Linear Algebra II. Notes 6 25th November 2010

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1 MTH6140 Liear Algebra II Notes 6 25th November Quadratic forms A lot of applicatios of mathematics ivolve dealig with quadratic forms: you meet them i statistics (aalysis of variace) ad mechaics (eergy of rotatig bodies), amog other places. I this sectio we begi the study of quadratic forms. 6.1 Quadratic forms For almost everythig i this chapter, we assume that the characteristic of the field K is ot equal to 2. This meas that 2 0 i K, so that the elemet 1/2 exists i K. Of our list of stadard fields, this oly excludes F 2, the itegers mod 2. (For example, i F 5, we have 1/2 = 3.) A quadratic form is a fuctio which, whe writte out i coordiates, is a polyomial i which every term has total degree 2 i the variables. For example, Q(x,y,z) = x 2 + 4xy + 2xz 3y 2 2yz z 2 is a quadratic form i three variables. We will meet a formal defiitio of a quadratic form later i the chapter, but for the momet we take the followig. Defiitio 6.1 A quadratic form i variables x 1,...,x over a field K is a polyomial i=1 A i j x i x j j=1 i the variables i which every term has degree two (that is, is a multiple of x i x j for some i, j), ad each A i j belogs to K. 1

2 I the above represetatio of a quadratic form, we see that if i j, the the term i x i x j comes twice, so that the coefficiet of x i x j is A i j + A ji. We are free to choose ay two values for A i j ad A ji as log as they have the right sum; but we will always make the choice so that the two values are equal. That is, to obtai a term cx i x j, we take A i j = A ji = c/2. (This is why we require that the characteristic of the field is ot 2.) Ay quadratic form is thus represeted by a symmetric matrix A with (i, j) etry A i j (that is, a matrix satisfyig A = A ). This is the third job of matrices i liear algebra: Symmetric matrices represet quadratic forms. We thik of a quadratic form as defied above as beig a fuctio from the vector space K to the field K. It is clear from the defiitio that x 1 Q(x 1,...,x ) = v Av, where v =.. x Now if we chage the basis for V, we obtai a differet represetatio for the same fuctio Q. The effect of a chage of basis is a liear substitutio v = Pv o the variables, where P is the trasitio matrix betwee the bases. Thus we have so we have the followig: v Av = (Pv ) A(Pv ) = (v ) (P AP)v, Propositio 6.1 A basis chage with trasitio matrix P replaces the symmetric matrix A represetig a quadratic form by the matrix P AP. As for other situatios where matrices represeted objects o vector spaces, we make a defiitio: Defiitio 6.2 Two symmetric matrices A,A over a field K are cogruet if A = P AP for some ivertible matrix P. Propositio 6.2 Two symmetric matrices are cogruet if ad oly if they represet the same quadratic form with respect to differet bases. Our ext job, as you may expect, is to fid a caoical form for symmetric matrices uder cogruece; that is, a choice of basis so that a quadratic form has a particularly simple shape. We will see that the aswer to this questio depeds o the field over which we work. We will solve this problem for the fields of real ad complex umbers. 2

3 6.2 Reductio of quadratic forms Eve if we caot fid a caoical form for quadratic forms, we ca simplify them very greatly. Theorem 6.3 Let Q be a quadratic form i variables x 1,...,x, over a field K whose characteristic is ot 2. The by a suitable liear substitutio to ew variables y 1,...,y, we ca obtai Q = α 1 y α 2 y α y 2 for some α 1,...,α K. Proof Our proof is by iductio o. We call a quadratic form which is writte as i the coclusio of the theorem diagoal. A form i oe variable is certaily diagoal, so the iductio starts. Now assume that the theorem is true for forms i 1 variables. Take where A i j = A ji for i j. Q(x 1,...,x ) = i=1 A i j x i x j, j=1 Case 1: Assume that A ii 0 for some i. By a permutatio of the variables (which is certaily a liear substitutio), we ca assume that A Let The we have y 1 = x 1 + A 11 y 2 1 = A 11 x i=2 i=2 (A 1i /A 11 )x i. A 1i x 1 x i + Q (x 2,...,x ), where Q is a quadratic form i x 2,...,x. That is, all the terms ivolvig x 1 i Q have bee icorporated ito A 11 y 2 1. So we have Q(x 1,...,x ) = A 11 y Q (x 2,...,x ), where Q is the part of Q ot cotaiig x 1 mius Q. By iductio, there is a chage of variable so that ad so we are doe (takig α 1 = A 11 ). Q (x 2,...,x ) = 3 i=2 α i y 2 i,

