Linear Algebra (XXVIII) Yijia Chen 1. Quadratic Forms Definition 1.1. Let A be an n × n symmetric matrix over K, i.e., A = aij i,j2[n] where aij = aji 0 1 x1 B . C for all i, j 2 [n]. Then for the variable vector x = @ . A xn 0 1 0 1 a11 ··· a1n x1 0 x ··· x B . .. C B . C x Ax = 1 n @ . A @ . A an1 ··· ann xn = aij · xi · xj i,Xj2[n] 2 = aii · xi + 2 aij · xi · xj. 1 i<j n iX2[n] 6X6 is a quadratic form. Definition 1.2. A quadratic form x0Ax is diagonal if the matrix A is diagonal. Definition 1.3. Let A and B be two n × n-matrices. If there exists an invertible C with B = C0AC, then B is congruent to A. Lemma 1.4. The matrix congruence is an equivalence relation. More precisely, for all n × n-matrices A, B, and C, (i) A is congruent to A, (ii) if A is congruent to B, then B is congruent to A as well, and (iii) if A is congruent to B, B is congruent to C, then A is congruent to C. Lemma 1.5. Let A and B be two n × n-matrices. If B is obtained from A by the following elementary operations, then B is congruent to A. (i) Switch the i-th and j-th rows, then switch the i-th and j-th columns, where 1 6 i < j 6 j 6 n. (ii) Multiply the i-th row by k, then multiply the i-th column by k, where i 2 [n] and k 2 K n f0g. (iii) Multiply the i-th row by k and add the result to the j-th row, then multiply the i-th column by k and add the result to the j-th column, where i, j 2 [n] with i 6= j and k 2 K. Proof: (i) Recall that switching the i-th and j-th rows of the matrix A is equivalent to multiplying A on its left by the elementary matrix Pij. Similarly the column switching corresponds to multiplying the same Pij on A’s right. Thus 0 B = PijAPij = PijAPij. 1 1 Furthermore, Pij is invertible . Thus B is congruent to A. (ii) and (iii) can be shown in exactly the same fashion. 2 Theorem 1.6. Every symmetric matrix is congruent to a diagonal matrix. We first prove a technical lemma. Lemma 1.7. Let A be an n × n symmetric nonzero matrix. Then there exists an invertible matrix C such that 0 (C AC)11 6= 0. 0 0 Recall that (C AC)11 denotes the element on the first row and the first column of the matrix C AC. Proof: If (A)11 6= 0, then we are done by taking C = In. Otherwise, assume that (A)ii 6= 0 for some 2 6 i 6 n, then we switch the first and the i-th rows, and then switch the first and the i-th columns. Thereby, we move (A)ii to the first row and the first column. The result then follows from Lemma 1.5 (i). Now we are left with the case that all diagonal elements of A are 0. Since A is symmetric and nonzero, there exist some 1 6 i < j 6 n with Aij 6= 0. We add the i-th row to the j-th row, and then add the i-th column to the j-th column. Let B the resulting matrix. Thereby (B)jj = (A)jj + (A)ij + (A)ii + (A)ji = 2 · Aij 6= 0. B is congruent to A by Lemma 1.5 (ii). So we can apply the above second case and in addition use Lemma 1.5 (iii). 2 Proof of Theorem 1.6: Assume that an n × n-matrix A is symmetric, we need to show that A is congruent to a diagonal matrix. The case for n = 1 is trivial. So let n > 2. By Lemma 1.7 and Lemma 1.5 we can assume without loss of generality 0 1 a11 a12 ··· a1n Ba21 a22 ··· a2n C A = B C with a 6= 0. B . .. C 11 @ . A an1 an2 ··· ann We multiply the first row by -a21=a11 and add it to the second row, then multiply the first column by -a21=a11 = -a12=a11 and add it to the second column. This yields a matrix of the form 0 1 a11 0 ··· a1n B 0 - ··· - C B C . B . .. C @ . A an1 - ··· ann Then we do the same to the third row and the third column, and so on. Eventually, we multiply the first row by -an1=a11 and add it to the n-th row, then multiply the first column by -an1=a11 = -a1n=a11 and add it to the n-th column. The resulting matrix has the form a 0 11 , 0 B where B is (n - 1) × (n - 1) and symmetric. 