Evaluation Maps and Interpolation Isomorphisms

From a Polynomial to Its Values

A polynomial can be represented by its coefficients, but it can also be identified by its values at selected points. If the points are chosen suitably, these two descriptions contain exactly the same information: every permitted list of values comes from one polynomial, and no two different permitted polynomials give the same list. This correspondence is the evaluation isomorphism, and interpolation gives a direct way to reverse it.

Let \(n\in\mathbb{N}\). Write \(\mathcal{P}_{n-1}(\mathbb{R})\) for the real vector space of polynomials of degree at most \(n-1\), including the zero polynomial. Choose distinct real numbers \(a_1,\dots,a_n\). For \(p\in\mathcal{P}_{n-1}(\mathbb{R})\), its values at these points form a column vector in \(\mathbb{R}^n\).

The map is linear because evaluating a sum or a scalar multiple gives the corresponding sum or scalar multiple of the values. Explicitly, for \(p,q\in\mathcal{P}_{n-1}(\mathbb{R})\) and \(c\in\mathbb{R}\), \(E(p+q)=E(p)+E(q)\) and \(E(cp)=cE(p)\). The key question is whether every vector of values is attained, and whether it is attained only once.

Why Distinct Values Determine a Polynomial

We first establish the root-counting fact that ensures uniqueness. A root of \(p\) means a real number \(r\) such that \(p(r)=0\).

Proof. We use induction on \(d\). If \(d=0\), the polynomial is a nonzero constant, so it has no roots. Now suppose \(d\geq1\) and the claim holds for degree \(d-1\). If \(p\) has no roots, the claim already holds. Otherwise let \(r\) be a root. For each integer \(k\geq1\), the identity \(x^k-r^k=(x-r)(x^{k-1}+x^{k-2}r+\cdots+r^{k-1})\) shows that \(x^k-r^k\) is divisible by \(x-r\). Applying this identity to each term of \(p(x)-p(r)\), and using \(p(r)=0\), gives \(p(x)=(x-r)q(x)\) for a polynomial \(q\) of degree \(d-1\). Every root \(s\ne r\) of \(p\) satisfies \(0=p(s)=(s-r)q(s)\), so \(q(s)=0\). By the induction hypothesis, \(q\) has at most \(d-1\) distinct roots. Including \(r\), \(p\) has at most \(d\) distinct roots. \(\square\)

In particular, a polynomial in \(\mathcal{P}_{n-1}(\mathbb{R})\) that vanishes at all \(n\) distinct nodes must be the zero polynomial. The degree restriction matters: without it, a nonzero polynomial such as \((x-a_1)\cdots(x-a_n)\) vanishes at all \(n\) nodes.

Constructing the Interpolation Polynomials

For each node \(a_i\), we want a polynomial that equals \(1\) at \(a_i\) and \(0\) at every other node. The distinctness assumption ensures that the denominators in the following definition are nonzero.

Each \(L_i\) has degree at most \(n-1\). At \(x=a_i\), every numerator equals its denominator, so \(L_i(a_i)=1\). If \(k\ne i\), the factor with \(j=k\) has numerator \(a_k-a_k=0\), so \(L_i(a_k)=0\). Thus \(L_i(a_k)\) equals \(1\) when \(i=k\), and \(0\) otherwise.

Proof. Let \(p(x)=\sum_{i=1}^{n}y_iL_i(x)\). Each summand has degree at most \(n-1\), so \(p\in\mathcal{P}_{n-1}(\mathbb{R})\). For each \(k\), the values of the Lagrange polynomials give \(p(a_k)=\sum_{i=1}^{n}y_iL_i(a_k)=y_k\). This proves existence and shows that every vector in \(\mathbb{R}^n\) is attained by \(E\). To prove uniqueness, suppose \(q\in\mathcal{P}_{n-1}(\mathbb{R})\) has the same values as \(p\) at every node. Then \(q-p\) has degree at most \(n-1\) and vanishes at \(n\) distinct nodes. The lemma implies \(q-p\) is the zero polynomial, so \(q=p\). Thus \(E\) is bijective. Its linearity was verified above, so it is an isomorphism. Finally, the displayed formula sends any value vector \((y_1,\dots,y_n)^{\top}\) to its unique polynomial; it therefore gives \(E^{-1}\). \(\square\)

The formula describes both directions of the correspondence. Evaluation sends a polynomial to its values. The inverse map takes a value vector and forms the weighted sum of the Lagrange polynomials. Because \(L_i\) has value \(1\) only at node \(a_i\), the coefficient \(y_i\) directly sets the prescribed value at that node without changing the values at the others.

Using the Inverse Map

Connection with Coefficient Coordinates

In the standard coefficient basis \((1,x,\dots,x^{n-1})\), a polynomial \(p(x)=c_0+c_1x+\cdots+c_{n-1}x^{n-1}\) has coefficient column \((c_0,\dots,c_{n-1})^{\top}\). Evaluation at \(a_1,\dots,a_n\) is represented in these coordinates by the matrix whose \(i\)th row is \((1,a_i,\dots,a_i^{n-1})\). This matrix sends coefficients to values. The interpolation formula constructs its inverse in coefficient coordinates: the \(i\)th column of that inverse consists of the coefficients of \(L_i\). This explains why interpolation is a concrete inverse map, not only an existence argument.

The proof also identifies the role of each hypothesis. The degree bound limits the polynomials to a space that cannot contain a nonzero polynomial vanishing at all the nodes. Distinctness ensures each denominator \(a_i-a_j\) is nonzero and makes each Lagrange polynomial take the required values. With either condition absent, the conclusion need not hold.

Common Pitfall: Repeated Nodes

A tempting mistake is to apply the Lagrange formula when two nodes coincide. For nodes \(a_1=a_2=1\), the denominator for \(L_1\) includes \(a_1-a_2=0\), so the proposed expression requires division by zero and is not defined. There is also a direct reason evaluation cannot be an isomorphism in this case: every evaluated vector has the form \((p(1),p(1))^{\top}\). For example, \((0,1)^{\top}\) cannot be obtained from any polynomial. The correction is to check distinctness before using the formula; this theorem concerns distinct nodes, not repeated ones.

A second useful check is to distinguish the input and output of the inverse. The input to \(E^{-1}\) is a column of values, not a column of coefficients. The Lagrange formula first uses those values as weights, then produces a polynomial. Substituting the resulting polynomial into each node is the final check that the values have been placed in the intended order.