Adding Polynomial Subspaces

What a Sum of Polynomial Subspaces Contains

A sum of polynomial subspaces collects every polynomial that can be formed by adding one polynomial from each subspace. To compute the sum, it is not enough to list some polynomials that can be made: we need a precise description of all the possibilities. Coefficients provide a direct method. They show which powers can occur in each summand, which powers can occur in the result, and how to build the two summands for a given candidate polynomial.

Work over \(\mathbb{R}\). For a nonnegative integer \(n\), let \(P_n\) denote the vector space of real polynomials of degree at most \(n\), together with the zero polynomial. Each polynomial in \(P_n\) has a unique expression \(a_0+a_1x+\cdots+a_nx^n\), where any of the coefficients may be zero. We will also use subspaces in which only even powers or only odd powers are allowed.

A decomposition need not be unique: different choices of \(u\) and \(w\) can sometimes give the same polynomial. For membership, the essential task is to find at least one valid pair. To describe the whole sum, we must also show that every polynomial in it has the stated form and that every polynomial of the stated form can actually be decomposed.

Sums with Degree Bounds

When both subspaces allow every power up to a degree bound, the larger bound determines the sum. More precisely, for nonnegative integers \(m\) and \(n\), \(P_m+P_n=P_{\max(m,n)}\). A polynomial from either space already belongs to the space with the larger degree bound; conversely, any polynomial in the larger space can be used as one summand, with zero as the other.

Proof. Set \(r=\max(m,n)\). If \(q\in P_m+P_n\), then \(q=u+w\) for some \(u\in P_m\) and \(w\in P_n\). Both \(u\) and \(w\) have degree at most \(r\), so their sum also has degree at most \(r\), and \(q\in P_r\). For the reverse inclusion, suppose \(q\in P_r\). If \(r=m\), then \(q\in P_m\) and \(0\in P_n\), so \(q=q+0\in P_m+P_n\). If \(r=n\), then \(0\in P_m\) and \(q\in P_n\), so \(q=0+q\in P_m+P_n\). These cases include \(m=n\), and prove the equality. \(\square\)

The coefficient split in this example gives a useful habit: before trying to find summands, write the candidate in powers of \(x\) and group each term according to which subspace permits it. If every term can be assigned, the grouped expressions give a decomposition. If a term cannot be assigned to either subspace, it may obstruct membership.

Sums with Even- and Odd-Power Restrictions

A polynomial with only even powers has the form \(a_0+a_2x^2+a_4x^4+\cdots\); a polynomial with only odd powers has the form \(a_1x+a_3x^3+a_5x^5+\cdots\). These descriptions concern the powers that appear, not whether the polynomial's degree is even or odd. For example, \(x^2+x^3\) has both an even-power term and an odd-power term.

When adding \(E_m\) and \(O_n\), even and odd terms have different sources: even terms can come from the first summand, and odd terms from the second. The degree bounds may differ, so a high even power is not automatically allowed just because an odd power of the same size is allowed, or vice versa.

Proof. First suppose \(q\in E_m+O_n\). Then \(q=u+w\) for some \(u\in E_m\) and \(w\in O_n\). The polynomial \(u\) has no even-power term above degree \(m\), and \(w\) has no odd-power term above degree \(n\). Also, \(u\) has no odd-power terms and \(w\) has no even-power terms. Therefore the coefficient of every even power \(k>m\) and every odd power \(k>n\) in \(q\) is zero. Conversely, suppose those coefficients of \(q\) are zero. Let \(u\) be the sum of all even-power terms of \(q\) with exponents at most \(m\), and let \(w\) be the sum of all odd-power terms of \(q\) with exponents at most \(n\). The assumed zero coefficients ensure that \(q\) has no other terms. Thus \(q=u+w\), with \(u\in E_m\) and \(w\in O_n\), so \(q\in E_m+O_n\). \(\square\)

Common Mistake: Assigning Terms by Degree Instead of Power

A frequent error is to decide which summand receives a term by looking only at the degree of the whole polynomial, or by treating a polynomial of odd degree as if all its terms were odd powers. For instance, \(x^2+x^3\) has degree \(3\), but it is not an element of \(O_3\), because its \(x^2\) term is an even power. The restriction is on each power appearing, not just on the highest power.

Here is a wrong decomposition for the polynomial \(q(x)=1+x^2+x^3\) in \(E_2+O_3\): assigning \(q\) itself to the odd-power summand claims \(q\in O_3\). That is incorrect because \(O_3\) contains no \(x^2\) term. The correction is to group by the powers allowed in each set:

The corrected summands add back to \(1+x^2+x^3\). Whenever a decomposition seems uncertain, check both conditions separately: each proposed summand must belong to its assigned subspace, and their sum must equal the candidate polynomial.

How to Check a Sum Description

There are two complementary checks. To establish that a proposed set describes the whole sum, show that any sum of allowed elements satisfies the proposed coefficient restrictions. Then take an arbitrary polynomial satisfying those restrictions and construct its two summands by grouping the permitted terms. The first direction prevents the description from being too large; the second prevents it from being too small.

For a particular candidate, the same logic becomes a practical procedure:

This coefficient-based approach makes explicit what a sum of polynomial subspaces permits. Degree bounds control how high a power can occur; parity restrictions control which powers can occur; and a decomposition certifies that a particular polynomial is in the sum. The zero polynomial causes no exception: all its coefficients are zero, and it can always be written as \(0+0\).