= Solution
Set $\theta_k=k\pi/(M+1)$ and extend the proposed vector by $v_0=v_{M+1}=0$. For $v_m=i^m\sin(m\theta_k)$, the angle-addition identity gives
$$
v_{m+1}-v_{m-1}=i^{m+1}[\sin((m+1)\theta_k)+\sin((m-1)\theta_k)]=2i\cos\theta_k\,v_m.
$$
Hence the <Toeplitz antisymmetric tridiagonal matrix> has eigenvalues
$$
\boxed{\lambda_k=a+2ib\cos\frac{k\pi}{M+1},\qquad1\le k\le M.}
$$
The real sine vectors are eigenvectors of the real symmetric tridiagonal matrix with both off-diagonals one, with distinct eigenvalues $2\cos\theta_k$; thus they are orthogonal and form a basis. Multiplication by the unitary diagonal matrix $\operatorname{diag}(i,i^2,\ldots,i^M)$ preserves orthogonality. Their squared norms are $(M+1)/2$, so multiplying by $\sqrt{2/(M+1)}$ gives a common <orthonormal basis> for all $a,b$, even when some eigenvalues coincide.
Let $K$ have superdiagonal one, subdiagonal minus one and diagonal zero. The prescribed endpoint values remove the exterior unknowns, and the scheme is
$$
\boxed{B=I-\frac\mu4K,\qquad C=I+\frac\mu4K,\qquad Bu^{n+1}=Cu^n.}
$$
In the common orthonormal eigenbasis, the two eigenvalues are $1\mp i(\mu/2)\cos\theta_k$. Every eigenvalue of $B$ is nonzero for real $\mu$, and the amplification matrix $Q=B^{-1}C$ has eigenvalues
$$
q_k=\frac{1+i(\mu/2)\cos\theta_k}{1-i(\mu/2)\cos\theta_k},\qquad |q_k|=1.
$$
Thus $Q$ is a <unitary matrix> and $\|u^{n+1}\|_2=\|u^n\|_2$ exactly. The same holds for the mesh-weighted norm. This matrix argument proves $\boxed{\text{stability for every real }\mu\text{, hence every physical }\mu\ge0}$, without a Fourier stability calculation.
\b[The printed mesh ratio has a consistency error:] advection requires $\mu=\Delta t/\Delta x$, while the PDF prints $\Delta t/(\Delta x)^2$. Indeed the centered spatial difference is $2\Delta x\,u_x+O((\Delta x)^3)$, so division of the scheme by $\Delta t$ gives leading right-hand coefficient $\mu\Delta x/\Delta t$. The printed choice makes this $1/\Delta x$ rather than one. The stability conclusion is valid for the algebraic scheme with either definition, but does not fix this inconsistency. The continuous first-order advection problem also ordinarily specifies only the inflow endpoint, here $x=1$; zero data at both endpoints restrict admissible initial data. For example the initially boundary-zero profile $\sin\pi x$ develops nonzero outflow $u(0,t)=\sin\pi t$ before inflow changes reach that boundary. The matrix norm calculation concerns the stated two-endpoint discretization and does not establish well-posedness for arbitrary continuous initial data.
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