Polynomials as a Vector Space
Consider the set $\mathbb{P}_n$, the real polynomials of degree $\le n$, expressed with $n+1$ scalar coefficients. With standard polynomial addition and scalar multiplication, $\mathbb{P}_n$ satisfies all vector space axioms: addition is associative and commutative, the zero polynomial is the additive identity, each polynomial has an additive inverse, scalar multiplication has 1 as identity and is associative, and both distributivity laws hold. Thus $\mathbb{P}_n$ is a vector space over the reals.
Basis and Linear Independence
Linear independence in $\mathbb{P}_n$ follows the usual definition: a set $\{p_1, \dots, p_k\}$ is linearly independent if the only linear combination equal to the zero polynomial has all coefficients zero. The fundamental set $\{1, x, x^2, \dots, x^n\}$ is linearly independent because each polynomial is uniquely determined by its coefficients, and this set spans the entire space by definition. This is the monomial basis.
Verifying a Candidate Basis
To check if an arbitrary set forms a basis, we use the same matrix techniques as for $\mathbb{R}^n$. As a concrete example, suppose we want to test whether the set of polynomials $p_1 = 4x^3 - 2x^2 + 3$, $p_2 = x^3 - 2x^2 + 5$, $p_3 = x^3 - 1$, and $p_4 = 3x^3 - 5x^2 + 2x + 1$ is a basis for $\mathbb{P}_3$.
For linear independence, set a linear combination equal to zero and group by powers of $x$. Each coefficient must vanish — in matrix form, this reduces to the identity matrix, which forces all combination coefficients to be zero, so the set is linearly independent. For spanning, express an arbitrary polynomial $a_3 x^3 + a_2 x^2 + a_1 x + a_0$ as a combination of the candidate basis. The coefficient matrix is identical to the previous one, so a unique solution always exists — $\mathbb{P}_3$ is spanned.
Defining an Inner Product
Since polynomials are functions, we can equip $\mathbb{P}_n$ with an integral-based inner product. But most polynomials are not square-integrable over infinite intervals, so we restrict to finite bounds, say $[-1, 1]$, and set the weight function to 1. The inner product becomes
$$\langle p, q \rangle = \int_{-1}^{1} p(x) q(x)\, dx.$$
This satisfies the inner product space conditions. Conjugate symmetry follows from commutativity of real multiplication. Linearity in the first argument is a direct consequence of the linearity of integration. For positive-definiteness: $\langle p, p \rangle \ge 0$ because $p^2(x) \ge 0$. If the integral equals zero, the continuous non-negative function $p^2$ must be zero everywhere — since a polynomial can only be zero on an interval if it is the zero polynomial, positive-definiteness holds. Hence $\mathbb{P}_n$ is an inner product space under this definition.
Orthogonality
Two polynomials are orthogonal with respect to this inner product iff $\langle p, q \rangle = 0$. Despite intuition from the monomial basis, these basis polynomials are not orthogonal with respect to the $[-1,1]$ integral inner product. For example,
$$\langle 1, x \rangle = \int_{-1}^{1} x\, dx = 0, \quad \text{but} \quad \langle x, x^2 \rangle = \int_{-1}^{1} x^3\, dx = 0, \quad \text{whereas} \quad \langle 1, x^2 \rangle = \int_{-1}^{1} x^2\, dx = \frac{2}{3} \neq 0.$$
Other polynomial sets, such as the Legendre polynomials, do form orthogonal families under this inner product, but that topic lies beyond the present scope.
