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

Question

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

Answer 1

There is a general trick that applies to this case.

Assume a matrix $A=(a_{i,j})$ is such that there exists a function $\psi$ such that $$ a_{i,j}=\sum_{k|i,k|j}\psi(k) $$ for all $i,j$.

Then $$ \det A=\psi(1)\psi(2)\cdots\psi(n). $$ To see this, consider the matrix $B=(b_{i,j})$ such that $b_{i,j}=1$ if $i|j$ and $b_{i,j}=0$ otherwise. Note that $B$ is upper-triangular with ones on the diagonal, so its determinant is $1$.

Now let $C$ be the diagonal matrix whose diagonal is $(\psi(1),\ldots,\psi(n))$.

A matrix product computation shows that $$ A=B^tCB\quad\mbox{hence}\quad \det A=(\det B)^2\det C=\psi(1)\cdots\psi(n). $$

Now going back to your question. Consider Euler's totient function $\phi$. It is well-known that $$ m=\sum_{k|m}\phi(k) $$ so $$ a_{i,j}=gcd(i,j)=\sum_{k|gcd(i,j)}\phi(k)=\sum_{k|i,k|j}\phi(k). $$

Applying the general result above, we find: $$ \det A=\phi(1)\phi(2)\cdots\phi(n). $$

Answer 2

This is a nice result. We have $$ \det(A)= \phi(1)\phi(2)\dots\phi(n), $$ where $\phi$ is Euler's phi function, $\phi(n)$ being the number of positive integers $i\le n$ that are relatively prime with $n$. It satisfies $$ \sum_{j\mid n}\phi(j)=n, $$ which we use below.

Let's write $a_n$ for the determinant of the $n\times n$ version of $A$. The sequence $a_1,a_2,a_3,\dots$ begins $$ 1, 1, 2, 4, 16, 32, 192, \dots $$ which OEIS catalogs as $A001088$.

A cute short proof that only uses basic linear algebra (LDU decomposition) appears in a recent note,

Warren P. Johnson. An $LDU$ Factorization in Elementary Number Theory, Math. Mag. 76 (5), (2003), 392–394. MR1573717.

What one shows is that $$ A=L\Phi L^T $$ where $L$ is the $n\times n$ matrix whose $i,j$ entry is $1$ if $j$ divides $i$, and is $0$ otherwise, and $\Phi$ is the diagonal matrix whose $i,i$ entry is $\phi(n)$. Note that $L$ is lower diagonal, with $1$s as entries along its main diagonal, so $\det(L)=\det(L^T)=1$, and the result follows once we prove that $A=L\Phi L^T$, as claimed. Johnson calls this Le Paige's result, who established it to compute $\det(A)$ (originally found by H. J. S. Smith around 1875, according to the references at the OEIS site).

To prove Le Paige's result, simply expand $L\Phi L^T$, and note that its $i,j$ entry is $$ \sum_{k=1}^n L_{ik}\Phi_{kk}(L^T)_{kj}=\sum_{k=1}^n L_{ik}\phi(k) L_{jk}=\sum_{k\mid i,k\mid j}\phi(k) =\sum_{k|{\rm gcd}(i,j)} \phi(k)={\rm gcd}(i,j)=A_{ij}, $$ where I use $B_{ij}$ to denote the $i,j$ entry of the matrix $B$.

Answer 3

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.