1 In fact, PijPij = In. 2 By induction hypothesis, there exists an invertible D such that D0BD is diagonal. Then 1 0 a 0 1 0 a 0 11 = 11 0 D0 0 B 0 D 0 D0BD is diagonal. Since D is invertible, 1 0 1 0 = I . 0 D 0 D-1 n 1 0 a 0 Thus is invertible as well. So by Lemma 1.4, A is congruent to the diagonal 11 . 0 D 0 D0BD 2 1.1. Sylvester’s Law of Inertia. Let A be an n × n-matrix over R. By Theorem 1.6 there exists an invertible n × n-matrix C over R such that C0AC is diagonal. Without loss of generality, we can assume that for some d1,..., dr 6= 0 0 1 d1 ··· 0 S 0 C0AC = with S = B . .. C . 0 0 @ . A 0 ··· dr We can further assume that for some p 2 [r] d1,..., dp > 0 and dp+1,..., dr < 0. In terms of quadratic form, 0 0 2 2 y (C AC)y = d1x1 + ··· drxr. By further taking p p z1 := d1y1, ··· , zp := dpyp, p p zp+1 := -dp+1y1, ··· , zr := -dryr, and zr+1 := yr+1, ··· , zr := yn, we obtain a very simple quadratic from 2 2 2 2 z1 + ··· + zp - zp+1 - ··· - zr. The question arises whether the numbers r and p are uniquely determined by the original matrix A. Theorem 1.8 (Sylvester’s Law of Inertia). Assume that an n × n symmetric matrix A over R is congruent to two diagonal matrices 0 1 0 1 Ip1 0 0 Ip2 0 0 A1 = @ 0 -Ir1-p1 0A and A1 = @ 0 -Ir2-p2 0A . 0 0 0 0 0 0 Then r1 = r2 = rank(A) and p1 = p2. We give a proof which is very different from the one in the textbook, which might offer more insights. To that end, some preparations are in order. Lemma 1.9. Let A be an m × n-matrix. Then for every invertible m × m-matrix B rank(A) = rank(BA). Similarly, for every invertible n × n-matrix C, we have rank(A) = rank(AC). 3 Proof: By the Rank-Nullity Theorem n = Rank(A) + dim N(A), and n = Rank(A) + dim N(BA). n Thus it suffices to show N(A) = N(BA), i.e., for every x 2 K , Ax = 0 () BAx = 0. The direction from left to right is trivial. For the converse, assume that Ax = b 6= 0 and Bb = 0. But this means that the column vectors of B are linearly dependent, contradicting that B is invertible, or equivalently, column rank(B) = rank(B) = n. The second property can be proved similarly by observing the row vectors of C. 2 Corollary 1.10. Any two congruent matrices have the same rank. Proof: Assume A is congruent to B, i.e., there exists an invertible C with B = C0AC. Then Lemma 1.9 implies that rank(B) = rank(C0AC) = rank(AC) = rank(A). 2 The proofs of the next two lemmas are rather routine from the definitions. Lemma 1.11. Let A be an n × n-matrix of K. Then the mapping x 7! Ax n n is an isomorphism from K to K if and only if A is invertible. Lemma 1.12. Let ' : V ! V be an isomorphism (i.e., an automorphism of V). Furthermore, let W be a subspace of V. (i) '(W) is a subspace of V. (ii) If W is finite-dimensional, then dim(W) = dim '(W). Proof of Lemma 1.8: Corollary 1.10 implies r1 = rank(A1) = rank(A) = rank(A2) = r2. Next, we show that p1 = p2. Observe that by Lemma 1.4, the two matrices A1 and A2 are congruent. Hence, there is an invertible n × n-matrix C such that 0 A1 = C A2C. (1) For i 2 [2] let 0 W := x ··· x 0 ··· 0 x ,..., x 2 ⊆ n. i 1 pi 1 pi R R In particular, for any x 2 Wi n f0g, 0 x Aix > 0, and furthermore dim(Wi) = pi. Therefore, we aim to show dim(W1) = dim(W2). 4 Consider two further subspaces 0 N := 0 ··· 0 x ··· x x ,..., x 2 ⊆ n. i pi+1 n pi+1 n R R Then for every x 2 N1 we have 0 x A1x 6 0, (2) n n and additionally dim(N1) = n - p1. Let ' : R ! R be defined by '(x) := C-1x. n By Lemma 1.11 ' is an isomorphism, and then Lemma 1.12 implies '(W2) is a subspace of R with dim '(W2) = dim(W2) = p2.