= Solution
Let $D\in\mathbb R^{M\times M}$ be the <Toeplitz antisymmetric tridiagonal matrix> with $D_{m,m+1}=1$, $D_{m,m-1}=-1$, and zero diagonal. Fixed homogeneous endpoint values give the homogeneous update
$$
\left(I-\frac\mu4D\right)u^{n+1}
=\left(I+\frac\mu4D\right)u^n.
$$
The hint gives the eigenvalues
$$
d_j=2i\cos\frac{j\pi}{M+1},
\qquad 1\leq j\leq M.
$$
Because each $d_j$ is purely imaginary, $1-(\mu/4)d_j\ne0$ for every real $\mu$, so the left matrix is invertible. The two matrices are polynomials in $D$ and therefore have the same orthonormal eigenvectors. The amplification matrix
$$
Q=\left(I-\frac\mu4D\right)^{-1}
\left(I+\frac\mu4D\right)
$$
has eigenvalues
$$
q_j=
\frac{1+i(\mu/2)\cos(j\pi/(M+1))}
{1-i(\mu/2)\cos(j\pi/(M+1))}.
$$
The numerator and denominator are complex conjugates, so $|q_j|=1$. Moreover, the common orthonormal eigenbasis makes $Q$ a <normal matrix>. By the <matrix 2-norm of a normal matrix>,
$$
\lVert Q^n\rVert_2=\rho(Q^n)=1
$$
for every $n$. Thus the <Crank-Nicolson centered-advection scheme on a finite interval> is unconditionally stable:
$$
\boxed{\text{every }\mu>0.}
$$
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