The Determinant of a Direct Product ©Philip Pennance1 Version: {January 1986. 1. Introduction. a basis for W: For i = 1; : : : m, j = 1 : : : n de- Many results in matrix algebra can be regarded fine ∗ ∗ as statements about linear mappings between vi ⊗ wj : V × W ! R vector spaces. This viewpoint is used to derive by the formula for the determinant of the direct vi ⊗ wj(f; g) = f(vi)g(wj) product of two matrices. then (exercise) the maps vi ⊗ wj form a basis m×n 2. Let R be the vector space of real m×n ma- for V ⊗ W: trices. If A 2 Rm×m and B 2 Rn×n the direct product (A × B) is defined by 7. Definition. Let α : V ! V and β : W ! W be linear (A × B)ij;kl = AikBjl maps. The tensor product of α with β is the linear map where the rows and columns are specified by the lexicographically ordered pairs i; j and k; l: α ⊗ β : V ⊗ V ! W ⊗ W 3. Example In the case n; m = 2 the elements of the direct product are arranged: defined by its action on the above basis by 2 3 (A × B)11;11 (A × B)11;12 (A × B)11;21 (A × B)11;22 α ⊗ β(vi ⊗ wj) = α(vi) ⊗ β(wj): 6 (A × B)12;11 (A × B)12;12 (A × B)12;21 (A × B)12;22 7 6 (A × B) (A × B) (A × B) (A × B) 7 4 21;11 21;12 21;21 21;22 5 (By an abuse of notation the same symbol is (A × B)22;11 (A × B)22;12 (A × B)22;21 (A × B)22;22 used for the tensor product of linear maps, and the tensor product of vector spaces). 4. Exercise. Check that: 8. Lemma. 2 a B a B ··· a B 3 11 12 1m Let A 2 Rm×m and B 2 Rn×n: If A and B 6 a B a B ··· a B 7 6 21 22 2m 7 are the matrices of the linear maps α and β on A × B = 6 . 7 4 . 5 vector spaces V , W of dimensions m, n with re- a B a B ··· a B m1 m2 mm spect to bases v1; v2; : : : vm, and w1; w2; : : : wm for V and W respectively, then the matrix of 5. Definition. Let V , W be vector spaces of di- α ⊗ B in the basis vi ⊗ wj is the direct product mensions m, n and let V ∗, W ∗ be the cor- α × B: responding dual spaces The tensor product Proof. Applying α⊗β to the basis vi ⊗wj with V ⊗ W is the vector space of bilinear maps lexicographic ordering by pairs (i; j): V ∗ × W ∗ ! R: A ⊗ B(vi ⊗ wj) = Avi ⊗ Bwj m n X X 6. A basis for V ⊗ W = arivr ⊗ bsjws r=1 s=1 Let m n X X v1; v2; : : : vm = aribsjvr ⊗ ws r=1 s=1 be a basis for V and m n X X = (A × B) v ⊗ w w ; w ; : : : ; w rs;ij r s 1 2 n r=1 s=1 1http://pennance.us 1 m×n m×n 9. Claim. Similarly, if ρB : R ! R is right multi- Let A 2 Rm×m and B 2 Rn×n then plication by B, then det(A × B) = (det A)n(det B)m IV ⊗ β Proof. Let α and β be as in the lemma. Write V ⊗ W −! V ⊗ W ? ? α ⊗ β = (α ⊗ IW ) ◦ (IV ⊗ β) ? ? yγ yγ m×n m×n where IV and IW are the identity maps in V , R −! R W respectively to conclude that, ρB det(A × B) = det(α ⊗ β) = det[(α ⊗ I ) ◦ (I ⊗ β)] W V commutes. From the latter diagram it follows = det(α ⊗ IW ) det(IV ⊗ β) that Since I ⊗ B is block diagonal with each block m m det(ρ ) = det(I ⊗ β) = [det(B)] equal to B have that det(IV ⊗ β) = (det B) : B V To obtain the determinant of A × IW observe m×n m×n and from the former that if λA : R ! R is left multiplica- tion by the matrix A, and γ : V ⊗W ! Rn×m, det(α ⊗ IW ) = det(λA): the isomorphism defined by γ(vi ⊗ wj) = Eij where Eij is the matrix with 1 in position (i; j) and 0 elsewhere, then the following diagram Let commutes. T : Rm×n ! Rn×m α ⊗ IW V ⊗ W −! V ⊗ W be the transpose map T (A) = AT : Then ? ? ? ? yγ yγ m×n m×n R −! R λA = T ◦ ρ(TA) ◦ T λA Thus To see this notice that −1 γ ◦ (α ⊗ I) ◦ γ (Eij) 2 det(A ⊗ IW ) = (det T ) det(ρA): = γ ◦ (α ⊗ I)(vi ⊗ wj) = γ(α(vi) ⊗ I(wj)) Observe that T can be written as a composi- m X tion of mappings Tij, 1 ≤ i ≤ m, 1 ≤ j ≤ n, = γ arivr ⊗ wj i 6= j ,where Tij interchanges aij and aji only. r=1 Furthermore, the matrix of each Tij with re- m X spect to the standard bases in Rm×n, Rn×m = a E ri rj differs from the identity matrix by a single col- r=1 2 3 umn interchange. Hence j det T j = 1: and 0 ··· 0 a1j ··· 0 6 0 ··· 0 a ··· 0 7 6 1j 7 n = 6 . 7 det(α ⊗ IW ) = det(ρA) = (det A) : 4 . 5 0 ··· 0 a ··· 0 1j It follows that = AEij = λAEij det(A × B) = (det A)n(det B)m 2.