Tensor product of two $n\times n$ orthogonal matrices with determinant $+1$

Question

If two square matrices A and B are two $n\times n$ orthogonal matrices with determinant unity i.e., $\det A=\det B=+1$ and $A^TA=B^TB=I$, will the tensor product $C\equiv A\otimes B$ also be orthogonal and have determinant $+1$? Can we understand this without restricting to special choices for $n$ such as $n=2,3$ etc

Answer 1

All of these properties can be understood by viewing the tensor product of matricies as a Kronecker product; I recommend looking into this.

As was suggested by @ Semiclassical we will use the fact that $$(A\otimes B)(C \otimes D)=(AC) \otimes (BD)$$ provided that the products are well defined. Take and $A,B \in O(n)$ As you pointed out, $AA^T=A^TA=BB^T=B^TB=I$ We consider $$(A\otimes B)(A^T \otimes B^T)$$ Transpose works nicely with the Kronecker product (I will leave it to you to check that): $$(A \otimes B)^T=(A^T \otimes B^T)$$ So we see that $(A \otimes B)\in O(n)$ As for the determinant we can use another property of the Kronecker product: $$det((A \otimes B))=det(A)^n det(B)^n$$