Evaluation Functionals and Their Row Matrices

Evaluation as a Linear Map

A polynomial can be described either by its formula or by its coefficients in a chosen basis. Evaluation at a fixed number turns that polynomial into a real number. The matrix of this operation is especially simple in the monomial basis: its entries are successive powers of the evaluation point. This gives a direct way to compute a value from a coefficient column, while keeping distinct the polynomial itself, its coordinates, and the output scalar.

Let \(n\) be a nonnegative integer. Write \(P_n(\mathbb{R})\) for the real vector space of polynomials of degree at most \(n\), including the zero polynomial. Its monomial basis is the ordered list \(\mathcal{B}_n=(1,x,\dots,x^n)\). Thus every \(p\in P_n(\mathbb{R})\) has a unique expression \(p(x)=c_0+c_1x+\cdots+c_nx^n\), and its coordinate vector is the column \([p]_{\mathcal{B}_n}=(c_0,c_1,\dots,c_n)^{\top}\). The codomain is \(\mathbb{R}\), with its standard basis \(\mathcal{C}=(1)\).

The map depends on the chosen number \(a\): evaluating at \(a=2\) and evaluating at \(a=-1\) are generally different operations. The polynomial space is finite-dimensional, of dimension \(n+1\), and the output space is one-dimensional. Accordingly, the matrix of this map in the specified bases has one row and \(n+1\) columns.

Proof. To prove linearity, take any \(p,q\in P_n(\mathbb{R})\) and any \(r,s\in\mathbb{R}\). Write \(p(x)=\sum_{j=0}^{n}c_jx^j\) and \(q(x)=\sum_{j=0}^{n}d_jx^j\). The polynomial \(rp+sq\) belongs to \(P_n(\mathbb{R})\), and its value at \(a\) is \((rp+sq)(a)=r p(a)+s q(a)\), by the definitions of polynomial addition, scalar multiplication, and substitution. Therefore \(\operatorname{ev}_a(rp+sq)=r\operatorname{ev}_a(p)+s\operatorname{ev}_a(q)\) for every \(p,q,r,s\), which is the definition of a linear map. \(\square\)

Deriving the Row Matrix

The matrix of a linear map has as its columns the coordinate vectors of the images of the domain basis vectors. Here each basis vector is a monomial. Its image under evaluation at \(a\) is a real number, whose coordinate in the codomain basis \((1)\) is that same number: \(\operatorname{ev}_a(x^j)=a^j\). The column for \(x^j\) is therefore the one-entry column \([a^j]\). Putting these columns in the order of \(\mathcal{B}_n\) gives the row matrix below.

Proof. For each \(j=0,\dots,n\), the \(j+1\)st domain basis vector is \(x^j\). Its image is \(\operatorname{ev}_a(x^j)=x^j\big|_{x=a}=a^j\), so its coordinate in \(\mathcal{C}=(1)\) is the one-entry column \([a^j]\). By the definition of a matrix representation, placing these image coordinates as columns, in basis order, produces the row \(\begin{bmatrix}a^0&a^1&\cdots&a^n\end{bmatrix}\). Since \(a^0=1\), this is the stated matrix. Multiplying that row by the coefficient column gives \(c_0a^0+c_1a^1+\cdots+c_na^n\). The polynomial expansion \(p(x)=\sum_{j=0}^{n}c_jx^j\), evaluated at \(x=a\), gives exactly the same sum. The coordinate of a real number in the basis \((1)\) is the number itself, so the product is the coordinate of \(\operatorname{ev}_a(p)\), as claimed. This reasoning also applies when \(n=0\): the matrix is \(\begin{bmatrix}1\end{bmatrix}\) and evaluation leaves a constant polynomial’s value unchanged. \(\square\)

The formula depends on the bases as well as on the map. In particular, the row lists the images of the monomials in their specified order. Its first entry is \(1\), the image of the constant basis polynomial \(1\), not \(a\): the monomial \(1\) is \(x^0\), and \(a^0=1\), including when \(a=0\). The coefficient vector is a column, so multiplying the \(1\times(n+1)\) row by it returns a \(1\times1\) output coordinate.

Computing Values from Coefficient Columns

What the Row Encodes—and a Common Error

The row matrix stores the values of the basis polynomials at the chosen point. It does not store the coefficients of the input polynomial, nor does it act on the polynomial formula without first expressing that polynomial in the domain basis. The multiplication works because the coordinate column records the coefficients in precisely the same order as the entries of the row: constant term first, then the coefficient of \(x\), continuing through \(x^n\).

This description is useful whenever a value constraint must be computed from polynomial coordinates. For a fixed \(a\), the equation \(p(a)=b\) becomes a single row-matrix equation in the coefficient column: the row \(\begin{bmatrix}1&a&\cdots&a^n\end{bmatrix}\) multiplied by that column must equal \(b\). The point \(a\) determines the row, while the coefficients determine the input column. Keeping those roles separate makes the representation easy to interpret and check.

Common mistake: starting the powers at \(a^1\). For the first example, an incorrect row \(\begin{bmatrix}2&4&8&16\end{bmatrix}\) would give \(2(3)+4(-1)+8(2)+16(1)=6-4+16+16=34\). This row incorrectly assigns \(a\) to the basis polynomial \(1\), shifting every power one place. The correct first basis vector is \(1=x^0\), so the first entry must be \(2^0=1\); the correct row is \(\begin{bmatrix}1&2&4&8\end{bmatrix}\), and its product is \(17\), as direct substitution confirms.

A reliable procedure is to write the basis order, list the value of each basis polynomial at \(a\), and only then form the row. Next, write every coefficient in the matching position of a column, using zero for absent powers. Finally, multiply and substitute directly into the original polynomial as a check. These steps guard against both a power-index shift and a coefficient-order mismatch.