Make indistinguishable vectors into one vector

Try

Suppose two measurements in R² are regarded as equivalent when both coordinates are shifted by the same amount. Then (2,7), (5,10), and (−1,4) carry the same difference of five. The invisible direction is U=span((1,1)). A quotient vector is the entire class v+U, not a selected point on its line. Draw three parallel lines y−x=c and label them by c. The quotient records which line contains the measurement.

Predict

Predict whether the first coordinate x alone can label a quotient vector. It cannot: adding (1,1) changes x while leaving the class unchanged. The difference y−x can label it because the common shift cancels. This is the basic well-definedness test: a rule acting on equivalence classes must give the same answer for every representative. A formula may look linear in coordinates yet fail this prior requirement.

Experiment

Use representatives v=(2,7) and w=(4,1). Their classes have coordinates 5 and −3. Adding representatives gives (6,8), whose difference is 2, the sum of the class coordinates. Replace v by (12,17) and repeat: the resulting representative (16,18) still has difference 2. Make this comparison for scalar multiplication by −2 as well. Your calculation should display both the changed representative and the unchanged class label, rather than merely announcing that the operation works.

Understand

For a subspace U of V, define v∼w when v−w∈U. Reflexivity, symmetry and transitivity follow from zero, additive inverses and addition in U. Define (v+U)+(w+U)=(v+w)+U and a(v+U)=av+U. If representatives change by elements of U, so do their sums and scalar multiples, establishing well-definedness. The vector-space axioms then descend from V. Extend a basis of U by v₁,…,vᵣ to a basis of V. The cosets of these added vectors span V/U; a relation among them would put their combination in U and contradict the full basis independence. Thus dim(V/U)=dim V−dim U.

Apply

For a linear map T, its kernel describes exactly which input differences are invisible. Define the induced map from V/ker T to im T by sending v+ker T to T(v). It is well-defined, linear and onto. If its output is zero, v lies in the kernel and its class is the zero class, so it is one-to-one. This proves the first isomorphism theorem. Apply it to differentiation on P₃: polynomials differing by a constant give the same derivative, and the quotient by constants is isomorphic to P₂. Find the class of one antiderivative of 3t²−2; every other antiderivative must belong to that same class.

For a concrete quotient basis in P₂ modulo constants, use the classes of t and t². Their linear combination is the zero class exactly when at+bt² is constant, forcing a=b=0. They span because the constant term disappears from every polynomial. This check distinguishes the zero coset, which contains many representatives, from the zero polynomial, which is only one representative.

Recover every polynomial compatible with the observations

Over the real numbers, sample a cubic at 0, 1 and 2 with target values 1, 0 and 1. The quadratic p₀(x)=(x−1)² fits. So does every polynomial p₀(x)+c x(x−1)(x−2), because the added term vanishes at each observed node. The unknown constant c is not an interpolation error: it describes information that the three measurements cannot see. We will prove this is the whole solution family, not merely a way to find some solutions.

Let F be a field, let x₁ through xᵣ be distinct field elements, and let P_d(F) be formal polynomials of degree at most d, where d≥r−1. The evaluation map E sends p to its list of r values. It is linear. For each i, the Lagrange polynomial

is defined because every denominator is nonzero and hence invertible in the field. It equals one at xᵢ and zero at every other selected node. Therefore p₀=Σ yᵢℓᵢ has the requested values and degree at most r−1, proving surjectivity of E. This proof works in positive characteristic; it does not divide by an integer unless that integer has been checked nonzero in the field.

Set H(x)=product of (x−xᵢ) over all distinct nodes. A polynomial vanishing at every node is divisible by H. To see this without assuming the conclusion, divide by the first linear factor using the factor theorem. At any remaining node the removed factor is nonzero, so the quotient still vanishes there. Repeat through all nodes. Conversely every multiple of H vanishes at them. Hence ker E consists exactly of H times polynomials of degree at most d−r. When d=r−1 this means only the zero polynomial; when d≥r, its basis is H, xH, through x^(d−r)H, giving dimension d+1−r. Independence follows because a nonzero polynomial times H cannot be zero in a polynomial ring over a field.

Two polynomials have the same observation vector precisely when their difference lies in this kernel. Every fiber is therefore the affine set p₀+ker E. If the target vector is nonzero it is not itself a vector subspace, since it omits the zero polynomial. The induced map from P_d(F)/ker E to F^r is an isomorphism: the quotient forgets exactly the invisible polynomial changes. For a finite field of q elements, each nonempty fiber contains q^(d+1−r) polynomials. Counting interpolants is not counting possible data vectors, which number q^r.

Repeated nodes and finite fields change what is identifiable

If listed nodes repeat, equal field inputs must have equal target values. Conflicting targets give an empty fiber. Otherwise merge equal nodes, apply the construction to the r distinct inputs, and retain only r independent measurements. In characteristic two, the requested integers 0,1,2 reduce to 0,1,0. Their distinct appearance as integers does not justify dividing by their difference after reduction. Duplicate observations do not add derivative conditions; Hermite interpolation would require separately specified derivatives.

On a field F_q, evaluating at all q elements identifies each polynomial of degree less than q uniquely, but not arbitrary formal polynomials. The kernel on the full polynomial ring is the ideal generated by H_q=product over a∈F_q of (x−a). Each field element satisfies a^q=a: for nonzero a, multiplication by a permutes the nonzero elements, and cancellation of their product gives a^(q−1)=1. Thus the monic polynomials H_q and x^q−x have the same q roots and degree, so they are equal. Formal polynomials represent the same function exactly when their difference is divisible by x^q−x.

Every function from F_q to F_q has a unique representative of degree less than q by Lagrange interpolation. There are q^q such functions and q^q such representatives. Higher-degree polynomials can be distinct while agreeing at every field element. The inspector’s F₃ example exhibits this: changing the coefficient of x³−x changes the polynomial but leaves its entire field-value table fixed. Over F₅, the same three observed nodes leave additional unobserved function values free.

For prerequisite formulas see polynomial interpolation and polynomial division. The present calculation concerns exact fibers over a stated field, not floating-point interpolation stability or a fitted-data model.

Continue

Quotients describe information loss without choosing a preferred representative. A complement can identify a quotient with an ordinary subspace, but that identification depends on the chosen complement. The next lesson studies scalar-valued measurements, called linear functionals, which help express which information survives. Do not transfer a geometric angle or length to a quotient automatically: a vector-space quotient has addition and scaling, while a norm or inner product needs an additional compatible construction.

Review Linear Algebra I when a prerequisite needs practice. Continue to Describe spaces through linear measurements.