Polynomials as Coefficient Lists
Polynomials provide a useful example of a vector space whose elements are not ordinary coordinate columns, even though their operations can be checked by tracking coefficients. The key distinction is between the space of all real polynomials, where each polynomial has some finite degree, and a space with one fixed degree bound. Both are vector spaces under the usual operations, but the reason that the degree-bounded spaces remain closed under those operations deserves care.
We work over the scalar field \(\mathbb{R}\), with polynomial variable \(x\). A real polynomial is a finite expression \(p(x)=a_0+a_1x+\cdots+a_mx^m\), where each coefficient \(a_k\) is real. Equivalently, we can write \(p(x)=\sum_{k=0}^{m}a_kx^k\). Two polynomials are equal precisely when their coefficients at every power of \(x\) are equal. We may extend a coefficient list by zeros: for example, a polynomial of degree at most \(m\) can be regarded as having coefficient \(a_k=0\) for every \(k>m\). The zero polynomial, denoted \(0\), has every coefficient equal to zero.
These definitions are coefficientwise: add coefficients at matching powers, or multiply every coefficient by the scalar. They agree with the usual algebraic rules for polynomials. To verify that \(\mathbb{R}[x]\) is a real vector space, we must check that these operations stay within the set and satisfy the real vector space axioms. In detail, addition must be closed, commutative, and associative; there must be a zero element and an additive inverse for each polynomial; scalar multiplication must be closed; and the two distributive laws, scalar associativity, and the identity-scalar law must hold.
Verifying the Axioms in the Full Polynomial Space
Proof. Let \(p,q,r\in\mathbb{R}[x]\), and let \(a,b\in\mathbb{R}\). Write their coefficients as \(p_k,q_k,r_k\), taking a coefficient to be zero wherever the corresponding power does not occur. Each polynomial has only finitely many nonzero coefficients. The sum \(p+q\) has coefficients \(p_k+q_k\), which can be nonzero only at powers appearing in \(p\) or \(q\). Thus \(p+q\) is a polynomial, proving closure under addition. For every \(k\), \((p+q)_k=p_k+q_k=q_k+p_k=(q+p)_k\), so \(p+q=q+p\). Also, \(((p+q)+r)_k=(p_k+q_k)+r_k=p_k+(q_k+r_k)=(p+(q+r))_k\), so addition is associative. The polynomial \(0\), whose coefficients are all zero, satisfies \((p+0)_k=p_k+0=p_k\) for every \(k\), hence \(p+0=p\). The polynomial \(-p\) with coefficients \(-p_k\) is a polynomial and satisfies \((p+(-p))_k=p_k-p_k=0\) for every \(k\); therefore \(p+(-p)=0\).
For scalar multiplication, \(ap\) has coefficients \(ap_k\), which are nonzero at only finitely many powers, so \(ap\) is a polynomial. At every power \(k\), the real-number laws give \((a(p+q))_k=a(p_k+q_k)=ap_k+aq_k=(ap+aq)_k\), and \(((a+b)p)_k=(a+b)p_k=ap_k+bp_k=(ap+bp)_k\). Thus both distributive laws hold. The same coefficient check gives \((a(bp))_k=a(bp_k)=(ab)p_k=((ab)p)_k\), so scalar multiplication is associative, and \((1p)_k=1p_k=p_k\), so multiplication by the scalar \(1\) leaves \(p\) unchanged. Equality of every coefficient establishes each asserted polynomial equality. These are all the vector space axioms over \(\mathbb{R}\). \(\square\)
This verification does not require a fixed maximum degree. A polynomial has finitely many nonzero coefficients, and adding two finite coefficient lists or scaling one of them still gives a finite coefficient list. The position of the highest nonzero coefficient may vary from one polynomial to another.
Worked Example: Adding and Scaling Polynomials
Let \(p(x)=2x^3-x^2+4\) and \(q(x)=-x^3+3x^2+2x-1\). Compute \(p+q\) and \(2p-q\) by matching coefficients.
