= Solution
For $\theta_k=k\pi/(M+1)$, extend the proposed <vector> by $v_0=v_{M+1}=0$. The sine addition identity gives $v_{i-1}+v_{i+1}=2\cos\theta_k\,v_i$. Thus row-by-row multiplication yields
$$
\boxed{Av^{(k)}=(a+2b\cos\theta_k)v^{(k)},\qquad
\lambda_k=a+2b\cos\frac{k\pi}{M+1}.}
$$
These <vectors> do not depend on $a,b$. To prove they form a basis even when <eigenvalues> of $A$ coincide, apply the calculation first to the <symmetric matrix> with $a=0,b=1$, whose <eigenvalues> $2\cos\theta_k$ are distinct. Its $M$ nonzero <eigenvectors> are orthogonal and form a basis. Their squared lengths are $(M+1)/2$, so multiplying each by $\sqrt{2/(M+1)}$ gives a common <orthonormal eigenbasis> for every <matrix> in the family.
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