Polynomial Subspaces From Coefficient Constraints

Polynomial Conditions and Their Ambient Space

A polynomial restriction is meaningful only after its ambient space has been specified. For example, “the coefficient of \(x^3\) is zero” describes a subset of the polynomials of degree at most \(5\), but it also describes a subset of all real polynomials. We will work over \(\mathbb{R}\). For \(n\geq 0\), let \(P_n(\mathbb{R})\) be the vector space of real polynomials of degree at most \(n\), together with the zero polynomial. Each \(p\in P_n(\mathbb{R})\) has a unique expression \(p(x)=a_0+a_1x+\cdots+a_nx^n\), where any of the coefficients may be zero.

A subset \(S\) of a vector space \(V\) is a subspace if it contains the zero vector and is closed under addition and scalar multiplication. For polynomial subsets, the zero vector is the zero polynomial. The subspace test therefore asks: does the zero polynomial satisfy the restriction, and do sums and real scalar multiples of polynomials satisfying it still satisfy it?

Zero-coefficient restrictions are homogeneous: adding two zero coefficients gives zero, and multiplying zero by any real scalar still gives zero. A prescribed nonzero coefficient behaves differently, since scalar multiplication changes its value. The examples below develop this distinction and show how parity restrictions fit the same coefficient-based approach.

Zero-Coefficient Conditions Give Subspaces

Proof. The zero polynomial has every coefficient equal to zero, so it belongs to \(S\). Suppose \(p(x)=\sum_{k=0}^{n}a_kx^k\) and \(q(x)=\sum_{k=0}^{n}b_kx^k\) belong to \(S\). For each \(k\in I\), \(a_k=b_k=0\), so the coefficient of \(x^k\) in \(p+q\) is \(a_k+b_k=0\). Thus \(p+q\in S\). If \(c\in\mathbb{R}\), the coefficient of \(x^k\) in \(cp\) is \(ca_k=0\) for each \(k\in I\); hence \(cp\in S\). The zero polynomial and both closure properties are established, so \(S\) is a subspace. \(\square\)

This result applies whether the restriction removes one power or many. Any coefficients not named in the restriction remain unrestricted. In particular, requiring the coefficient of \(x^4\) to vanish does not require the coefficient of \(x^2\) or the constant coefficient to vanish.

Parity Restrictions as Coefficient Patterns

For a polynomial, having only even powers means that every coefficient of an odd power is zero. Having only odd powers means that every coefficient of an even power, including the constant coefficient, is zero. Thus each restriction is a particular collection of zero-coefficient conditions. The terminology also agrees with the identities \(p(-x)=p(x)\) and \(p(-x)=-p(x)\), respectively, as the next result verifies directly.

Proof. First consider \(p(-x)=p(x)\). Substituting \(-x\) gives \(p(-x)=\sum_{k=0}^{n}(-1)^ka_kx^k\). If all odd-power coefficients vanish, then every remaining exponent is even, so \((-1)^k=1\) for every nonzero term and \(p(-x)=p(x)\). Conversely, if \(p(-x)=p(x)\), uniqueness of polynomial coefficients gives \((-1)^ka_k=a_k\) for each \(k\). For odd \(k\), this says \(-a_k=a_k\), so \(2a_k=0\). Since the scalars are real, \(a_k=0\). Now consider \(p(-x)=-p(x)\). If all even-power coefficients vanish, each remaining exponent is odd, so every term changes sign and the identity follows. Conversely, equality of coefficients in \(p(-x)=-p(x)\) gives \((-1)^ka_k=-a_k\) for each \(k\). For even \(k\), this says \(a_k=-a_k\), hence \(2a_k=0\) and \(a_k=0\). Both equivalences are proved. \(\square\)

Because these parity conditions are zero-coefficient restrictions, they define subspaces of \(P_n(\mathbb{R})\) by the zero-coefficient theorem. The identities provide another way to check a particular polynomial, but the coefficient description makes closure especially transparent.

Prescribing the Value at Zero

If \(p(x)=a_0+a_1x+\cdots+a_nx^n\), then \(p(0)=a_0\): every positive power of zero is zero. Thus the condition \(p(0)=0\) is exactly the zero-coefficient condition \(a_0=0\), so it defines a subspace. In contrast, fixing \(p(0)\) to a nonzero value excludes the zero polynomial.

Proof. If \(c=0\), then \(p(0)=0\) is equivalent to \(a_0=0\), so \(S_0\) is a subspace by the zero-coefficient theorem. If \(c\ne0\), the zero polynomial has value \(0\) at zero, not \(c\), so the zero polynomial does not belong to \(S_c\). A subspace must contain the zero vector; therefore \(S_c\) is not a subspace. These two cases prove the stated equivalence. \(\square\)

Common Mistake: Nonzero Values Do Not Survive Scaling

A tempting but incorrect argument is: “If \(p(0)=2\), then multiplying \(p\) by a scalar still gives a polynomial with value \(2\) at zero.” The calculation above shows the error. For \(v(x)=2+x\), \(v(0)=2\), but \((2v)(0)=2v(0)=4\), not \(2\). More generally, if \(p(0)=c\), then \((ap)(0)=a\,p(0)=ac\). When \(c\ne0\), choosing \(a=0\) gives \(0\), so the prescribed nonzero value cannot be preserved for every scalar.

The reliable procedure is to identify the coefficients controlled by the condition, then apply the subspace test. Requiring specified coefficients to be zero is preserved under sums and real scalar multiples. Requiring a coefficient to equal a specified nonzero number is not. In the special case of evaluation at zero, this distinction is exactly the difference between \(p(0)=0\) and \(p(0)=c\ne0\).