Positive definiteness and vectorization

Question

If matrix $A$ is positive definite, what properties does its vectorization column vector has? Here vectorization means $\operatorname{vec}(A)$, that is to make it into a single column by list first column, then 2nd column, then 3rd.

Answer 1

There is unfortunately not known easy interesting exact condition on $\operatorname{vec}(A)$ besides the trivial ones; e.g. diagonal matrices.

There is one in the sense that a symmetric matrix $A$ is positive definite if and only if all its principal minors are positive. This translates into a nonlinear system of inequalities in the entries of $\operatorname{vec}(A)$, but this is not something which is particularly useful.

If you are just interested in sufficient conditions. You can state one in terms of strict diagonal dominance which states that $A$ is positive definite if the diagonal entries of $A$ are positive and $A1>0$ where $1$ is a vector of ones, and the inequality is component-wise. This directly translates into conditions on some of the entries of $\operatorname{vec}(A) but nothing is gained here.

We may also look at the product $A=BB^T$ for some square matrix $B$, which yields

$$\operatorname{vec}(BB^T)=(B\otimes B)\operatorname{vec}(I),$$

but one can see that $B\otimes B$ may fail to be symmetric and, even if it were, may also fail to be positive definite.

Finally, it is easy to construct a counterexample to the fact that $\operatorname{vec}(A)=B\operatorname{vec}(D)$ with $B,D$ are symmetric and positive definite and such that $A$ is symmetric implies that that $A$ is also positive definite. Note that $B$ needs to have a very specific structure here to guarantee that $A$ is symmetric.

So, let $D=I$ and

$$B=\begin{bmatrix} 1 & -2 & 0 & 0\\ -2 & 4 & 0 &0\\ 0 & 0 & 8 & -2\\ 0 & 0 & -2 & 1 \end{bmatrix}.$$

Then, we have that $$B\operatorname{vec}(D)=\begin{bmatrix} 1\\ -2\\ -2\\ 1 \end{bmatrix},$$

which yields to $$A=\begin{bmatrix} 1 & -2\\ -2 & 1 \end{bmatrix},$$

which is not positive definite.