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Given the tensor product of a Invertable Square Positive-Definite Vandermonde Matrix $a$

$$\mathbf{a} = \ \left( \begin{array}{ccc} 1 & 1 & \ldots & 1^{D-1} \\ 1 & 2 & \ldots & 2^{D-1} \\ \vdots & & \ddots & \vdots \\ 1 & D & \ldots & D^{D-1} \\ \end{array} \right)$$

with itself: $b = a\otimes a$.

Is b invertable?

And perhaps a bit more generally, if a is an invertable positive-definite matrix, but not necessarily a Vandermonde one, would $b$ be invertable?

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Yes, the inverse of tensor product of invertible matrices is tensor product of inverse matrices. That is, $$b^{-1}=(a\otimes a)^{-1}=a^{-1}\otimes a^{-1}$$ To verify, $$(a\otimes a)(a^{-1}\otimes a^{-1})=(aa^{-1})\otimes(aa^{-1})=I_n\otimes I_n=I_{2n}$$ (The other direction is similar)

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