T. S. Blyth & E. F. Robertson · First-year university · Study material generated by AI from this edition. Not affiliated with the publisher.
Chapter 8 · pp. 109–117
This chapter turns matrices into machines that take two vectors and return a scalar. We see how such a form changes when the basis changes, and how a real quadratic form can always be reduced to a sum of plus and minus squares. The counts of those squares, the rank and the signature, tell us when a form is positive definite and how to classify critical points in calculus.
An matrix over a field defines a bilinear form on an -dimensional space by , where and are coordinate columns. Expanding gives a sum of terms . This is linear in when is fixed, and linear in when is fixed. The entry equals , so relative to the chosen basis the matrix of the form is itself. For on we get the dot product . (p. 110)
Each entry contributes its own term. For a symmetric real matrix with diagonal and off-diagonal entries (positions 1,2), (positions 1,3), (positions 2,3), the form is
Symmetry of the matrix means and share one coefficient. If the matrix is not symmetric, the form need not be. A matrix with a single 1 in position (1,2) gives just , which differs from . (p. 110)
Suppose is the matrix of relative to the basis , and is the transition matrix to a new basis , so . Bilinearity gives
which is the entry of . So the new matrix is . For example, with and , we get . (pp. 110–111)
A frequent mistake is to write , the rule for a linear map. A bilinear form takes two vectors, so the transition matrix acts on both arguments, which gives on the left and on the right. The two rules agree only when is orthogonal, since then . (pp. 110–111)
This motivates a definition: is congruent to if for some invertible . Congruence is an equivalence relation. It is reflexive (take ), symmetric (use ) and transitive (multiply the matrices ). Matrices that represent the same bilinear form in different bases are exactly the congruent matrices. (p. 111)
A bilinear form is symmetric if for all . This happens exactly when its matrix is symmetric. The dot product is an example. (p. 111)
A quadratic form on is a map for which some symmetric bilinear form satisfies for all . In other words, we feed the same vector into both slots. For example, equals with . Similarly, has matrix . (pp. 111–112)
The theory is developed over . Most applications are real, and in fields of characteristic 2 (where ) we cannot divide by 2, which the next formulas need. (p. 112)
Let be symmetric on a real space and . Then
The second follows from expanding , which uses symmetry. Since , we have . Subtracting this from leaves , which is the third identity. (p. 112)
A consequence is that every real quadratic form comes from exactly one symmetric bilinear form. If and both give , the formula expresses each of them by the same quantity, so . That is why we can speak of the matrix of a quadratic form. (p. 112)
To find that matrix, put each squared-term coefficient on the diagonal. Split each cross-term coefficient of equally between positions and . For the matrix is , and the bilinear form is . (p. 113)
Two symmetric matrices represent the same quadratic form in different ordered bases if and only if they are congruent, by the rule . So the aim is to find a simple representative of each congruence class. (p. 113)
Every real symmetric matrix is congruent to a unique matrix with ones, minus ones and the remaining entries zero. (pp. 113–114)
Existence. Diagonalize orthogonally, so with all . Then scale with a diagonal matrix that has entries in the nonzero places and 1 elsewhere. This gives , and is invertible. (pp. 113–114)
Uniqueness. Suppose and are congruent and . Congruent matrices have equal rank, so , which forces . Let be the -dimensional subspace on which for every nonzero . Let be the -dimensional subspace on which . Let be the image of under the isomorphism given by , so . Then , so and share a nonzero vector. That vector would give a form value both negative and non-negative, which is impossible. Hence and . (pp. 114–115)
Sylvester's theorem restates this for quadratic forms. On an -dimensional real space there is a basis in which
and and do not depend on the basis. Different bases can give different-looking formulas, but always the same number of positive and negative squares. (p. 115)
Because and are fixed, we can define the rank as (the number of nonzero squares) and the signature as . For we have , , so the rank is 3 and the signature is 1. (p. 115)
Completing the squares. Group the terms variable by variable and write them as squares of linear expressions. For :
The signs are , so the rank is 3 and the signature is 1. (pp. 115–116)
If no squared term exists, as in , first substitute , , and then complete squares. The result has one positive and one negative square, for instance in the original variables. So , the rank is 2 and the signature is 0. (p. 116)
Using matrices. Find an orthogonal with diagonal. Put , so . Then
The diagonal entries of are the eigenvalues of . The number of positive eigenvalues gives and the number of negative ones gives . Rescaling then gives unit coefficients, for example when the eigenvalue signs are . (p. 115)
A quadratic form is positive definite if for every nonzero . For this says the matrix is positive definite. In canonical form, all squares are present and all have plus signs, so and . For example is positive definite on , but is not, because . (p. 116)
Do not assume that rank equal to signature is enough. That condition only says , so . We also need full rank . The form on has rank 1 and signature 1, yet it vanishes at . (p. 116)
After orthogonal diagonalization, with the the eigenvalues. This is positive for all nonzero vectors exactly when every . For the eigenvalues are 3, 3, 6, so the form becomes and is positive definite. (pp. 116–117)
At a point where both first partial derivatives of vanish, the leading change in comes from the quadratic form . Its normal form decides the geometry:
