MTH.B405; Sect. 1 (20190721) 2 1 Bilinear Forms Bilinear forms and quadratic forms. Definition 1.3. A symmetric bilinear form on the vector space A Review of Linear Algebra. V is a map q : V V R satisfying the following: × → Definition 1.1. A square matrix P of real components For each fixed x V , both is said to be an• orthogonal matrix if tPP = P tP = I • ∈ t q(x, ): V y q(x, y) R and holds, where P denotes the transposition of P and I is · ∋ 7→ ∈ the identity matrix. q( , x): V y q(y, x) R · ∋ 7→ ∈ A square matrix A is said to be (real) symmetric matrix are linear maps. • t if A = A holds. For any x and y V , q(x, y) = q(y, x) holds. • ∈ Fact 1.2. The eigenvalues of a real symmetric matrix are The quadratic form associated to the symmetric bilinear form q • real numbers, and the dimension of the corresponding eigenspace is a mapq ˜: V x q(x, x) R. ∋ 7→ ∈ coincides with the multiplicity of the eigenvalue. Lemma 1.4. A quadratic form determines the symmetric bi- linear form. In other words, two symmetric bilinear forms with Real symmetric matrices can be diagonalized by orthogonal common quadratic form coincide with each other. • matrices. In other words, for each real symmetric matrix A, there exists an orthogonal matrix P satisfying Proof. Let q be a symmetric bilinear form andq ˜ the quadratic form associated to it. Since 1 t P − AP = P AP = diag(λ1, . , λn), q˜(x + y) = q(x + y, x + y) where diag(... ) denotes the diagonal matrix with diago- = q(x, x) + q(x, y) + q(y, x) + q(y, y) nal components “... ”. In particular, λ1, . , λn are the =q ˜(x) + 2q(x, y) +q ˜(y) eigenvalues of A counted with their multiplicity.{ } holds for each x, y V , we have ∈ In this section, V denotes an n-dimensional vector space over 1 R (n < ). q(x, y) = q˜(x + y) q˜(x) q˜(y) . ∞ 2 − − 09. April, 2019. Revised: 16. April, 2019 Hence the symmetric bilinear( form q is determined) byq ˜. 3 (20190721) MTH.B405; Sect. 1 MTH.B405; Sect. 1 (20190721) 4 By virtue of Lemma 1.4, a symmetric bilinear form itself is Lemma 1.6. For a symmetric bilinear form q on V , there exists often called a quadratic form. the unique n n symmetric matrix A satisfying × Example 1.5. For an n n symmetric matrix A = (aij) with t × q(x, y) = qA(xˆ, yˆ) = xˆAyˆ, real components and column vectors xˆ, yˆ Rn, we set ∈ ( ) n n t n (1.1) qA : R R (xˆ, yˆ) xˆAyˆ R, where xˆ and yˆ R are the components of x, y with respect to × ∋ 7−→ ∈ ∈ [v1,..., vn], respectively where txˆ the column vector obtained by transposing xˆ. Then n Proof. Setting A = (aij) by aij := q(vi, vj), the conclusion qA is a symmetric bilinear form on R . In particular, qI is the n follows. In fact, canonical inner product of R , where I is the identity matrix. Conversely, for each symmetric bilinear form q in n, there R n n n exists a symmetric matrix A such that q = qA. In fact, setting q(x, y) = q x v , y v = x y q(v , v ) a := q(e , e ), A = (a ) satisfies q = q , where [e ] is the i i j j i j i j ij i j ij A j i=1 j=1 i,j=1 canonical basis of n. ∑ ∑ ∑ R n t = xiyjaij = xˆAyˆ = qA(xˆ, yˆ), Matrix representation of quadratic forms. Take a basis i,j=1 ∑ [v1,..., vn] of the vector space V . Then t t where xˆ = (x1, . , xn) and yˆ = (y1, . , yn). x 1 We call the matrix A in Lemma 1.6 the representative matrix . n (1.2) V x xˆ = . R , ∋ 7−→ . ∈ of q with respect to the basis [vj]. xn Lemma 1.7. Take two bases [v1,..., vn], [w1,..., wn] of V x ij 1 and let U = (u ) GL(n, R) be the basis change matrix: x = x v + + x v = [v ,..., v ] . ∈ 1 1 ··· n n 1 n . x [w1,..., wn] = [v1,..., vn]U n n n gives an isomorphism of vector spaces. We call xˆ R the i.e., wj = uijvi (j = 1, . , n), ∈ i=1 component of x with respect to the basis [vj]. ∑ 5 (20190721) MTH.B405; Sect. 1 MTH.B405; Sect. 1 (20190721) 6 where GL(n, R) denotes the general linear group, that is, the set Remark 1.10. A positive (resp. negative) quadratic form is non- of n n regular matrix of real components. Let A (resp. A˜) be degenerate. In fact, if q(x, y) = 0 holds for all y, q(x, x) = 0 the representative× matrix of a symmetric bilinear form q with holds. On the other