Kernel and Image of Polynomial Differentiation

The Differentiation Map on Polynomial Spaces

For a polynomial, differentiation changes its coefficients and usually lowers its degree. To understand differentiation as a linear map, it is useful to ask two precise questions: which inputs have derivative zero, and which polynomials can occur as derivatives? These questions depend on specifying both the domain and codomain. In this tutorial all scalars are real, and the notation \(P_n\) means the vector space of real polynomials of degree at most \(n\), including the zero polynomial.

Fix an integer \(n\geq 1\). Every polynomial in \(P_n\) has a unique expression \(a_0+a_1x+\cdots+a_nx^n\), where the coefficients \(a_0,\dots,a_n\) are real. Differentiation sends such a polynomial to \(a_1+2a_2x+\cdots+na_nx^{n-1}\), which belongs to \(P_{n-1}\). Thus the relevant map is \(D_n:P_n\to P_{n-1}\), defined by \(D_n(p)=p'\). The subscript records the domain and codomain: the differentiation rule is the same for each \(n\), but the map’s specified input and output spaces change.

The differentiation rules imply that \(D_n\) is linear: for \(p,q\in P_n\) and real numbers \(a,b\), \((ap+bq)'=ap'+bq'\). The kernel and image therefore refer to a linear map between two particular polynomial spaces. Our goal is to determine those sets exactly, not merely to list some examples of inputs and outputs.

What Differentiation Does to Degree

A nonzero polynomial \(p\) is called constant if it has the form \(a_0\), and nonconstant if at least one coefficient of a positive power of \(x\) is nonzero. If a nonconstant polynomial has highest nonzero term \(a_mx^m\), then \(m\geq1\) and \(a_m\ne0\). Differentiation changes that term to \(ma_mx^{m-1}\). Since \(ma_m\ne0\) over \(\mathbb{R}\), this term remains present and is the highest-power term of the derivative.

Proof. Write \(p(x)=a_0+a_1x+\cdots+a_mx^m\), where \(a_m\ne0\) and \(m\geq1\). Then \(p'(x)=a_1+2a_2x+\cdots+ma_mx^{m-1}\). The coefficient \(ma_m\) is nonzero, so the derivative has a nonzero term of degree \(m-1\); all its other terms have lower degree. Hence its degree is \(m-1\). If \(p(x)=a_0\), the differentiation rule gives \(p'(x)=0\). \(\square\)

This degree change explains why \(P_{n-1}\) is a natural codomain for \(D_n\): no polynomial in \(P_n\) can have a derivative of degree \(n\). The proposition alone does not tell us whether every polynomial of degree at most \(n-1\) is reached. That requires constructing an input for an arbitrary proposed output.

The Kernel Consists Exactly of Constants

A constant polynomial is sent to zero, so every constant belongs to \(\ker D_n\). The converse follows by comparing coefficients: a polynomial with zero derivative cannot have any nonzero coefficient on a positive power of \(x\). This includes the zero polynomial, which is itself a constant polynomial (with constant term \(0\)).

Proof. First let \(p\in\ker D_n\). By the definition of the kernel, \(p'=0\). Write \(p(x)=a_0+a_1x+\cdots+a_nx^n\). Its derivative is \(p'(x)=a_1+2a_2x+\cdots+na_nx^{n-1}\). Since this is the zero polynomial, every coefficient is zero. In particular, \(ka_k=0\) for each \(k=1,\dots,n\). Each such \(k\) is a nonzero real number, so \(a_k=0\) for every \(k=1,\dots,n\). Therefore \(p(x)=a_0\) is constant. Conversely, if \(p(x)=a_0\), then \(p'(x)=0\), so \(p\in\ker D_n\). Both inclusions hold, giving the stated equality. \(\square\)

Every Polynomial in the Codomain Is an Output

To determine the image, begin with an arbitrary \(q\in P_{n-1}\), not just a particular example. Write its coefficients as \(q(x)=b_0+b_1x+\cdots+b_{n-1}x^{n-1}\). For each term \(b_jx^j\), the polynomial \(\frac{b_j}{j+1}x^{j+1}\) differentiates to \(b_jx^j\). The denominator \(j+1\) is never zero, so this construction is valid for every term. Adding these terms produces an input in \(P_n\).

Proof. Every value \(D_n(p)\) belongs to \(P_{n-1}\), by the definition of the map, so \(\operatorname{im}D_n\subseteq P_{n-1}\). For the reverse inclusion, take any \(q\in P_{n-1}\) and write \(q(x)=\sum_{j=0}^{n-1}b_jx^j\). Define \(p(x)=\sum_{j=0}^{n-1}\frac{b_j}{j+1}x^{j+1}\). Its highest possible power is \(x^n\), so \(p\in P_n\). Differentiating term by term gives \(D_n(p)(x)=\sum_{j=0}^{n-1}\frac{b_j}{j+1}(j+1)x^j=\sum_{j=0}^{n-1}b_jx^j=q(x)\). Thus every \(q\in P_{n-1}\) is in \(\operatorname{im}D_n\), proving \(P_{n-1}\subseteq\operatorname{im}D_n\). The two inclusions establish equality. \(\square\)

The Smallest Domain Case

The condition \(n\geq1\) includes \(n=1\), when the domain is \(P_1\) and the codomain is \(P_0\), the constant polynomials. There is no exceptional failure in this case: differentiating \(a_0+a_1x\) gives \(a_1\). The kernel is obtained when \(a_1=0\), and any constant \(c\in P_0\) is reached by differentiating \(cx\).

A Common Coefficient Error

When constructing an input for a desired derivative, raising the power without adjusting the coefficient is not enough. For the target \(q(x)=2-3x+4x^2+5x^3\), a tempting but incorrect choice is \(r(x)=2x-3x^2+4x^3+5x^4\). Its derivative is

which is not \(q(x)\). The error is multiplying each coefficient by the new exponent during differentiation without first compensating for that multiplication. The correction is to divide the coefficient of \(x^j\) by \(j+1\) when assigning the coefficient of \(x^{j+1}\). For instance, the term \(-3x\) in the target requires \(-\frac{3}{2}x^2\) in the input, because \(\left(-\frac{3}{2}x^2\right)'=-3x\).

The kernel and image answer complementary questions about this map. The kernel identifies exactly the information differentiation loses: adding or changing a constant does not affect the derivative. The image result says that, with the domain \(P_n\) and codomain \(P_{n-1}\), there is no further restriction on the output polynomial. The domain and codomain matter: this conclusion is about these specified polynomial spaces and follows from the degree bound and the explicit coefficient construction.