Bilinear Forms - MIT Mathematics

1 Bilinear FormsEitan 28, 2005We may begin our discussion of Bilinear Forms by looking at a special case that we arealready familiar with. Given a vector spaceVover a fieldF, the dot product between twoelementsXandY(represented as column vectors whose elements are inF) is the mapV V Fdefined by:< X, Y >=XT Y=x1y1+..+xnynThe property of the dot product which we will use to generalize to Bilinear Forms is bilinearity:the dot product is a linear function from V to F if one of the elements is a vector space over a field F. Abilinear formBonVis a function oftwo variablesV V Fwhich satisfies the following axioms:B(v1+v2, w) =B(v1, w) +B(v2, w)(1)B(f v, w) =f B(v, w)(2)B(v, w1+w2) =B(v, w1) +B(v, w2)(3)B(v, f w) =f B(v, w)(4)When working with linear transformations, we represent our transformation by a squarematrixA.

2 Similarly, given a square matrix B, we may define a Bilinear form for allv, w Vas follows:B(v, w) =vT BwThis form satisfies the axioms because of the distributive laws and the ability to pull outa scalar in matrix multiplication. We can also see that given a Bilinear formB, after fixing abasis the matrix representingBis unique. Given a basis{b1, .., bn}ofV, the matrix definedby Bi,j=B(bi, bj) is the unique matrix representing the formB. To see this, we can actuallyBilinear Forms2compute the value of the Bilinear form for arbitraryv, w V. Since{bi}is a basis forV,we havev= ivibiandw= iwibi, wherevi, wi F.

3 ThenB(v, w) =B( ivibi, jvjbj) = i,jviB(bi, bj)wj=vT Bwwhere v and w are represented as column matrices whose elements the matrix of a Bilinear form is dependent on the choice of basis, we might wonderhow this matrix is affected by a change of basis. Given our original basis{bi}and a newbasis{ci}, we can express our new basis vectors as linear combinations of the old ones:cj= ipi,jbiwherePis our invertible change of basis matrix. To compute the entries forthe new matrix B for the Bilinear form , we compute the values of the form on the new basisvectors: B i,j=B(ci, cj) =B( kpk,ibk, lpl,jbl) = k,lpk,iB(bk, bl)pl,j={PT BP}i,j.

4 So thenew matrix for the Bilinear transform is B =PT BPGiven a Bilinear form , we would like to be able to classify it by just a single element ofour field F to be able to read certain properties of the form . The determinant of the matrix Bof our form would be a good mapping to use except for the fact that the matrix of theform, and hence the determinant associated with the form , is dependent on the choice ofbasis. Since we know how the matrix changes when the basis is changed, we can look at howthe determinant changes. If we change bases with associated matrix P, then the new matrixisPT BPso that the new determinant is det(PT BP) = det(PT)det( B)det(P) = det(P)2det( B) (since det(P)=det(PT)).

5 Since any invertible matrix can serve as a change of basismatrix, and we can construct an invertible matrix to have any nonzero determinant we wish(make it diagonal with all 1s except one diagonal element set to be the desired determinant),the set of determinants that we can associate with a given Bilinear form are all multiples ofeachother by a square element of F, and conversely given any square element of F, the setmust be closed under multiplication by this element. So we can define a new quantity, calledthediscriminantof a Bilinear form , to be associated with the set of determinants we canobtain from matrices associated with the Bilinear be more precise, we define the subgroup of square elements in the multiplicativegroup ofF:F 2={f2:f F }We can see that this is a normal subgroup inF because conjugating a square by an element produces a new element which is also a square:f a2f 1= (f af 1)2.

6 So the discriminant is actually a mapping from the set of Bilinear formsonVto the quotient group we produce using the normal subgroup of squares (or possiblythe element 0 which is not in the multiplicative group). The discriminant is defined in sucha way that it is independent of the choice of a Bilinear form on a vector spaceVand Bbe a matrix associated withBilinear Forms3the form . ThediscriminantofBis defined to bediscr(B) = 0if det B= 0det BF 2 F /F call a Bilinear formBnondegenerateif the discriminant ofBis nonzero. To be ableto apply the properties of the discriminant and nondegeneracy, we must first understandorthogonality.

7 Given vectorsv, w Vwe say thatvisorthogonaltow(denotedv w) ifB(v, w) = 0. We would like to describe the vectors in a vector space that are orthogonal toeverything else in that vector space. This set of vectors is referred to as the radical. Sinceorthogonality is not necessarily a commutative relation, we need to be more specific. Givena vector spaceVand a Bilinear formB, we define theleftandright radicalsas follows:radL(V) ={v V:B(v, w) = 0, w V}radR(V) ={v V:B(w, v) = 0, w V}Proposition Bilinear formBon a vector spaceVis nondegenerate radR(V) =0 radL(V) = 0 ProofI will only prove the first equivalence because the proof for the second is similar.

8 Assume there is a nonzero elementXof radR(V) andBis nondegenerate and fix a bases{b1, .., bn} BX6= 0 there exists somew Vsuch thatB(w, X)6= 0 X6 radR(V)which is a contradiction. Assume B not nondegenerate and fix a basis{b1, .., bn} det( B)= 0 there is anontrivial solution to the matrix equationAX= 0 wT BX= 0 for allw V X radR(V) radR(V) has a nonzero in the definition of a radical we include elements ofVthat are orthogonal to allother elements ofV, we may be more specific and seek only elements ofVthat are orthogonalto all elements of some subsetSofV: L(S) ={v V:B(v, w) = 0, w S} R(S) ={v V:B(w, v) = 0, w S}It would be nice if we did not have to deal with all of the distinguishing between left andright orthogonality.

9 A Bilinear formBsuch thatB(v, w) = 0 B(w, v) = 0 for allv, w Vis calledreflexive. Given a reflexive Bilinear form and a subsetSofV, we maywrite L(S) = R(S) =S . IfWis a subspace ofVthen we callW theorthogonalcomplementofW. We define theradicalof a subspaceWofVto beradW=W W andcallWanondegenerate subspaceif radW= Forms4 Proposition a reflexive Bilinear form on a vector spaceV, andWbe a non-degenerate subspace ofV. ThenV=W W .ProofWe know thatW W = 0 becauseWis nondegenerate. So all we need to showis that togetherWandW spanV dimV=dimW+dimW.

10 Letn=dimVandk=dimWand fix a basis{v1, .., vk}ofWand extend to a basis ofVwith{vk+1, .., vn}.Given arbitraryv W , we may writev=c1v1+..+cnvnandv W B(vi, v) =0 1 i k jB(vi, vj)cj= 0 j B(i, j)cj= 0 vis in the null space of ak nmatrix which is the top k rows of the matrix B dimW n k dimW+dimW =dimW dimW n+n k=n W W = Bilinear formBon a vector spaceVis calledsymmetricifB(v, w) =B(w, v)for allw, v VWe can see that the matrix Bof a symmetric Bilinear form must itself be symmetric bytaking standard basis vectorseiandej:B(ei, ej) =B(ej, ei) eTi Bej=eTj Bei Bi,j= Bj,i B= BTDefinitionA Bilinear formBon a vector spaceVis calledalternate(orskew-symmetricifcharF6 = 2) ifB(v, v) = 0 for allw, v VIf we have an alternate formBand take arbitraryv, w Vthen 0 =B(v+w, v+w) =B(v, v) +B(v, w) +B(w, v) +B(w, w) =B(v, w) +B(w, v) B(v, w) = B(w, v).