Condition for a matrix to be positive definite

Question

I have $T = X^TX \otimes Z^TZ$ matrix where $X \in \mathbb{R}^{n \times k}$ and $Z \in \mathbb{R}^{k \times n}$. Is it possible to find conditions on $X$ and $Z$ such that $T \succeq \mu I$ in other words $T$ is a positive definite matrix?

Answer 1

Claim: Suppose $A,B$ are square matrices. Then $A\otimes B$ has full rank if and only if $A,B$ has full rank.

Proof: Follows immediately from $\det(A\otimes B)=\det(A)^m\det(B)^n$. QED.

Claim: $X^TX\otimes Z^TZ=(X\otimes Z)^T(X\otimes Z)$ for any matrix $X,Z$ of any size.

Proof: Follows from the identities $(A\otimes B)(C\otimes D)=(AC)\otimes (BD)$ and $(A\otimes B)^T=A^T\otimes B^T$. QED.

Let $X\in\mathbb{R}^{n\times k},Z\in\mathbb{R}^{k\times n}$. Combining the two claims, $X^TX\otimes Z^TZ$ is positive semi-definite, and it is invertible iff both $X^TX$, $Z^TZ$ are invertible. In other words, $X^TX\otimes Z^TZ$ is positive-definite iff $n=k$ and $X,Z$ are invertible.