Real and Complex Vector Spaces

Definition In this course the field of scalars $\mathbb{F}$ is either the real numbers $\mathbb{R}$ or the complex numbers $\mathbb{C}$. A result stated over $\mathbb{F}$ holds for both, and the field is named whenever the answer depends on it. A complex number has the form $z = a + bi$ with $a, b \in \mathbb{R}$, where $i^2 = -1$. Its real part is $\operatorname{Re} z = a$ and its imaginary part is $\operatorname{Im} z = b$, and two complex numbers are equal exactly when both parts agree. Sums and products follow the usual rules of algebra, with $i^2$ replaced by $-1$,

Both $\mathbb{R}$ and $\mathbb{C}$ are fields, number systems in which addition and multiplication obey the usual rules of algebra and every nonzero number has a reciprocal. A vector space over $\mathbb{F}$ allows vectors to be multiplied only by elements of $\mathbb{F}$, so linear combinations, spans, independence, subspaces and dimension are all taken with coefficients in $\mathbb{F}$.

Definition The complex conjugate of $z = a + bi$, with $a, b \in \mathbb{R}$, is $\overline{z} = a - bi$. Since $z\overline{z} = (a + bi)(a - bi) = a^2 + b^2$, the modulus of $z$ is the nonnegative real number

For all $z, w \in \mathbb{C}$ the rules $\overline{z + w} = \overline{z} + \overline{w}$, $\overline{zw} = \overline{z}\,\overline{w}$ and $\lvert zw \rvert = \lvert z \rvert \lvert w \rvert$ hold, and $\lvert z \rvert = 0$ exactly when $z = 0$. A complex number $z$ is real exactly when $\overline{z} = z$. For $z \ne 0$ the reciprocal is

Definition A vector space over $\mathbb{F}$ is a set $V$ with an addition, which gives a vector $u + v \in V$ for all $u, v \in V$, and a scalar multiplication, which gives a vector $av \in V$ for all $a \in \mathbb{F}$ and $v \in V$, such that the following hold for all $u, v, w \in V$ and $a, b \in \mathbb{F}$.

• Addition is commutative and associative, $u + v = v + u$ and $(u + v) + w = u + (v + w)$.
• There is a vector $0 \in V$ with $v + 0 = v$ for every $v \in V$.
• Every $v \in V$ has an additive inverse $-v \in V$ with $v + (-v) = 0$.
• Scalar multiplication satisfies $a(bv) = (ab)v$ and $1v = v$.
• Both distributive laws hold, $a(u + v) = au + av$ and $(a + b)v = av + bv$.

A statement proved from these rules alone holds in every vector space over $\mathbb{F}$, whether its vectors are columns, matrices, polynomials or functions.

Definition Let $V$ be a vector space over $\mathbb{F}$. A subset $U \subseteq V$ is a subspace if it is a vector space over $\mathbb{F}$ under the operations of $V$. The subspace test says that $U$ is a subspace exactly when $U$ is nonempty and

The condition is sufficient. Since $U$ is nonempty it contains some vector $u$, and taking $v = u$ with $a = b = 0$ puts $0$ in $U$. For any $u, v \in U$, the choice $a = -1$ and $b = 0$ puts $(-1)u = -u$ in $U$. The choice $a = b = 1$ gives closure under addition, and the choice $b = 0$ with any $a$ gives closure under scaling. The remaining axioms hold in $U$ because they hold in $V$ for the same operations. Conversely, a subspace is nonempty, since it contains a zero vector, and it contains every such combination, since it is closed under both operations. The word nonempty cannot be dropped, because the empty set satisfies the closure condition vacuously but contains no zero vector.

Definition A basis of a vector space $V$ over $\mathbb{F}$ is a list of vectors that spans $V$ and is linearly independent, both with coefficients in $\mathbb{F}$. When $V$ has a finite basis, the dimension $\dim_{\mathbb{F}} V$ is the number of vectors in a basis, and the lesson Basis Exchange and Dimension proves that this number does not depend on the basis. Every vector space over $\mathbb{C}$ is also a vector space over $\mathbb{R}$, by allowing only real scalars. If $v_1, \dots, v_n$ is a basis of $V$ over $\mathbb{C}$, then $v_1, iv_1, \dots, v_n, iv_n$ is a basis of $V$ over $\mathbb{R}$. Indeed, a complex coefficient $a_j + b_j i$ on $v_j$, with $a_j, b_j$ real, contributes the same as the real coefficients $a_j$ on $v_j$ and $b_j$ on $iv_j$. So these $2n$ vectors span $V$ over $\mathbb{R}$. Because $v_1, \dots, v_n$ are independent over $\mathbb{C}$, a real combination of the $2n$ vectors is zero only when every $a_j + b_j i$ is zero, that is, when every $a_j$ and every $b_j$ is zero. Hence $\dim_{\mathbb{R}} V = 2\dim_{\mathbb{C}} V$ for every finite-dimensional complex vector space $V$, and in particular

Definition For an integer $n \ge 0$, the set $\mathcal{P}_n$ consists of all polynomials

with coefficients $a_0, \dots, a_n \in \mathbb{F}$, including the zero polynomial, whose coefficients are all $0$. Two polynomials are equal when their coefficients agree. Polynomials are added and multiplied by scalars coefficient by coefficient, so matching $p$ with its coefficient column $(a_0, \dots, a_n)$ turns these operations into the operations of $\mathbb{F}^{n+1}$, and $\mathcal{P}_n$ inherits every vector space axiom. The polynomials $1, t, \dots, t^n$ form a basis, so $\dim \mathcal{P}_n = n + 1$. By contrast, the polynomials of degree exactly $n$ do not form a subspace, since the zero polynomial is not among them.