= Solution
With the endpoint values zero, move the new-level differences to the left and keep the old-level differences on the right. Then $B$ is tridiagonal with diagonal $1+\mu$ and off-diagonal entries $-\mu/2$, while $C$ has diagonal $1-\mu$ and off-diagonal entries $\mu/2$. Thus \b[$Bu^{n+1}=Cu^n$].
Part (a) supplies their common <orthonormal eigenbasis>. Their <eigenvalues> are $1+\mu(1-\cos\theta_k)$ and $1-\mu(1-\cos\theta_k)$. For $\mu\geq0$, $B$ is invertible, and the step <matrix> $B^{-1}C$ has <eigenvalues>
$$
\eta_k=\frac{1-\mu(1-\cos\theta_k)}{1+\mu(1-\cos\theta_k)},\qquad |\eta_k|\leq1.
$$
Since the eigenbasis is orthonormal, $\|(B^{-1}C)^n\|_2=\max_k|\eta_k|^n\leq1$, uniformly in the number of steps and mesh. Hence the method is \b[unconditionally stable for every $\mu\geq0$], which includes all physical positive time steps. For negative $\mu$, every nonzero modal parameter has modulus larger than one whenever its denominator is nonzero; if a denominator vanishes the step is undefined. Thus the algebraic stable range is exactly $[0,\infty)$. This proof uses simultaneous <matrix> diagonalization, not Fourier stability analysis.
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