Linear Algebra II (pdf, 500 kB)

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28 8.14. Lemma. Let φ be a bilinear form on the finite-dimensional vector space V, represented (w.r.t. some basis) by the matrix A. Then (1) φ is symmetric if and only if A ⊤ = A; (2) φ is skew-symmetric if and only if A ⊤ + A = 0; (3) φ is alternating if and only if A ⊤ + A = 0 and all diagonal entries of A are zero. (4) φ is non-degenerate if and only if det A = 0. Proof. Let v1, . . . , vn be the basis of V. Since aij = φ(vj, vi), the implications “⇒” in the first three statements are clear. On the other hand, assume that A ⊤ = ±A. Then x ⊤ Ay = (x ⊤ Ay) ⊤ = y ⊤ A ⊤ x = ±y ⊤ Ax , which implies “⇐” in the first two statements. For the third statement, we compute φ(v, v) for v = x1v1 + · · · + xnvn: n n φ(v, v) = aijxixj = aiix 2 i + (aij + aji)xixj = 0 , i,j=1 i=1 1≤i
Proof. We have to check a number of things. First, U ∩U ⊥ = {0} since v ∈ U ∩U ⊥ implies φ(v, u) = 0 for all u ∈ U, but φ is non-degenerate on U, so v must be zero. Second, U + U ⊥ = V : let v ∈ V, then U ∋ u ↦→ φ(v, u) is a linear form on U, and since φ is non-degenerate on U, by Cor. 8.9 there must be u ′ ∈ U such that φ(v, u) = φ(u ′ , u) for all u ∈ U. This means that φ(v − u ′ , u) = 0 for all u ∈ U, hence v − u ′ ∈ U ⊥ , and we see that v = u ′ + (v − u ′ ) ∈ U + U ⊥ as desired. So we have V = U ⊕ U ⊥ . The last statement is clear, since by definition, φ is zero on U × U ⊥ . Here is a first and quite general classification result for symmetric bilinear forms: they can always be diagonalized. 8.18. Lemma. Assume that char(F ) = 2, let V be an F -vector space and φ a symmetric bilinear form on V. If φ = 0, then there is v ∈ V such that φ(v, v) = 0. Proof. If φ = 0, then there are v, w ∈ V such that φ(v, w) = 0. Note that we have 0 = 2φ(v, w) = φ(v, w) + φ(w, v) = φ(v + w, v + w) − φ(v, v) − φ(w, w) , so at least one of φ(v, v), φ(w, w) and φ(v + w, v + w) must be nonzero. 8.19. Theorem. Assume that char(F ) = 2, let V be a finite-dimensional F -vector space and φ a symmetric bilinear form on V. Then there is a basis v1, . . . , vn of V such that φ is represented by a diagonal matrix with respect to this basis. Equivalently, every symmetric matrix A ∈ Mat(n, F ) is congruent to a diagonal matrix. Proof. If φ = 0, there is nothing to prove. Otherwise, we proceed by induction on the dimension n. Since φ = 0, by Lemma 8.18, there is v1 ∈ V such that φ(v1, v1) = 0 (in particular, n ≥ 1). Let U = L(v1), then φ is non-degenerate on U. By Prop. 8.17, we have an orthogonal splitting V = L(v1) ⊕ U ⊥ . By induction (dim U ⊥ = n − 1), U ⊥ has a basis v2, . . . , vn such that φ| U ⊥ ×U ⊥ is represented by a diagonal matrix. But then φ is also represented by a diagonal matrix with respect to the basis v1, v2, . . . , vn. 8.20. Remark. The entries of the diagonal matrix are not uniquely determined. For example, we can always scale the basis elements; this will multiply the entries by arbitrary nonzero squares in F . But this is not the only ambiguity. For example, we have 2 0 1 −1 1 0 1 1 = . 0 2 1 1 0 1 −1 1 On the other hand, the number of nonzero entries is uniquely determined, since it is the rank of the matrix, which does not change when we multiply on the left or right by an invertible matrix. 8.21. Example. Let us see how we can find a diagonalizing basis in practice. Consider the bilinear form on F 3 (with char(F ) = 2) given by the matrix ⎛ ⎞ 0 1 1 A = ⎝1 0 1⎠ . 1 1 0 Following the proof above, we first have to find an element v1 ∈ F 3 such that v ⊤ 1 Av1 = 0. Since the diagonal entries of A are zero, we cannot take one of 29
Page 1 and 2: Linear Algebra II Course No. 100 22
Page 3 and 4: that D = Q −1 AQ is diagonal. Sin
Page 5 and 6: If r = 0, then deg(r) < deg(MA), bu
Page 7 and 8: 3. The Structure of Nilpotent Endom
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Page 11 and 12: we see that v = a1(f) p1(f) (v)
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Page 15 and 16: Proof. Here is a sketch that gives
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Page 19 and 20: and for linear maps f1 : V1 → V2,
Page 21 and 22: 6.19. Theorem. Let V be a finite-di
Page 23 and 24: 7.3. Remark. A norm on a real vecto
Page 25 and 26: 8.3. Lemma and Definition. A biline
Page 27: so q(x) = 9 − 36x + 30x 2 , and p
Page 31 and 32: 8.25. Theorem (Classification of Sy
Page 33 and 34: 9.2. Definition. Let V be a complex
Page 35 and 36: 9.6. Theorem (Gram-Schmidt Orthonor
Page 37 and 38: 9.11. Theorem. Let V and W be two i
Page 39 and 40: 9.17. Proposition. Let V and W be t
Page 41 and 42: 10.4. Lemma. Let V be a finite-dime
Page 43 and 44: eigenvalues 1 and 6. This already t
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Page 47 and 48: which is of the required form. For
Page 49 and 50: 12.12. Lemma. Hom(U, Hom(V, W ))
Page 51 and 52: 12.16. Lemma. Let U, V, W be vector
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Page 55 and 56: 13.6. Definition. Let x ∈ V ⊗n
Page 57 and 58: (2) Let W ⊂ V ⊗n be the subspac
Page 59: 13.14. Lemma and Definition. Let f

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