Question

Hello, all! Could somebody draw a proof-sketch of next expression from tensor algebra on matrices over finite fields: determinant of tensor product $A~ \times ~B$ of $n \times n$-matrix $A$ over finite field $GF(q)$ on $m \times m$-matrix $B$ over finite field $GF(q)$ is $\det(A)^m \cdot \det(B)^n$.

Please, give me a link or reference if it is online or in some book. Thank you.

Answer 1

Darij's first comment could be made into an answer as follows.

Darij advised to write

$$A \otimes B = (A \circ I_n) \otimes (I_m \circ B) = (A \otimes I_m) \circ (I_n \otimes B)$$

where the second equation follows from functoriality of the tensor product. Here both $A \otimes I_m$ and $I_n \otimes B$ are square matrices of size $m n \times m n$. Since the determinant from such matrices to the scalar field is a monoid homomorphism, the determinant of the last expression is

$$\det(A \otimes I_m) \det(I_n \otimes B)$$

so we are left to determine the two determinants above. Since these are similar, we do the first. We may express an $m$-dimensional vector space $k^m$ as a direct sum of 1-dimensional vector spaces, so

$$A \otimes I_{k^m} = A \otimes (I_k \oplus \ldots \oplus I_k) = (A \otimes I_k) \oplus \ldots \oplus (A \otimes I_k)$$

because tensor products preserve direct sums. This is just $A \oplus \ldots \oplus A$. This matrix consists of $m$ blocks of $A$, so its determinant is $\det(A)^m$, and we are done.