The Ramanujan $\tau$-function satisfies $$\tau(mn)=\sum_{d|m,n}d^{11}\mu(d)\tau(m/d)\tau(n/d)\tag{1}$$for any two natural numbers m,n , where $\mu(n) $ is the Moebius function .
Let $A$ be the $n\times n$ matrix whose (i,j) entry is $\tau(ij)$ . Show that the determinant of A is given by $n!^{11}\mu(1)\mu(2) \cdot...\cdot\mu(n)$ . In particular, the determinant is zero if $n \geq4 .$
Well , I do not understand why (1) implies that I can write $A=LDL^T$ , where D is a diagonal matrix whose i-th diagonal entry is $i^{11}\mu(i)$ and $L$ is the lower triangular matrix whose (i, j) entry is given by $\tau(j/i)$ if $i|j$ and zero otherwise .
If $L$ is lower triangular, and $D$ is diagonal, then $$(LDL^T)_{i,j}=\sum_{k=1}^{\min\{i,j\}}L_{i,k}D_{k,k}L_{j,k}.$$ With your $L$ and $D$, the $d$-th term of the sum is zero if neither $d\mid i$ nor $d\mid j$. Otherwise... we arrive exactly at $(1)$ with $m=i$ and $n=j$.