Classifying Maps from Polynomial Values and Derivatives

From Polynomial Inputs to Two Numbers

A polynomial in \(\mathcal P_2(\mathbb R)\) has the form \(p(x)=c_0+c_1x+c_2x^2\), where \(c_0,c_1,c_2\in\mathbb R\). A rule that returns two values from such a polynomial defines a map into \(\mathbb R^2\). Its behavior depends on what information the two outputs collect: evaluating at distinct points gives two separate constraints, while evaluating twice at the same point repeats the same constraint. A derivative can supply a different kind of information from a value, even when both are taken at the same point.

We will classify maps of the forms \(p\mapsto(p(a),p(b))\) and \(p\mapsto(p(a),p'(b))\), for real \(a,b\). The space \(\mathcal P_2(\mathbb R)\) consists of all real polynomials of degree at most two, including the zero polynomial. It has basis \((1,x,x^2)\), so its dimension is \(3\). The codomain \(\mathbb R^2\) has dimension \(2\).

For these maps, linearity follows from the rules for sums and scalar multiples of polynomials: \((cp+dq)(a)=cp(a)+dq(a)\), and \((cp+dq)'(b)=cp'(b)+dq'(b)\). Thus both evaluation and derivative evaluation are linear, and pairing two such quantities gives a linear map into \(\mathbb R^2\). The classification question is then to find which inputs map to \((0,0)\), and which output pairs can occur.

Two Point Evaluations

Proof. For \(p,q\in\mathcal P_2(\mathbb R)\) and \(c,d\in\mathbb R\), evaluating \(cp+dq\) at \(a\) and \(b\) gives \(c(p(a),p(b))+d(q(a),q(b))\), so \(E_{a,b}\) is linear. If \(a\ne b\), any desired output \((u,v)\in\mathbb R^2\) is attained by the polynomial \(p(x)=u\frac{x-b}{a-b}+v\frac{x-a}{b-a}\). Substitution gives \(p(a)=u\) and \(p(b)=v\), proving that the image is all of \(\mathbb R^2\).

To determine the kernel when \(a\ne b\), write any polynomial in the form \(p(x)=A+B(x-a)+C(x-a)^2\). The condition \(p(a)=0\) gives \(A=0\). The condition \(p(b)=0\) then gives \(B(b-a)+C(b-a)^2=0\). Since \(b-a\ne0\), this is equivalent to \(B=-C(b-a)\). Therefore \(p(x)=C(x-a)(x-b)\), and every such polynomial vanishes at both points. This proves the stated kernel formula.

If \(a=b\), both coordinates of \(E_{a,a}(p)\) equal \(p(a)\), so the image consists exactly of pairs \((u,u)\): every such pair occurs, for example from the constant polynomial \(p(x)=u\). The kernel is precisely the set of polynomials with \(p(a)=0\). Writing \(p(x)=A+B(x-a)+C(x-a)^2\), this condition is \(A=0\); hence the kernel is spanned by the two independent polynomials \(x-a\) and \((x-a)^2\), and has dimension \(2\). In both cases the kernel contains a nonzero polynomial: when \(a\ne b\), use \((x-a)(x-b)\); when \(a=b\), use \(x-a\). Thus neither map is injective. The image description shows that the map is surjective precisely in the distinct-point case. \(\square\)

A Value Together with a Derivative

Proof. Linearity follows because \((cp+dq)(a)=cp(a)+dq(a)\) and \((cp+dq)'(b)=cp'(b)+dq'(b)\). To prove surjectivity, take any \((u,v)\in\mathbb R^2\) and choose \(p(x)=u+v(x-a)\). Then \(p(a)=u\), while \(p'(x)=v\) and hence \(p'(b)=v\). Thus every output pair occurs.

For the kernel, express \(p\) about \(b\) as \(p(x)=A+B(x-b)+C(x-b)^2\). Its derivative is \(p'(x)=B+2C(x-b)\), so \(p'(b)=B\). The condition \(p'(b)=0\) gives \(B=0\). The other kernel condition becomes \(p(a)=A+C(a-b)^2=0\), so \(A=-C(a-b)^2\). Consequently \(p(x)=C((x-b)^2-(a-b)^2)=C(x-a)(x-(2b-a))\). Each polynomial of this form has value zero at \(a\) and derivative zero at \(b\), so this expression describes the entire kernel. It contains a nonzero polynomial, for example the one with \(C=1\); therefore the map is not injective. When \(a=b\), the two factors agree and the kernel is the multiples of \((x-a)^2\). \(\square\)

Reading Injectivity and Surjectivity Correctly

These examples show why the two properties must be checked separately. A map is injective exactly when its kernel contains only the zero input; all the maps classified here fail that test because each has a nonzero polynomial in its kernel. Surjectivity asks a different question: can every target pair occur? Distinct-point evaluations and every value-derivative map are surjective, while repeated evaluation is not.

The kernel dimensions agree with the Rank-Nullity Theorem: for a linear map from a finite-dimensional domain, the dimension of the domain equals the dimension of the kernel plus the dimension of the image. Here \(\mathcal P_2(\mathbb R)\) has dimension \(3\). For distinct evaluations and for value-derivative maps, the kernel has dimension \(1\), leaving an image of dimension \(2\). For repeated evaluations, the kernel has dimension \(2\), leaving an image of dimension \(1\). This matches the explicit image descriptions; it does not replace the work of identifying which output pairs actually occur.

Common mistake: treating surjectivity as injectivity. Since \(T(p)=(p(1),p'(1))\) is surjective, one might incorrectly conclude that it is injective as well. But the nonzero polynomial \(q(x)=(x-1)^2\) satisfies \(T(q)=(0,0)\): its value at \(1\) is zero and its derivative \(2x-2\) also vanishes at \(1\). Thus two different inputs, \(q\) and the zero polynomial, have the same output. Surjectivity guarantees an input for every target; it does not guarantee that this input is unique.