Two Polynomial Maps with Different Constraints
Evaluation and integration both take polynomials to simpler, explicitly described outputs. Evaluation records the value of a polynomial at one chosen point. Integration from \(0\) produces an antiderivative with a prescribed value at \(0\). Their formulas make it possible to determine exactly which inputs map to zero and which outputs can occur. Those two sets—the kernel and the image—also show directly whether either map is one-to-one or onto.
Throughout, \(n\) is a nonnegative integer and \(\mathbb{R}\) is the scalar field. The space \(P_n\) consists of the real polynomials of degree at most \(n\), including the zero polynomial. Thus each \(p\in P_n\) has a unique expression \(p(x)=a_0+a_1x+\cdots+a_nx^n\). A map is injective if distinct inputs always have distinct outputs, and is surjective if every element of its codomain is the output of some input.
Both maps are linear. For evaluation, \((c p+d r)(a)=c p(a)+d r(a)\) for \(p,r\in P_n\) and \(c,d\in\mathbb{R}\). For integration, linearity of the integral gives \(J(c p+d r)=cJ(p)+dJ(r)\). In coefficients, if \(p(x)=\sum_{k=0}^{n}a_kx^k\), then
This formula also verifies that \(J(p)\) belongs to \(P_{n+1}\) and that \(J(p)(0)=0\). The lower endpoint in the definition is important: it fixes the integration constant rather than allowing an arbitrary one.
Evaluation: Values at a Point
Proof. By definition, \(p\in\ker(\operatorname{ev}_a)\) exactly when \(p(a)=0\). If \(n=0\), then \(p(x)=a_0\) is constant, and \(p(a)=a_0=0\) forces \(p=0\); hence the kernel is \(\{0\}\). Now suppose \(n\geq1\). For each integer \(k\geq1\), the identity \(x^k-a^k=(x-a)(x^{k-1}+x^{k-2}a+\cdots+a^{k-1})\) shows that \(x^k-a^k\) is divisible by \(x-a\). If \(p(x)=\sum_{k=0}^{n}a_kx^k\) and \(p(a)=0\), then \(p(x)=p(x)-p(a)=\sum_{k=1}^{n}a_k(x^k-a^k)\). Each term on the right is divisible by \(x-a\), so \(p(x)=(x-a)r(x)\) for some \(r\in P_{n-1}\). Conversely, if \(p(x)=(x-a)r(x)\), then \(p(a)=0\). This proves the kernel formula in both directions. For the image, take any \(c\in\mathbb{R}\). The constant polynomial \(p(x)=c\) belongs to \(P_n\) and satisfies \(\operatorname{ev}_a(p)=c\), so the image is all of \(\mathbb{R}\). When \(n=0\), the kernel calculation shows that equal outputs require equal constant inputs, so the map is injective. When \(n\geq1\), the nonzero polynomial \(x-a\) belongs to \(P_n\) and maps to zero, just as the zero polynomial does; therefore the map is not injective. \(\square\)
Worked Example: Evaluation on P₃ at 2
Consider \(\operatorname{ev}_2:P_3\to\mathbb{R}\). Write an arbitrary input as \(p(x)=a_0+a_1x+a_2x^2+a_3x^3\). Evaluating gives \(\operatorname{ev}_2(p)=a_0+2a_1+4a_2+8a_3\). Thus \(p\) lies in the kernel precisely when \(a_0=-2a_1-4a_2-8a_3\). Substituting this value into \(p\), and factoring each difference of powers, gives
For a specific check, choose \(a_1=1\), \(a_2=-1\), and \(a_3=2\). The kernel condition gives \(a_0=-2+4-16=-14\), so \(p(x)=2x^3-x^2+x-14\). Its factorization and evaluation are
The factorization checks the kernel description for this input. The map is onto because any desired output \(c\) comes from the constant input \(p(x)=c\). It is not one-to-one: \(p(x)=x-2\) and the zero polynomial are distinct inputs with the same output \(0\).
Worked Example: Evaluation on P₀ at −1
Here \(P_0\) contains only constant polynomials. If \(p(x)=c\), then \(\operatorname{ev}_{-1}(p)=c\). For example, the input \(p(x)=-3\) gives \(\operatorname{ev}_{-1}(p)=-3\), and the constant output is unchanged. The only constant mapping to zero is \(p(x)=0\), so the kernel is \(\{0\}\). Every real output \(c\) is attained by the constant input \(p(x)=c\), so the image is \(\mathbb{R}\). Therefore this evaluation map is both injective and surjective. This edge case differs from evaluation on \(P_n\) when \(n\geq1\), where the nonzero polynomial \(x-a\) is always in the kernel.
