tensor product of matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:09:55Z http://mathoverflow.net/feeds/question/56997 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56997/tensor-product-of-matrices tensor product of matrices spk 2011-03-01T12:52:42Z 2011-03-02T02:21:02Z <p>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 $\det(A)^m \cdot \det(B)^n$.</p> <p>Please, give me a link or reference if it is online or in some book. Thank you.</p> http://mathoverflow.net/questions/56997/tensor-product-of-matrices/57004#57004 Answer by Todd Trimble for tensor product of matrices Todd Trimble 2011-03-01T14:01:14Z 2011-03-02T02:21:02Z <p>Darij's first comment could be made into an answer as follows. </p> <p>Darij advised to write </p> <p>$$A \otimes B = (A \circ I_n) \otimes (I_m \circ B) = (A \otimes I_m) \circ (I_n \otimes B)$$</p> <p>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 </p> <p>$$\det(A \otimes I_m) \det(I_n \otimes B)$$ </p> <p>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 </p> <p>$$A \otimes I_{k^m} = A \otimes (I_k \oplus \ldots \oplus I_k) = (A \otimes I_k) \oplus \ldots \oplus (A \otimes I_k)$$ </p> <p>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. </p>