The Determinant of a Direct Product ©Philip Pennance1 Version: –January 1986
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 ∈ Rm×m and B ∈ 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
(A × B)11,11 (A × B)11,12 (A × B)11,21 (A × B)11,22 α ⊗ β(vi ⊗ wj) = α(vi) ⊗ β(wj). (A × B)12,11 (A × B)12,12 (A × B)12,21 (A × B)12,22 (A × B) (A × B) (A × B) (A × B) 21,11 21,12 21,21 21,22 (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. a B a B ··· a B 11 12 1m Let A ∈ Rm×m and B ∈ Rn×n. If A and B a B a B ··· a B 21 22 2m are the matrices of the linear maps α and β on A × B = . . . . . . . . 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 ∈ Rm×m and B ∈ 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 umn interchange. Hence | det T | = 1. and 0 ··· 0 a1j ··· 0 0 ··· 0 a ··· 0 1j n = . . . . . det(α ⊗ IW ) = det(ρA) = (det A) . . . . . . 0 ··· 0 a ··· 0 1j It follows that = AEij = λAEij det(A × B) = (det A)n(det B)m
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