hand, q(x, x) > 0 (resp. < 0) when x = 0. ̸ respect to the basis [vj](resp. [wj]). Then it holds that Hence x = 0. ˜ t A = UAU. Signature of non-degenerate quadratic forms. Proof. Writing x and y as Proposition 1.11. A quadratic form q on V is positive definite (resp. positive semi-definite, negative definite, non-degenerate) x = [v1,..., vn]xˆ = [w1,..., wn]x˜, if and only if all eigenvalues of the representative matrix of q are y = [v1,..., vn]yˆ = [w1,..., wn]y˜, positive (resp. non-negative, negative, non-zero). This condition does not depend on choice of bases. we have Proof. Let [v ,..., v ] be a basis of V and A the representative xˆ = Ux˜, yˆ = Uy˜. 1 n matrix of q with respect to it. Since A is a symmetric matrix, t Hence q(x, y) = xˆAyˆ = tx˜tUAUy˜. there exists an orthogonal matrix P such that Non-degenerate quadratic forms. λ1 ... 0 t . P AP = Λ, Λ := . .. , Definition 1.8. A symmetric bilinear form (a quadratic form) . 0 . λ q on V is said to be n tPP =I = the identity matrix, positive definite (resp. positive semi definite) if q(x, x) > 0 • (resp. 0) holds for all x V 0 , where λ1,. , λn are eigenvalues of A, which are real numbers. ≧ ∈ \{ } Then, by setting [w1,..., wn] := [v1,..., vn]P , the representa- negative definite if q is positive definite, • − tive matrix of q with respect to [wj] is the diagonal matrix Λ. Denoting the components of vectors x,y with respect to [wj] by non-degenerate when “q(x, y) = 0 for all y V ” implies t t • “x = 0”. ∈ xˆ = (x1, . , xn) and yˆ = (y1, . , yn), respectively, n Example 1.9. An inner product (in the undergraduate Linear q(x, y) = λjxjyj. Algebra course) is nothing but a positive definite quadratic form. j=1 ∑ 7 (20190721) MTH.B405; Sect. 1 MTH.B405; Sect. 1 (20190721) 8 The conclusion is obtained by this equality. In fact, if q is pos- The pair (n+, n ) is called the signature of q. itive (resp. negative) definite, q(w , w ) = λ is positive (resp. − j j j Example 1.13. A positive (resp. negative) definite quadratic negative) for each j = 1, . , n. Hence all eigenvalues of A are form q on V has signature (n, 0) (resp. (0, n)), where n = dim V . positive (resp. negative). Conversely, if all eigenvalues are posi- tive (resp. negative), Theorem 1.14. Let (n+, n ) be the signature of a non-degenerate − n quadratic form q on V . Then n+ (resp. n ) is the number of − 2 q(x, x) = λj(xj) positive (resp. negative) eigenvalues of the representative matrix j=1 of q. In particular, n+ + n = n = dim V holds. ∑ − is positive (resp. negative). The conclusion for positive semi- Proof. As seen in the proof of Proposition 1.11 we may assume definite case is obtained in the same way. that the matrix representative with respect to the basis [wj] is a On the other hand, let q be a non-degenerate quadratic form diagonal matrix Λ, without loss of generality. Since all diagonal and assume λj = 0 for some j = 1, . , n. Then components of Λ are non-zero, we may assume that λ1, . , λt are negative, and λt+1, . , λn are positive. Then q is negative q(wj, wj) = λj = 0, ̸ definite on the subspace generated by w1,..., wt , and hence { } contradiction to non-degeneracy. Conversely, assume λj = 0 n ≧ t. On the other hand, q is positive definite on the subspace ̸ − (j = 1, . , n), and q(x, y) = 0 for all y V . Then generated by w ,..., w , and then n n t: ∈ { t+1 n} + ≧ − 0 = q(x, wj) = λjxj (1.3) n ≧ t, n+ ≧ n t. − − holds, which implies x = 0, for j = 1, . , n. Thus, x = 0.b j Here, by definition, there exists a subspace W+ (resp. W ) Hence q is non-degenerate. − of V such that q W+ (resp. q W ) is positive (resp. negative) | | − definite and dim W+ = n+ (resp. dim W = n ). Take a vector Let W be a linear subspace of the vector space V . Then a − − x W+ W . Then q(x, x) ≦ 0 and q(x, x) ≧ 0 hold, that is, symmetric bilinear form q : V V R on V induces a symmet- ∈ ∩ − × → q(x, x) = 0. Noticing q W is positive definite, x = 0. Hence ric bilinear form q W : W W R on W . | + | × → W+ W = 0 , and then we have Definition 1.12. For a non-degenerate quadratic form q on V , ∩ − { } we define n+ + n = dim W+ + dim W ≦ dim V = n.