Symmetric matrix with positive entries being invertible?

Let $u_1,\dots,u_n\in S^{n-1}$ be $n$ linearly independent unit vectors (thus a basis for $\mathbb{R}^n$). Let $a_{ij}:=u_i^\top u_j$. Then $a_{ii}=1$ and $a_{ij}=a_{ji}$.

My question is:

Is the following matrix always invertible $$A=(a_{ij}^2)_{1\leq i,j\leq n}$$

For $n=2$ it is, but for general $n$ I think there are counterexamples. Since the entrywise squareroot of $A$ is invertible, I guess we can start with some invertible(maybe unitary?) matrix $P$ with $P^2$(entrywise) not invertible, and then modify it to find the corresponding vectors. But I failed to modify it to find such unit vectors...

Solution 1:

Yes, $A$ is always invertible. Let $B$ be the matrix $[u_1 \cdots u_n]$. Then $B$ is always invertible, so $C := B^T B$ is (by a characterization) positive definite. It is known for positive definite matrices $X$ and $Y$ that $\det(X\ \square\ Y) \geq \det(X) \det(Y)$ where $\square$ is the so-called Hadamard product. When we take $X = Y = C$ then $A = C\ \square\ C$. We conclude that $\det(A) \geq \det(C)^2 = \det(B)^4 > 0$, so in particular $A$ is always invertible.

The condition that the vectors $u_i$ are unit length is not necessary.