We define the Redheffer matrix by $H_n = \left(h_{ij}\right)_{(i,j) \in \llbracket 1,n \rrbracket^2}$ where
$$h_{ij} = \begin{cases} 1 & \text{if } j = 1 \\ 1 & \text{if } i \text{ divides } j \text{ and } j \neq 1 \\ 0 & \text{otherwise.} \end{cases}$$
Using the expression
$$\chi_n(\lambda) = (\lambda - 1)^n - \sum_{k=1}^{\left\lfloor \log_2 n \right\rfloor} (\lambda - 1)^{n-k-1} S_k(n),$$
finally show that $H_n$ has 1 as an eigenvalue and that its multiplicity is exactly
$$n - \left\lfloor \log_2 n \right\rfloor - 1.$$