$ we note that the matrix elements of $\hat{B}$ are $$b_{ij}=2{\delta }_{ij}-{\delta }_{i+1j}-{\delta }_{i-1j},$$ meaning that we have the following set of eigenequations for component $i$ $$(\hat{B}\hat{x}{)}_i={\mu }_i{x}_i,$$ resulting in $$(\hat{B}\hat{x}{)}_i=\sum _{j=1}^n(2{\delta }_{ij}-{\delta }_{i+1j}-{\delta }_{i-1j})x_j=2x_i-x_{i+1}-x_{i-1}={\mu }_i{x}_i.$$

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