Integration: Polynomials That Vanish at Zero
Proof. Write \(p(x)=\sum_{k=0}^{n}a_kx^k\). The coefficient formula for \(J\) gives \(J(p)(x)=\sum_{k=0}^{n}\frac{a_k}{k+1}x^{k+1}\). If \(J(p)\) is the zero polynomial, uniqueness of polynomial coefficients implies \(a_k/(k+1)=0\) for every \(k\). Since \(k+1\neq0\), all \(a_k=0\), so \(p=0\). Conversely, \(J(0)=0\), proving \(\ker(J)=\{0\}\). Now let \(q\in\operatorname{im}(J)\). Then \(q=J(p)\) for some \(p\in P_n\), and the defining integral gives \(q(0)=\int_0^0p(t)\,dt=0\). This proves that every image element vanishes at zero. For the reverse inclusion, suppose \(q\in P_{n+1}\) and \(q(0)=0\). Write \(q(x)=\sum_{k=1}^{n+1}b_kx^k\), and define \(p(x)=\sum_{k=1}^{n+1}k b_kx^{k-1}\), which belongs to \(P_n\). Then \(J(p)(x)=\sum_{k=1}^{n+1}\frac{k b_k}{k}x^k=q(x)\). Hence every polynomial in \(P_{n+1}\) vanishing at zero lies in the image. Such polynomials are exactly the linear combinations of \(x,x^2,\dots,x^{n+1}\), by their coefficient form. The zero kernel means that \(J(p)=J(r)\) implies \(J(p-r)=0\), hence \(p-r=0\), so \(J\) is injective. Finally, the constant polynomial \(1\in P_{n+1}\) does not vanish at zero and therefore is not in the image. Thus \(J\) is not surjective. \(\square\)
Worked Example: Integrating from 0 on P₂
Let \(p(x)=2-3x+4x^2\in P_2\). Integrate each term from \(0\) to \(x\):
To check the output, differentiate: the derivative of \(2x\) is \(2\), the derivative of \(-\frac{3}{2}x^2\) is \(-3x\), and the derivative of \(\frac{4}{3}x^3\) is \(4x^2\). The result is \(2-3x+4x^2=p(x)\). Also \(J(p)(0)=0\), as required by the lower endpoint.
To see how to recover an input for a proposed image element, take \(q(x)=5x-2x^2+3x^3\in P_3\). Its coefficients give the input \(p(x)=5-4x+9x^2\in P_2\), since the coefficient of \(x^k\) in \(q\) is multiplied by \(k\) to form the coefficient of \(x^{k-1}\) in \(p\). Direct integration checks the claim:
This output has zero constant term, as every element of the image must. By contrast, the constant polynomial \(1\) cannot be an output of \(J\), since \(1(0)=1\).
Common Pitfall: An Antiderivative Is Not Always an Image
A common error is to say that \(J\) is onto because every polynomial has an antiderivative. This overlooks the fixed lower endpoint and the codomain. For example, an antiderivative of the constant polynomial \(1\) is \(x+C\), but no choice of input to \(J\) can produce the constant output \(1\): every output satisfies \(J(p)(0)=0\), whereas \(1(0)=1\). In fact, \(J(1)(x)=x\), not \(1\). The correction is to distinguish an unrestricted antiderivative, which may include any constant, from the specific anchored integral \(\int_0^x p(t)\,dt\).
For evaluation, the corresponding mistake is to assume that a zero output forces a zero input. At \(a=2\), the calculation \(\operatorname{ev}_2(x-2)=2-2=0\) is a counterexample: the input \(x-2\) is not the zero polynomial. The kernel records all inputs that map to zero, not just the zero input.
These examples show why specifying the domain and codomain matters. Evaluation always reaches every real number, but on a space containing nonconstant polynomials it loses information about the input. Integration retains enough information to be one-to-one, but its outputs must satisfy the value condition at zero and so do not fill the stated codomain.
Check Your Understanding
Use the definitions of kernel and image to justify each answer.
- For \(\operatorname{ev}_{-2}:P_3\to\mathbb{R}\), describe the kernel and give a nonzero polynomial in it.
- Is \(\operatorname{ev}_5:P_0\to\mathbb{R}\) injective, surjective, both, or neither? Explain.
- Find the input \(p\in P_2\) such that \(J(p)(x)=4x+3x^2-2x^3\), and check by integration.
- Give an element of \(P_4\) that is not in the image of \(J:P_3\to P_4\), and state why it cannot be an output.
- Explain why \(J:P_n\to P_{n+1}\) is injective using its kernel, without relying on a dimension argument.
