Let $f$ be an arithmetic function, $n \in \mathbb{N}^*$ and $g = f * \mu$. We denote $M = \left(m_{ij}\right)$ the matrix of $\mathcal{M}_n(\mathbb{C})$ with general term $m_{ij} = f(i \wedge j)$. We also define the divisor matrix $D = \left(d_{ij}\right)$ by: $$d_{ij} = \begin{cases} 1 & \text{if } j \text{ divides } i \\ 0 & \text{otherwise} \end{cases}$$ Let $M'$ be the matrix with general term $m_{ij}' = \begin{cases} g(j) & \text{if } j \text{ divides } i, \\ 0 & \text{otherwise.} \end{cases}$ Show that $M = M' D^\top$, where $D^\top$ is the transpose of $D$.
Let $f$ be an arithmetic function, $n \in \mathbb{N}^*$ and $g = f * \mu$. We denote $M = \left(m_{ij}\right)$ the matrix of $\mathcal{M}_n(\mathbb{C})$ with general term $m_{ij} = f(i \wedge j)$. We also define the divisor matrix $D = \left(d_{ij}\right)$ by:
$$d_{ij} = \begin{cases} 1 & \text{if } j \text{ divides } i \\ 0 & \text{otherwise} \end{cases}$$
Let $M'$ be the matrix with general term $m_{ij}' = \begin{cases} g(j) & \text{if } j \text{ divides } i, \\ 0 & \text{otherwise.} \end{cases}$
Show that $M = M' D^\top$, where $D^\top$ is the transpose of $D$.