Let the matrix $A=[a_{ij}]_{n×n}$ be defined by $a_{ij}=\gcd(i,j )$. How prove that $A$ is invertible, and compute $\det(A)$? Let $A=[a_{ij}]_{n×n}$ be the matrix defined by letting $a_{ij}$ be the rational number such that $$a_{ij}=\gcd(i,j ).$$ How prove that $A$ is invertible, and compute $\det(A)$? thanks in advance 

Let $F\colon \{1, \ldots, n\}\to \mathbb{R}$ be a function.  The $n\times n$ matrix $A=(a_{ij}) = (F(\gcd(i.j)))$ is symmetric. It it possible to give its Cholesky decomposition. Indeed, let $g\colon \{ 1, \ldots, n\}\to \mathbb{R}$ such that
$$F(m) = \sum_{d \mid m} G(d)$$
The function $g$ is uniquely determined by the equation
$$G(m) = \sum_{d\mid m} \mu(d) F(\frac{m}{d})$$F
where $\mu$ is the Mobius function. We have
$$F(\gcd(i.j)) = \sum_{d=1}^n u_{d i} u_{d j}\, G(d)$$
where $u_{d i} = 1$ if $d\mid i$ and $0$ otherwise.
We see that we have the equality
$$A = U^t \cdot D \cdot U$$
where $U$ is the upper triangular matrix $(u_{ij})$, and $D$ is the diagonal matrix $(G(1), \ldots, G(n))$.
We conclude that $\det A= G(1) \cdots G(n)$.
Note: if $F(m) = m^{\alpha}$, with $\alpha > 0$, then $G(m)$ is always positive ( easy to see using the Mobius inversion formula). We conclude that all the Hadamard positive powers of the matrix $A$ are positive definite, with determinants that can be calculated explicitly.
This method works for other partial orders on the set $\{1, \ldots, n\}$, in particular the standard order. The bordered determinants are therefore easy to calculate.
