Differentiation Matrices in Monomial and Shifted Bases

Differentiation as a Linear Operator

One linear transformation can have different matrices when its input and output are described using different bases. Differentiation on a polynomial space is a useful setting for seeing this distinction: the transformation itself is fixed, but the coordinates of a derivative depend on the chosen basis. We will compute those coordinates in the monomial basis and in a basis shifted from \(x\) to \(x-1\), then check the basis-change formula directly.

Work over \(\mathbb{R}\). Let \(P_2\) denote the vector space of real polynomials of degree at most two, including the zero polynomial. Define \(D:P_2\to P_2\) by \(D(f)=f'\). The derivative of a polynomial of degree at most two has degree at most one, so it still belongs to \(P_2\). Moreover, differentiation is linear: for \(f,g\in P_2\) and \(a,b\in\mathbb{R}\), \(D(af+bg)=aD(f)+bD(g)\).

We use the ordered bases \(\mathcal{B}=(1,x,x^2)\) and \(\mathcal{C}=(1,x-1,(x-1)^2)\). The matrix \([D]_{\mathcal{B}}\) has as its \(j\)th column the \(\mathcal{B}\)-coordinate vector of the derivative of the \(j\)th basis polynomial. The same column rule applies to \([D]_{\mathcal{C}}\). This rule follows from the definition of a matrix of a linear map.

Proof. Write an arbitrary \(\mathbf{v}\in V\) as \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\), so \([\mathbf{v}]_{\mathcal{B}}=(c_1,\dots,c_n)^{\top}\). By linearity, \(T(\mathbf{v})=c_1T(\mathbf{b}_1)+\cdots+c_nT(\mathbf{b}_n)\). The \(j\)th column \([T(\mathbf{b}_j)]_{\mathcal{B}}\), multiplied by \(c_j\) and summed over \(j\), is therefore exactly \([T(\mathbf{v})]_{\mathcal{B}}\). This is the matrix-vector product \([T]_{\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\), so the columns of the matrix must be the stated coordinate vectors. \(\square\)

The Matrix in the Monomial Basis

The factor \(2\) in the third column records the coefficient produced by differentiating \(x^2\). It is not an extra scaling convention: it appears because the derivative of the third basis vector is \(2x\). The column method keeps the basis and the output coordinates visible, which matters especially when the basis vectors are not monomials.

The Matrix in the Shifted Basis

The two matrices happen to have the same entries. In both bases, differentiating the first basis vector gives zero, the second gives the first, and the third gives twice the second. Equal entries here do not mean the bases are the same: the entries describe the same operator relative to different coordinate systems. The transition matrix will show precisely how those coordinate systems are related.

Changing Coordinates and Verifying Similarity

Let \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\) be the change-of-basis matrix that converts \(\mathcal{B}\)-coordinates into \(\mathcal{C}\)-coordinates. Its \(j\)th column is the \(\mathcal{C}\)-coordinate vector of the \(j\)th vector in \(\mathcal{B}\). To find those columns, express each monomial using \(x=1+(x-1)\): \(1=1\), \(x=1+(x-1)\), and \(x^2=1+2(x-1)+(x-1)^2\). Hence

The reverse change-of-coordinates matrix is its inverse. Solving for the monomials in terms of the shifted basis, or multiplying to check, gives

The coordinate direction determines the order of the matrices in the formula. First convert a vector's \(\mathcal{C}\)-coordinates to \(\mathcal{B}\)-coordinates, apply the matrix of \(D\) in \(\mathcal{B}\), and convert the result back to \(\mathcal{C}\)-coordinates.

Proof. Take any \(\mathbf{v}\in V\). Its coordinate vectors satisfy \([\mathbf{v}]_{\mathcal{C}}=\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\), so the reverse relation is \([\mathbf{v}]_{\mathcal{B}}=\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}[\mathbf{v}]_{\mathcal{C}}\). Apply \(T\) and express its output in \(\mathcal{C}\)-coordinates. By the definition of an operator matrix, \([T(\mathbf{v})]_{\mathcal{C}}=[T]_{\mathcal{C}}[\mathbf{v}]_{\mathcal{C}}\). Alternatively, convert the input to \(\mathcal{B}\)-coordinates, apply \([T]_{\mathcal{B}}\), and convert the output to \(\mathcal{C}\)-coordinates: \([T(\mathbf{v})]_{\mathcal{C}}=\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}[T]_{\mathcal{B}}\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}[\mathbf{v}]_{\mathcal{C}}\). These two expressions agree for every coordinate vector \([\mathbf{v}]_{\mathcal{C}}\). In particular, they agree when that vector is each standard coordinate vector, which shows that the matrices multiplying it are equal. This is the stated formula. \(\square\)

This computation verifies similarity: the two matrices represent the same linear map in different bases and are related by the required invertible change of coordinates. In this example, the matrices are equal as well. The equality is a special feature of the chosen bases and operator, not a replacement for the formula. In other basis changes, the entries need not remain equal.

Common Pitfall: Mixing the Polynomial with Its Coordinates

A common error is to differentiate a shifted basis polynomial correctly but then record its coordinates as though the basis were \((1,x,x^2)\). For example, \(D((x-1)^2)=2(x-1)\). A mistaken coordinate vector is \((0,0,2)^{\top}\), which would put the \(2\) in the third column's third entry. That vector represents \(2(x-1)^2\) in \(\mathcal{C}\), not \(2(x-1)\), so it is the wrong coordinate vector. In \(\mathcal{C}\), \(2(x-1)=0\cdot1+2\cdot(x-1)+0\cdot(x-1)^2\), and the correct coordinates are \((0,2,0)^{\top}\).

The distinction is also visible by checking a polynomial. The incorrect third column would send \((0,0,1)^{\top}\) to \((0,0,2)^{\top}\), claiming that the derivative of \((x-1)^2\) is \(2(x-1)^2\). But direct differentiation gives \(2(x-1)\), whose coordinates are \((0,2,0)^{\top}\). When forming an operator matrix, differentiate first and then express the result in the stated output basis.