A Polynomial Derivative Map in Two Bases

Differentiation as a Coordinate Calculation

A derivative is a familiar polynomial, but its coordinate column depends on the basis used to describe it. This distinction matters when a linear map has one basis in its domain and another in its codomain: each column must record the image of a domain basis vector using codomain coordinates. We will calculate differentiation on quadratic polynomials using both the standard basis and a shifted-power basis, then compare the matched-basis and mixed-basis matrices.

Work over \(\mathbb{R}\), and let \(P_2\) be the vector space of real polynomials of degree at most two, including the zero polynomial. Differentiation defines a map \(D:P_2\to P_2\), where \(D(p)=p'\). If \(p=a_0+a_1x+a_2x^2\), then \(D(p)=a_1+2a_2x\), which again belongs to \(P_2\). For \(p,q\in P_2\) and scalars \(a,b\in\mathbb{R}\), the coefficient formula gives \(D(ap+bq)=aD(p)+bD(q)\); thus this particular map is linear.

Use the ordered bases \(\mathcal{B}=(1,x,x^2)\) and \(\mathcal{C}=(1,x-1,(x-1)^2)\). The first is a basis because every element of \(P_2\) has a unique expression \(a_0+a_1x+a_2x^2\). For the second, put \(y=x-1\), so \(x=y+1\). Every polynomial in \(x\) of degree at most two can then be written uniquely as a polynomial in \(y\) of degree at most two. Replacing \(y\) by \(x-1\) gives a unique expression in \(1,x-1,(x-1)^2\), so \(\mathcal{C}\) is also a basis.

The definition gives the useful identity \([T(\mathbf{p})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{p}]_{\mathcal{B}}\). Indeed, if \([\mathbf{p}]_{\mathcal{B}}=(c_1,\dots,c_n)^\top\), then \(\mathbf{p}=\sum_{j=1}^n c_j\mathbf{b}_j\). By linearity, \(T(\mathbf{p})=\sum_{j=1}^n c_jT(\mathbf{b}_j)\), whose \(\mathcal{C}\)-coordinate column is the same linear combination of the matrix’s columns. This explains both how to build the matrix and how to use it.

The Matrix in Each Matched Basis

Start with \(\mathcal{B}\) as both the domain and codomain basis. Differentiate each basis polynomial, then write the result in \(\mathcal{B}\)-coordinates. The derivative of \(1\) is \(0\), the derivative of \(x\) is \(1\), and the derivative of \(x^2\) is \(2x\). Their coordinate columns are respectively \((0,0,0)^\top\), \((1,0,0)^\top\), and \((0,2,0)^\top\).

For \(\mathcal{C}\) as both bases, the basis vectors differentiate to \(0\), \(1\), and \(2(x-1)\). These are \(0\), the first basis vector, and twice the second basis vector. The coordinate columns are again \((0,0,0)^\top\), \((1,0,0)^\top\), and \((0,2,0)^\top\), so

Proof. In \(\mathcal{B}\), the derivatives of the ordered basis vectors are \(0\), \(1\), and \(2x\). Their \(\mathcal{B}\)-coordinate columns are \((0,0,0)^\top\), \((1,0,0)^\top\), and \((0,2,0)^\top\). In \(\mathcal{C}\), the derivatives of the ordered basis vectors are \(0\), \(1\), and \(2(x-1)\). Their \(\mathcal{C}\)-coordinate columns are exactly \((0,0,0)^\top\), \((1,0,0)^\top\), and \((0,2,0)^\top\). The definition of a coordinate matrix places these columns in order, so the two matrices are equal. \(\square\)

The equal entries are not a claim that coordinates are independent of the basis. They result from these particular derivative relations: each basis vector after the first differentiates to a multiple of the preceding basis vector. In a different basis, the same polynomial images might have different coordinate columns.

When the Domain and Codomain Bases Differ

A map can use different bases for inputs and outputs. In \([D]_{\mathcal{C}\leftarrow\mathcal{B}}\), for example, the domain basis vectors are \(1,x,x^2\), but every derivative must be expressed in \(\mathcal{C}\)-coordinates. The first two image columns are still \(0\) and \(1\). For the third, \(D(x^2)=2x=2+2(x-1)\), so its \(\mathcal{C}\)-coordinate column is \((2,2,0)^\top\). Therefore,

Switching which basis is used for the domain gives a different mixed-basis matrix. The domain vectors are now \(1,x-1,(x-1)^2\), and the output coordinates are in \(\mathcal{B}\). The first two image columns are \((0,0,0)^\top\) and \((1,0,0)^\top\). For the third, \(D((x-1)^2)=2(x-1)=-2+2x\), so its \(\mathcal{B}\)-coordinate column is \((-2,2,0)^\top\). Thus

More generally, the column rule for a mixed-basis matrix follows directly from linearity and basis expansion. If \(p=c_1\cdot1+c_2x+c_3x^2\), then \(D(p)=c_1D(1)+c_2D(x)+c_3D(x^2)\). Expressing each of these three images in the codomain basis and combining their coordinate columns produces \([D(p)]_{\mathcal{C}}=[D]_{\mathcal{C}\leftarrow\mathcal{B}}[p]_{\mathcal{B}}\). The same reasoning applies with the bases exchanged. It is the reason the domain basis determines the input columns while the codomain basis determines how the image columns are recorded.

Common Pitfall: Reusing Coordinates in the Wrong Basis

A frequent error is to write the third column of \([D]_{\mathcal{C}\leftarrow\mathcal{B}}\) as \((0,2,0)^\top\), reasoning that \(D(x^2)=2x\) has coefficient \(2\) on \(x\). That column uses \(\mathcal{B}\)-coordinates, not \(\mathcal{C}\)-coordinates. In \(\mathcal{C}\), \(x=1+(x-1)\), and therefore \(2x=2\cdot1+2(x-1)\), with coordinate column \((2,2,0)^\top\). The incorrect column would make the matrix output for the polynomial \(x^2\) equal to \(2(x-1)\), which is not its derivative \(2x\). Reconstructing the polynomial from the proposed output coordinates catches the mistake.

Keep the domain and codomain roles separate. The column index identifies a domain basis vector; the entries in that column are the coordinates of its image in the codomain basis. A matched-basis matrix is appropriate only when the same basis is chosen on both sides. A matrix can have the same size as another matrix and still describe a different coordinate conversion.