If $ B= c1^T \odot A$ is positive definite, is $A$ invertible?

Question

Let matrix $B \in \mathbb{R}^{n \times n}$ be defined as \begin{align} B= c1^T \odot A \end{align} where $c$ is a vector with positive entries, $1^T$ is a vector of all ones, and $A$ is some square matrix. Here $\odot$ denotes Hadamard product.

Now suppose that $B$ is 1) symmetric, 2) positive definite, 3) has non-negative elements, and 4) elements on the main diagonal are strictly positive.

Question: What can we say about $A$?

For example, it is immediate that $A$ has non-negative entries. However, can we say, for example, whether $A$ is invertible?

Answer 1

Since $B$ is invertible, we have $$n=\mathrm{Rank}(B)\leq \mathrm{Rank}(c1^T)\mathrm{Rank}(A)=\mathrm{Rank}(A)\leq n.$$ Therefore, we see $A$ is also invertible.