For the sum, the coefficients of \(x^3,x^2,x,\) and \(1\) are, respectively, \(2+(-1)=1\), \(-1+3=2\), \(0+2=2\), and \(4+(-1)=3\). Hence \(p(x)+q(x)=x^3+2x^2+2x+3\).
For \(2p-q\), first \(2p(x)=4x^3-2x^2+8\). Subtracting \(q(x)\) gives \[ (4x^3-2x^2+0x+8)-(-x^3+3x^2+2x-1) =5x^3-5x^2-2x+9. \] The coefficients at powers \(3,2,1,0\) are \(5,-5,-2,9\), so this is a real polynomial, as required by closure.
Check: Adding the coefficient lists directly for \(2p-q\) gives \((4-(-1),-2-3,0-2,8-(-1))=(5,-5,-2,9)\), which reproduces \(5x^3-5x^2-2x+9\).
The coefficient checks in the theorem establish more than closure: they show why the familiar laws hold. For instance, the distributive law \(a(p+q)=ap+aq\) reduces at each power to the real-number identity \(a(p_k+q_k)=ap_k+aq_k\). Equality of polynomials then follows from equality of their coefficients. This coefficient-by-coefficient reasoning is the central method for verifying the axioms in this setting.
Degree-Bounded Polynomial Spaces
A different space is obtained by fixing one nonnegative integer \(n\) and allowing powers only from \(x^0\) through \(x^n\). We define this set by its coefficients, so the zero polynomial is included without needing to assign it a degree.
Proof. Let \(p,q\in\mathcal{P}_n\), so \(p_k=q_k=0\) for every \(k>n\). The coefficient of \(x^k\) in \(p+q\) is \(p_k+q_k=0\) for \(k>n\), so \(p+q\in\mathcal{P}_n\). For any \(a\in\mathbb{R}\), the coefficient of \(x^k\) in \(ap\) is \(ap_k=0\) for \(k>n\), so \(ap\in\mathcal{P}_n\). The zero polynomial belongs to \(\mathcal{P}_n\), and if \(p\in\mathcal{P}_n\), then \(-p\in\mathcal{P}_n\) because its coefficients above \(n\) are also zero. The vector space axioms hold for these polynomials because the operations are the same as in \(\mathbb{R}[x]\), where the axioms have been verified; every expression in each axiom remains in \(\mathcal{P}_n\) by the closure checks just given. Therefore \(\mathcal{P}_n\) is a vector space over \(\mathbb{R}\). \(\square\)
The two spaces differ in the degree bounds they impose. Every element of \(\mathcal{P}_n\) is in \(\mathbb{R}[x]\), but \(\mathbb{R}[x]\) has no one fixed upper bound on the powers allowed: for any specified \(n\), the polynomial \(x^{n+1}\) is in \(\mathbb{R}[x]\) and not in \(\mathcal{P}_n\). This does not prevent \(\mathbb{R}[x]\) from being a vector space. Its elements individually have finite coefficient lists; the space simply does not require the same list length for every element.
Worked Example: Checking Operations in \(\mathcal{P}_3\)
Take \(p(x)=2x^3-x+1\) and \(q(x)=-2x^3+x^2+4\), both in \(\mathcal{P}_3\). Compute their sum and a scalar multiple, then check that neither result uses a power above \(x^3\).
Write the coefficients from powers \(3\) down to \(0\): \(p\) has \((2,0,-1,1)\) and \(q\) has \((-2,1,0,4)\). Adding gives \((2+(-2),0+1,-1+0,1+4)=(0,1,-1,5)\), so \[ (p+q)(x)=x^2-x+5. \] For the scalar multiple, \(3p\) has coefficients \((3\cdot2,3\cdot0,3\cdot(-1),3\cdot1)=(6,0,-3,3)\), so \[ (3p)(x)=6x^3-3x+3. \] Both results have zero coefficients at every power greater than \(3\), and therefore both belong to \(\mathcal{P}_3\).
Check: The \(x^3\) coefficient of \(p+q\) is \(2-2=0\), and the \(x^3\) coefficient of \(3p\) is \(6\). Their remaining coefficients are exactly those displayed, so the coefficient lists verify both calculations and the degree bound.