4 Case 2: All A ii are zero, but A i j 0 for some i j. Now x i x j = 1 4 ( (xi + x j ) 2 (x i x j ) 2), so takig x i = 1 2 (x i + x j ) ad x j = 1 2 (x i x j ), we obtai a ew form for Q which does cotai a o-zero diagoal term. Now we apply the method of Case 1. Case 3: All A i j are zero. Now Q is the zero form, ad there is othig to prove: take α 1 = = α = 0. Example 6.1 Cosider the quadratic form Q(x,y,z) = x 2 + 2xy + 4xz + y 2 + 4z 2. We have (x + y + 2z) 2 = x 2 + 2xy + 4xz + y 2 + 4z 2 + 4yz, ad so Q = (x + y + 2z) 2 4yz = (x + y + 2z) 2 (y + z) 2 + (y z) 2 = X 2 +Y 2 Z 2, where X = x + y + 2z, Y = y z, Z = y + z. Otherwise said, the matrix represetig the quadratic form, amely A = is cogruet to the diagoal matrix A = How do we fid a ivertible matrix P such that P AP = A? Here is how: If v is the vector cosistig of the origial variables, so v = vector cosistig of ew variables, so v = the argumet o page 2 of this chapter). X Y Z x y, ad v is the z, the P is defied by v = Pv (see 4

5 Now i the curret example we have X Y = Z so P is the iverse of the above matrix, i other words P = x y, z Thus ay quadratic form ca be reduced to the diagoal shape 1. α 1 x α x 2 by a liear substitutio. But this is still ot a caoical form for cogruece. For example, if y 1 = x 1 /c, the α 1 x1 2 = (α 1c 2 )y 2 1. I other words, we ca multiply ay α i by ay factor which is a perfect square i K. Over the complex umbers C, every elemet has a square root. Suppose that α 1,...,α r 0, ad α r+1 = = α = 0. Puttig { ( αi )x y i = i for 1 i r, x i for r + 1 i, we have Q = y y 2 r. We will see later that r is a ivariat of Q: however we do the reductio, we arrive at the same value of r. Over the real umbers R, thigs are ot much worse. Sice ay positive real umber has a square root, we may suppose that α 1,...,α s > 0, α s+1,...,α s+t < 0, ad α s+t+1,...,α = 0. Now puttig we get y i = { ( αi )x i for 1 i s, ( α i )x i for s + 1 i s +t, x i for s +t + 1 i, Q = x x 2 s x 2 s+1 x 2 s+t. Agai, we will see later that s ad t do t deped o how we do the reductio. [This is the theorem kow as Sylvester s Law of Iertia.] 5

