Differentiation Between Polynomial Spaces

Differentiation as a Map Between Polynomial Spaces

For a polynomial, differentiation lowers the degree unless the polynomial is constant, in which case its derivative is zero. That familiar operation can also be viewed as a function between two vector spaces. This viewpoint makes it possible to ask not only for the derivative of one polynomial, but also how differentiation acts on all polynomials in a specified domain, what outputs it can produce, and whether different inputs can have the same output.

Work over \(\mathbb{R}\), and let \(n\geq 1\). Write \(\mathcal{P}_n\) for the vector space of real polynomials of degree at most \(n\), including the zero polynomial. Thus \(\mathcal{P}_n\) has basis \((1,x,\dots,x^n)\). The codomain \(\mathcal{P}_{n-1}\) consists of polynomials of degree at most \(n-1\), with basis \((1,x,\dots,x^{n-1})\). Define \(D_n:\mathcal{P}_n\to\mathcal{P}_{n-1}\) by taking the usual polynomial derivative.

The coefficient rule follows by differentiating each term: the derivative of \(a_kx^k\) is \(ka_kx^{k-1}\) for \(k\geq1\), while the derivative of the constant term \(a_0\) is zero. In particular, the coefficient of \(x^j\) in the output is \((j+1)a_{j+1}\) for \(0\leq j\leq n-1\). The coefficient \(a_0\) does not affect the output.

Linearity and the Coefficient Matrix

Proof. Write \(p(x)=\sum_{k=0}^n a_kx^k\) and \(q(x)=\sum_{k=0}^n b_kx^k\). Their sum has coefficient \(a_k+b_k\) on \(x^k\), so \(D_n(p+q)=\sum_{k=1}^n k(a_k+b_k)x^{k-1}\). Distributing the factors and grouping the two sums gives \(\sum_{k=1}^n ka_kx^{k-1}+\sum_{k=1}^n kb_kx^{k-1}=D_n(p)+D_n(q)\). The polynomial \(cp\) has coefficient \(ca_k\) on \(x^k\), so \(D_n(cp)=\sum_{k=1}^n k(ca_k)x^{k-1}=c\sum_{k=1}^n ka_kx^{k-1}=cD_n(p)\). Both defining conditions for a linear map hold. \(\square\)

Use the ordered bases \((1,x,\dots,x^n)\) in the domain and \((1,x,\dots,x^{n-1})\) in the codomain. The matrix of \(D_n\) has \(n\) rows and \(n+1\) columns. Its \(j\)th column is the coordinate vector of the derivative of the \(j\)th domain basis vector. Since \(D_n(1)=0\) and \(D_n(x^k)=kx^{k-1}\) for \(1\leq k\leq n\), the first column is zero and each later column has a single nonzero entry.

In this display, the entry \(k\) is in row \(k\), column \(k+1\), for \(1\leq k\leq n\). If \([p]\) denotes the coordinate vector of \(p\) in the stated domain basis, and \([D_n(p)]\) its coordinate vector in the stated codomain basis, then the matrix represents the map by \([D_n(p)]=[D_n][p]\). The two coordinate vectors use different basis lengths: the input has \(n+1\) entries and the output has \(n\).

Kernel, Image, and What the Map Can Do

For a map \(T:V\to W\), its kernel is the set of inputs sent to the zero vector, \(\ker T=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}\}\). Its image is the set of outputs it actually attains, \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\). A map is injective if distinct inputs always have distinct outputs, and surjective if every element of its codomain is attained. For polynomial maps, the zero vector is the zero polynomial.

Proof. First, let \(p(x)=\sum_{k=0}^n a_kx^k\). It lies in \(\ker D_n\) exactly when \(D_n(p)\) is the zero polynomial. The coefficient of \(x^{k-1}\) in \(D_n(p)\) is \(ka_k\) for each \(1\leq k\leq n\). A polynomial is zero exactly when all its coefficients are zero, so \(D_n(p)=0\) exactly when \(ka_k=0\) for every such \(k\). Each integer \(k\geq1\) is nonzero in \(\mathbb{R}\), hence each equation forces \(a_k=0\). There is no condition on \(a_0\), which proves that the kernel consists precisely of constants.

Next, every output of \(D_n\) belongs to \(\mathcal{P}_{n-1}\) by its coefficient formula, so \(\operatorname{im}D_n\subseteq\mathcal{P}_{n-1}\). For the reverse inclusion, take any \(q(x)=b_0+b_1x+\cdots+b_{n-1}x^{n-1}\) in \(\mathcal{P}_{n-1}\), and form \(p(x)=b_0x+\frac{b_1}{2}x^2+\cdots+\frac{b_{n-1}}{n}x^n\). This is in \(\mathcal{P}_n\), and differentiating its terms gives \(D_n(p)=b_0+b_1x+\cdots+b_{n-1}x^{n-1}=q\). Thus every element of the codomain is attained and \(\operatorname{im}D_n=\mathcal{P}_{n-1}\). The kernel contains \(1\) and \(0\), which are distinct inputs with the same output \(0\), so \(D_n\) is not injective. Since every codomain element is attained, the map is surjective. \(\square\)

This proof uses the real field in two places: the coefficient \(k\) is nonzero, and division by \(k\) is possible. The theorem is stated over \(\mathbb{R}\), where both conditions hold. The explicit construction for a preimage also shows why surjectivity does not require injectivity: adding any constant to an input leaves its derivative unchanged.

Degree and a Second Coefficient Check

The degree behavior is another direct consequence of the coefficient rule. If \(p\) is nonconstant of degree \(m\), its leading term is \(a_mx^m\) with \(a_m\ne0\). Its derivative has leading term \(ma_mx^{m-1}\), and \(ma_m\ne0\) over \(\mathbb{R}\). Thus its derivative has degree \(m-1\). This statement does not apply to constant polynomials: their derivative is zero, which has no degree under the usual convention.

Common Pitfall: Shifting Coefficients the Wrong Way

A frequent error is to place the derivative coefficients in the same positions as the input coefficients. For instance, the incorrect calculation \(D_3(1+2x+3x^2)= (1,2,3,0)\) both keeps the wrong number of coefficients and fails to multiply by the exponents. The input has four possible coefficient positions in \(\mathcal{P}_3\), but the output has only three in \(\mathcal{P}_2\). The correct computation is \(D_3(1+2x+3x^2)=0+2+6x=2+6x\), with output coordinate vector \((2,6,0)^\top\) in the basis \((1,x,x^2)\).

The constant coefficient disappears because constants differentiate to zero; it does not move into the output. Each coefficient \(a_k\) for \(k\geq1\) instead becomes \(ka_k\) in the position for \(x^{k-1}\). Keeping the domain and codomain bases distinct helps prevent both the position error and the common mistake of treating the matrix as square.

Differentiation is therefore a concrete example of a linear map whose behavior is visible in its coefficient matrix: one input direction, the constant term, is lost, while every polynomial in the codomain can be produced. These facts are useful whenever a problem asks whether a target polynomial can be obtained as a derivative or whether a derivative determines its input uniquely.