Worked Example: A Polynomial Outside a Fixed Bound
Compare \(r(x)=4x^2-3x+2\) with the spaces \(\mathbb{R}[x]\), \(\mathcal{P}_2\), and \(\mathcal{P}_1\).
The polynomial \(r\) is a finite sum with real coefficients, so \(r\in\mathbb{R}[x]\). Its coefficients at powers above \(2\) are all zero, so \(r\in\mathcal{P}_2\). Its coefficient of \(x^2\) is \(4\ne0\), so it cannot be written using powers only from \(x^0\) through \(x^1\); hence \(r\notin\mathcal{P}_1\).
By contrast, \(s(x)=x^3\) belongs to \(\mathbb{R}[x]\), but \(s\notin\mathcal{P}_2\), because its coefficient at \(x^3\) is \(1\ne0\). This checks directly that a fixed degree bound excludes some polynomials even though all polynomials remain in the full space.
A Common Pitfall: The Degree Can Drop Under Addition
A tempting but incorrect rule is \(\deg(p+q)=\max(\deg p,\deg q)\) for all nonzero polynomials \(p,q\). The highest powers can cancel, so the degree of a sum may be smaller. For a concrete counterexample, take \(p(x)=x^2+x\) and \(q(x)=-x^2+1\). Both have a nonzero \(x^2\) coefficient, but \[ p(x)+q(x)=(x^2+x)+(-x^2+1)=x+1. \] The \(x^2\) coefficient is \(1+(-1)=0\), so the sum has no \(x^2\) term.
The correction is that if \(p\) and \(q\) have powers at most \(n\), then their sum also has powers at most \(n\): at every \(k>n\), both coefficients are zero, and their sum is zero. Cancellation may lower the highest power; it cannot create a nonzero coefficient above the larger of the powers already present. This is exactly why a fixed degree bound is preserved by addition, without requiring the degree of the sum to equal the maximum of the two input degrees.
A related point is that \(\mathcal{P}_n\) means degree at most \(n\), not degree exactly \(n\). For example, the zero polynomial and the constant polynomial \(5\) both belong to \(\mathcal{P}_2\), even though neither has a nonzero \(x^2\) coefficient. Describing \(\mathcal{P}_n\) through its allowed coefficients makes this inclusion precise and handles the zero polynomial naturally.
Why Both Polynomial Spaces Matter
The full space \(\mathbb{R}[x]\) is useful when polynomials of any finite degree must be allowed. A degree-bounded space \(\mathcal{P}_n\) is useful when the problem has one specified maximum power. In either case, the vector space verification has the same foundation: the zero polynomial supplies the additive identity, coefficientwise negation supplies additive inverses, and the real-number laws operate independently at each power of \(x\).
When checking a proposed polynomial space, keep two questions separate. First, are the operations the usual polynomial operations? Second, does the chosen set remain inside itself after addition and scalar multiplication? For \(\mathbb{R}[x]\), finiteness of each polynomial's coefficient list answers the closure question. For \(\mathcal{P}_n\), the coefficients above the fixed bound remain zero after either operation. Once closure is secured, the axioms follow from the coefficient calculations and the real-number laws.
Check Your Understanding
Use coefficient lists and the stated degree bounds to justify each answer.
- For \(p(x)=x^3-2x+1\) and \(q(x)=2x^3+x^2-4\), compute \(p+q\) and \(2p-q\).
- Explain why the zero polynomial belongs to \(\mathcal{P}_n\) for every nonnegative integer \(n\).
- For a fixed nonnegative integer \(n\), show from the coefficients that if \(p,q\in\mathcal{P}_n\), then \(p+q\in\mathcal{P}_n\).
- Give a polynomial in \(\mathbb{R}[x]\) that is not in \(\mathcal{P}_4\), and identify its nonzero coefficient above the permitted bound.
- Give an example where adding two polynomials with a nonzero \(x^2\) coefficient produces a sum with zero \(x^2\) coefficient.