6 6.3 Liear forms ad dual space Now we begi dealig with quadratic forms i a more abstract way. We begi with liear forms, that is, fuctios of degree 1. The defiitio is simple: Defiitio 6.3 Let V be a vector space over K. A liear form o V is a liear map from V to K, where K is regarded as a 1-dimesioal vector space over K: that is, it is a fuctio from V to K satisfyig f (v 1 + v 2 ) = f (v 1 ) + f (v 2 ), f (cv) = c f (v) for all v 1,v 2,v V ad c K. If dim(v ) =, the a liear form is represeted by a 1 matrix over K, that is, a row vector of legth over K. If f = [a 1 a 2... a ] represets a liear form, the for v = [x 1 x 2... x ] we have x 1 f (v) = [a 1 a 2... x a ] 2. = a 1x 1 + a 2 x a x. x Coversely, ay row vector of legth represets a liear form o K. Defiitio 6.4 Liear forms ca be added ad multiplied by scalars i the obvious way: ( f 1 + f 2 )(v) = f 1 (v) + f 2 (v), (c f )(v) = c f (v). So they form a vector space, which is called the dual space of V ad is deoted by V. Not surprisigly, we have: Propositio 6.4 If V is fiite-dimesioal, the so is V, ad dim(v ) = dim(v ). Proof We begi by observig that, if (v 1,...,v ) is a basis for V, ad a 1,...,a are ay scalars whatsoever, the there is a uique liear map f with the property that f (v i ) = a i for i = 1,...,. It is give by f (c 1 v c v ) = a 1 c a c, i other words, it is represeted by the row vector [a 1 a 2... a ], ad its actio o K is by matrix multiplicatio as we saw earlier. 6

7 Now let f i be the liear map defied by the rule that { 1 if i = j, f i (v j ) = 0 if i j. The ( f 1,..., f ) form a basis for V ; ideed, the liear form f defied i the precedig paragraph is a 1 f a f. This basis is called the dual basis of V correspodig to the give basis for V. Sice it has elemets, we see that dim(v ) = = dim(v ). We ca describe the basis i the precedig proof as follows. Defiitio 6.5 The Kroecker delta δ i j for i, j {1,...,} is defied by the rule that { 1 if i = j, δ i j = 0 if i j. Note that δ i j is the (i, j) etry of the idetity matrix. Now, if (v 1,...,v ) is a basis for V, the the dual basis for the dual space V is the basis ( f 1,..., f ) satisfyig f i (v j ) = δ i j. There are some simple properties of the Kroecker delta with respect to summatio. For example, δ i j a i = a j i=1 for fixed j {1,...,}. This is because all terms of the sum except the term i = j are zero. 6.4 Chage of basis Suppose that we chage bases i V from B = (v 1,...,v ) to B = (v 1,...,v ), with trasitio matrix P = P B,B. How do the dual bases chage? I other words, if B = ( f 1,..., f ) is the dual basis of B, ad (B ) = ( f 1,..., f ) the dual basis of B, the what is the trasitio matrix P B,(B )? The ext result aswers the questio. Propositio 6.5 Let B ad B be bases for V, ad B ad (B ) the dual bases of the dual space. The P B,(B ) = (P B,B ) 1. 7

8 Proof Use the otatio from just before the statemet of this Propositio. If P = P B,B has (i, j) etry p i j, ad Q = P B,(B ) has (i, j) etry q i j, we have ad so v i = f j = k=1 l=1 p ki v k, q l j f l, δ i j = f j(v i) = ( )( ) q l j f l p ki v i l=1 k=1 = = l=1 k=1 q k j p ki. k=1 q l j δ i j p ki Now q k j is the ( j,k) etry of Q, ad so we have I = Q P, whece Q = P 1, so that Q = ( P 1) = ( P ) 1, as required. 6.5 Quadratic ad biliear forms The formal defiitio of a quadratic form looks a bit differet from the versio we gave earlier, though it amouts to the same thig. First we defie a biliear form. Defiitio 6.6 (a) Let b : V V K be a fuctio of two variables from V with values i K. We say that b is a biliear form if it is a liear fuctio of each variable whe the other is kept costat: that is, b(v,w 1 + w 2 ) = b(v,w 1 ) + b(v,w 2 ), b(v,cw) = cb(v,w), with two similar equatios ivolvig the first variable, amely b(v 1 + v 2,w) = b(v 1,w) + b(v 2,w), b(cv,w) = cb(v,w). A biliear form b is symmetric if b(v,w) = b(w,v) for all v,w V. 8

