Monomial Bases for Polynomial Spaces

The Polynomial Space \(P_n\)

Polynomials have a natural ordered list of building blocks: \(1,x,x^2,\dots\). Restricting the degree gives a finite list that can describe every polynomial in the space. The central point is not merely that a polynomial can be expanded using these powers, but that the expansion is unique. Together, spanning and linear independence establish that the list is a basis.

Throughout, scalars are real. For an integer \(n\geq 0\), let \(P_n\) denote the vector space of real polynomials of degree at most \(n\), together with the zero polynomial. A polynomial in this space has the form \(a_0+a_1x+\cdots+a_nx^n\), where each coefficient \(a_k\) is real. Coefficients may be zero, including coefficients of powers that do not appear in a written expression. The zero polynomial belongs to every \(P_n\).

A basis of a vector space is a list that both spans the space and is linearly independent. To span \(P_n\), every polynomial in \(P_n\) must be expressible as a linear combination of the monomials. To be linearly independent, the only linear combination of those monomials that equals the zero polynomial must have all coefficients zero. These are separate requirements, and the proof checks each one.

Why the Monomials Form a Basis

Proof. First, the list spans \(P_n\). Let \(p(x)\in P_n\). By the definition of \(P_n\), there are real coefficients \(a_0,\dots,a_n\) such that \(p(x)=a_0+a_1x+\cdots+a_nx^n\). This is the linear combination \(a_0\cdot 1+a_1x+\cdots+a_nx^n\) of members of \(\mathcal{B}\), so every element of \(P_n\) lies in the span of \(\mathcal{B}\).

Second, the list is linearly independent. Suppose \(c_0\cdot 1+c_1x+\cdots+c_nx^n\) is the zero polynomial. The coefficients of a polynomial are its defining coefficients, and two polynomials are equal precisely when their coefficients at every power agree. The zero polynomial has coefficient \(0\) at every power. Therefore \(c_0=0,c_1=0,\dots,c_n=0\). The only linear combination that gives the zero polynomial is the trivial one. The list is both spanning and linearly independent, so it is a basis. This argument also covers \(n=0\): the list consists of \(1\), and \(c_0\cdot 1\) is the zero polynomial only when \(c_0=0\). \(\square\)

The proof explains why the coefficients are the representation: the spanning part gives an expansion, while the independence part rules out any different coefficient list producing the same polynomial. This is specific here to polynomial coefficients; it does not require choosing or reducing a matrix.

Reading a Representation

When finding \([p]_{\mathcal{B}}\), match each coefficient to its power of \(x\), starting with the constant term. If a power is missing, its coefficient is zero. If the polynomial has degree less than \(n\), append zero entries until there is one coefficient for each member of the basis list. This keeps the representation length fixed at \(n+1\).

Arithmetic with Coefficient Representations

Once the order of the entries is fixed, polynomial addition and scalar multiplication can be carried out coefficient by coefficient. This is useful when a problem gives polynomials in basis form, or asks for the representation of a sum. The rule follows directly by collecting the coefficients of each power.

Proof. Add the displayed expressions for \(p\) and \(q\), grouping terms with the same power: \[ p(x)+q(x) =(a_0+b_0)+(a_1+b_1)x+\cdots+(a_n+b_n)x^n. \] Thus the coefficient at each power \(x^k\) is \(a_k+b_k\). For scalar multiplication, distribute \(c\) over the terms: \[ cp(x)=(ca_0)+(ca_1)x+\cdots+(ca_n)x^n. \] Thus the coefficient at \(x^k\) is \(ca_k\). These expressions give the stated representations in the prescribed monomial order. \(\square\)

For example, if two polynomials in \(P_2\) have representations \(\begin{bmatrix}1&2&-1\end{bmatrix}^{\top}\) and \(\begin{bmatrix}3&-2&4\end{bmatrix}^{\top}\), their sum has representation \(\begin{bmatrix}1+3&2+(-2)&-1+4\end{bmatrix}^{\top}=\begin{bmatrix}4&0&3\end{bmatrix}^{\top}\). The zero in the middle records cancellation of the \(x\)-terms, not a missing coordinate.

Common Pitfall: Reversing the Coefficient Order

A common error is to write coefficients from the highest power down while using a basis list ordered from the constant term up. For \(p(x)=3-2x+5x^3\) in \((1,x,x^2,x^3)\), the column \(\begin{bmatrix}5&0&-2&3\end{bmatrix}^{\top}\) is wrong: it assigns \(5\) to \(1\) and \(3\) to \(x^3\). Reconstructing that column gives \(5-2x^2+3x^3\), which is not \(p(x)\). The correction is to follow the list exactly, giving \(\begin{bmatrix}3&-2&0&5\end{bmatrix}^{\top}\), whose reconstruction is \(3-2x+5x^3\).

A second version of the same mistake is to omit a zero coefficient. A representation in \(P_3\) must have four entries because its basis list has four members. For instance, \(1+x^3\) has representation \(\begin{bmatrix}1&0&0&1\end{bmatrix}^{\top}\), not a two-entry column. The two zeros preserve the positions assigned to \(x\) and \(x^2\).

Using the Basis in Practice

The monomial basis provides a standard way to describe each polynomial in \(P_n\) by a finite column of real numbers. That makes polynomial calculations compatible with familiar vector calculations: coefficients can be compared, added, and scaled in a fixed order. The basis also gives a direct way to check an answer. Expand the proposed coefficient column against \(1,x,\dots,x^n\); if the resulting polynomial is the original one, the representation is correct.

Keep the scope of the representation in view. The basis list ends at \(x^n\), so it represents polynomials in \(P_n\), not every polynomial of arbitrary degree. If a polynomial contains a nonzero term with power greater than \(n\), it does not belong to \(P_n\) and cannot be represented by this list. By contrast, a polynomial of lower degree belongs to \(P_n\) and uses zero coefficients in the unused higher positions.