Two vectors in, one scalar out
Let V be a finite-dimensional vector space over a field K. A bilinear form is a map B : V × V → K that is linear in each argument separately: B(au + a'u', w) = a·B(u, w) + a'·B(u', w), and the same on the right. It is a special bilinear map — the codomain is the ground field itself. The dot product on ℝⁿ is the prototype: B(x, y) = x₁y₁ + ⋯ + xₙyₙ. But there are many others, and the whole point of this track is to classify them.
Choose a basis e₁, …, eₙ of V. Because B is bilinear, it is completely determined by the n² numbers aᵢⱼ = B(eᵢ, eⱼ). Collect them into a matrix A = (aᵢⱼ), the Gram matrix of B in this basis. Then for column vectors x, y of coordinates, B(x, y) = xᵀ A y. The form and its matrix are interchangeable once a basis is fixed — but the matrix depends on the basis, and that dependence is the heart of the story.
Symmetry and the antisymmetric companion
A form is symmetric if B(u, w) = B(w, u) for all u, w — equivalently, its Gram matrix satisfies Aᵀ = A. It is alternating if B(v, v) = 0 for all v, which forces B(u, w) = −B(w, u). Away from characteristic 2 every form splits uniquely as symmetric plus alternating: B = ½(B + Bᵀ) + ½(B − Bᵀ). This track is about the symmetric half; the alternating half is the world of the alternating form and symplectic geometry. Symmetric is where squares and signatures live.
Now the crucial move: how does the Gram matrix change when we change basis? If the new basis is e'ⱼ = Σᵢ Pᵢⱼ eᵢ with P the (invertible) change-of-basis matrix, then the new coordinates relate by x = P x'. Substituting into B(x, y) = xᵀ A y gives B = (Px')ᵀ A (Py') = x'ᵀ (Pᵀ A P) y'. So the new Gram matrix is A' = Pᵀ A P — not P⁻¹ A P. This is *congruence*, not similarity.
Same form, two bases, two Gram matrices.
Let B(x, y) = x1*y1 + x1*y2 + x2*y1 + 2*x2*y2 on R^2.
In the standard basis its Gram matrix is
A = [1, 1; 1, 2] (symmetric: A^T = A)
Change basis with P = [1, -1; 0, 1] (so e1' = e1, e2' = -e1 + e2).
Congruence:
A' = P^T A P
= [1, 0; -1, 1] [1, 1; 1, 2] [1, -1; 0, 1]
= [1, 0; -1, 1] [1, 0; 1, 1]
= [1, 0; 0, 1]
The SAME form is now the plain dot product. We "completed the square":
B = (x1 + x2)^2 + x2^2 in the new coordinates.
Note det A = 1 and det A' = 1: congruence multiplies det by (det P)^2 = 1.
Similarity P^{-1} A P would have given [3, 1; -2, 0] -- NOT symmetric.Congruence is the equivalence we care about
Two symmetric matrices A and A' are congruent if A' = PᵀAP for some invertible P. This is exactly the relation “same form, different basis”. So classifying symmetric bilinear forms up to change of basis means classifying symmetric matrices up to congruence. Notice congruence cannot change the rank (P is invertible) and multiplies the determinant by a nonzero square (det P)². Over the reals it also cannot change the sign of the determinant — a first hint of the signature waiting in Guide 3.