9 (b) Let Q : V K be a fuctio. We say that Q is a quadratic form if (i) Q(cv) = c 2 Q(v) for all c K, v V, ad (ii) the fuctio b defied by b(v,w) = Q(v + w) Q(v) Q(w) is a biliear form o V. (Note that the biliear form b is symmetric!) If we thik of the prototype of a quadratic form as beig the fuctio x 2, the the first equatio says (cx) 2 = c 2 x 2, while the secod has the form (x + y) 2 x 2 y 2 = 2xy, ad 2xy is the prototype of a biliear form: it is a liear fuctio of x whe y is costat, ad vice versa. Note that the formula b(x,y) = Q(x + y) Q(x) Q(y) (which is kow as the polarisatio formula) says that the biliear form is determied by the quadratic form Q. Coversely, if we kow the symmetric biliear form b, the we have 2Q(v) = 4Q(v) 2Q(v) = Q(v + v) Q(v) Q(v) = b(v,v), so that Q(v) = 1 2b(v,v), ad we see that the quadratic form is determied by the symmetric biliear form. So these are equivalet objects. 6.6 Caoical forms for complex ad real forms Fially, i this sectio, we retur to quadratic forms (or symmetric matrices) over the real ad complex umbers, ad fid caoical forms uder cogruece. Recall that two symmetric matrices A ad A are cogruet if A = P AP for some ivertible matrix P; as we have see, this is the same as sayig that they represet the same quadratic form relative to differet bases. Theorem 6.6 Ay complex symmetric matrix A is cogruet to a matrix of the form [ Ir ] O O O for some r. Moreover, r = rak(a), ad so if A is cogruet to two matrices of this form the they both have the same value of r. 9

10 Proof We already saw that A is cogruet to a matrix of this form. Moreover, if P is ivertible, the so is P, ad so as claimed. r = rak(p AP) = rak(a) The ext result is Sylvester s Law of Iertia. Theorem 6.7 Ay real symmetric matrix A is cogruet to a matrix of the form I s O O O I t O O O O for some s,t. Moreover, if A is cogruet to two matrices of this form, the they have the same values of s ad of t. Proof Agai we have see that A is cogruet to a matrix of this form. Arguig as i the complex case, we see that s + t = rak(a), ad so ay two matrices of this form cogruet to A have the same values of s + t. Moreover, by restrictig to a subspace o which A is ivertible, we may assume without loss of geerality that s +t =. Suppose that two differet reductios give the values s,t ad s,t respectively, with s +t = s +t =. Suppose (i order to obtai a cotradictio) that s < s. Now let Q be the quadratic form represeted by A. The we are told that there are liear fuctios y 1,...,y ad z 1,...,z of the origial variables x 1,...,x of Q such that Q = y y 2 s y 2 s+1 y 2 = z z 2 s z2 s +1 z2. Now cosider the equatios y 1 = 0,...,y s = 0,z s +1 = 0,...z = 0 regarded as liear equatios i the origial variables x 1,...,x. The umber of equatios is s + ( s ) = (s s) <. Accordig to a lemma from much earlier i the course (we used it i the proof of the Exchage Lemma!), the equatios have a o-zero solutio. That is, there are values of x 1,...,x, ot all zero, such that the variables y 1,...,y s ad z s +1,...,z are all zero. Sice y 1 = = y s = 0, we have for these values Q = y 2 s+1 y 2 < 0. But sice z s +1 = = z = 0, we also have Q = z z 2 s > 0. But this is a cotradictio. So we caot have s < s. Similarly we caot have s < s either. So we must have s = s, as required to be proved. 10

11 We saw that s +t is the rak of A. Defiitio 6.7 The umber s t is kow as the sigature of A. Of course, both the rak ad the sigature are idepedet of how we reduce the matrix (or quadratic form); ad if we kow the rak ad sigature, we ca easily recover s ad t. 11