Vectors and Vector Spaces

Attempt these from memory before reading.

• What two operations define a vector space?
• How many axioms are there, and what are the two identity elements?
• Is the empty set a vector space? Why not?
• What is the smallest subspace any vector space has?
• Give three examples of vector spaces that are not $\mathbb{R}^n$.

Check your answers against the text. Prompt 3 is the one that stops people writing nonsense: an empty set has no zero vector, and the axioms cannot even be stated without one.

A definition in mathematics is often presented as a decree. The definition of a vector space is better read as the answer to a question: what is the smallest structure that makes linear combinations meaningful?

Everything on this page exists so that the phrase "$a v + b w$" has a unique answer that stays in the set. That requires two operations (add the vectors, scale by a scalar), associativity and commutativity so that order does not matter, identities so that “add nothing” and “scale by one” do something sensible, and distributivity so that scaling distributes over the two kinds of addition. The axioms are not a list to memorise — they are the minimum conditions under which the phrase you already wanted to write is well defined.

Why this is worth the abstraction. The same eight axioms are satisfied by arrows in space, by polynomials, by continuous functions, by solutions of a differential equation, and by $n$-tuples of any field. Nothing in the theory refers to what the vectors are — only to what you can do with them. One proof covers all of them, which is the entire point of axiomatising.

The non-obvious example is the most useful one. The set of solutions of a homogeneous linear differential equation is a vector space: adding two solutions gives a solution, and scaling one gives a solution. That single observation is why the solution space of a homogeneous ODE has a dimension, and why “general solution” means “linear combination of independent solutions”. Without the vector space axioms, that sentence has no meaning.

Exercise. Check that the set of polynomials of degree at most 3 is a vector space, and find its dimension. Then check that the polynomials of degree exactly 3 are not, and identify which axiom fails.

A vector space over a field $F$ ( $\mathbb{R}$ or $\mathbb{C}$) is a set $V$ equipped With two operations:

1. Vector addition: $+ : V \times V \to V$
2. Scalar multiplication: $\cdot : F \times V \to V$

Satisfying the following axioms for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and all $\alpha, \beta \in F$:

1. Commutativity: $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$
2. Associativity of addition: $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$
3. Additive identity: There exists $\mathbf{0} \in V$ such that $\mathbf{v} + \mathbf{0} = \mathbf{v}$
4. Additive inverse: For each $\mathbf{v}$ There exists $-\mathbf{v}$ such that $\mathbf{v} + (-\mathbf{v}) = \mathbf{0}$
5. Compatibility of scalar multiplication: $\alpha(\beta \mathbf{v}) = (\alpha\beta)\mathbf{v}$
6. Identity element of scalar multiplication: $1 \cdot \mathbf{v} = \mathbf{v}$
7. Distributivity over vector addition: $\alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v}$
8. Distributivity over scalar addition: $(\alpha + \beta)\mathbf{v} = \alpha\mathbf{v} + \beta\mathbf{v}$

Intuition. The abstract definition captures the algebraic structure shared by diverse objects: Geometric arrows, polynomials, functions, matrices. The axioms encode exactly what we need for Linear combinations to behave reasonably.

Example 1. $\mathbb{R}^n$ with component-wise addition and scalar multiplication is a vector space Over $\mathbb{R}$.

Example 2. The set $\mathcal{P}_n$ of all polynomials of degree at most $n$ with real coefficients, With the usual polynomial addition and scalar multiplication, is a vector space over $\mathbb{R}$. Its dimension is $n + 1$ With standard basis $\{1, x, x^2, \ldots, x^n\}$.

Example 3. The set $C[a,b]$ of all continuous real-valued functions on $[a,b]$ With point-wise Addition and scalar multiplication, is a vector space over $\mathbb{R}$. This space is Infinite-dimensional.

Example 4. The set $\mathcal{M}_{m \times n}(\mathbb{R})$ of all $m \times n$ real matrices is a Vector space over $\mathbb{R}$.

Example 5 (Function spaces). The set $\mathcal{F}(\mathbb{R}, \mathbb{R})$ of all functions $f : \mathbb{R} \to \mathbb{R}$ is a vector space over $\mathbb{R}$ under point-wise addition $(f + g)(x) = f(x) + g(x)$ and scalar multiplication $(\alpha f)(x) = \alpha \cdot f(x)$. The spaces $C^k(\mathbb{R})$ of $k$-times continuously differentiable functions and $L^2[a,b]$ of square-integrable functions are important subspaces of $\mathcal{F}(\mathbb{R}, \mathbb{R})$.

Example 6 (Sequence spaces). The set $\ell^2$ of all real sequences $(a_1, a_2, a_3, \ldots)$ With $\sum_{n=1}^{\infty} a_n^2 \lt \infty$ is a vector space over $\mathbb{R}$. This is the Infinite-dimensional analogue of $\mathbb{R}^n$ and is fundamental in functional analysis.

A subspace $W$ of a vector space $V$ is a subset $W \subseteq V$ that is itself a vector space under the same operations.

Theorem 1.1 (Subspace Criterion). A non-empty subset $W \subseteq V$ is a subspace if and only If for all $\mathbf{u}, \mathbf{v} \in W$ and all $\alpha \in F$:

1. $\mathbf{u} + \mathbf{v} \in W$ (closed under addition)
2. $\alpha \mathbf{u} \in W$ (closed under scalar multiplication)

Proof. If $W$ is a subspace, closure is immediate from the definition. Conversely, if $W$ is non-empty and closed under both operations, pick $\mathbf{u} \in W$. Then $-\mathbf{u} = (-1)\mathbf{u} \in W$ By closure under scalar multiplication, and $\mathbf{u} + (-\mathbf{u}) = \mathbf{0} \in W$ by closure Under addition. The remaining axioms are inherited from $V$. $\blacksquare$

Proposition 1.2 (Closure under Linear Combinations). If $W$ is a subspace of $V$ Then $W$ is Closed under all finite linear combinations: for all $\mathbf{v}_1, \ldots, \mathbf{v}_k \in W$ and all $\alpha_1, \ldots, \alpha_k \in F$

Proof. We proceed by induction on $k$. For $k = 1$, $\alpha_1 \mathbf{v}_1 \in W$ by closure under Scalar multiplication. Assume the result holds for $k - 1$ vectors. Then

By the inductive hypothesis, $\alpha_1 \mathbf{v}_1 + \cdots + \alpha_{k-1} \mathbf{v}_{k-1} \in W$ And $\alpha_k \mathbf{v}_k \in W$ by closure under scalar multiplication. Their sum is in $W$ by Closure under addition. $\blacksquare$

Example 7. The set of all solutions to the homogeneous equation $A\mathbf{x} = \mathbf{0}$ forms a Subspace of $\mathbb{R}^n$ Called the null space of $A$.

Problem. Determine whether each of the following subsets of $\mathbb{R}^3$ is a subspace.

(a) $W_1 = \{(x, y, z) \in \mathbb{R}^3 : x + 2y - z = 0\}$

(b) $W_2 = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 = 1\}$

(c) $W_3 = \{(x, y, z) \in \mathbb{R}^3 : x = 0 \mathrm{~and~} y = z\}$

Problem. Let $U = \{(x, y, z) \in \mathbb{R}^3 : z = 0\}$ (the $xy$-plane) and $W = \{(x, y, z) \in \mathbb{R}^3 : x = 0\}$ (the $yz$-plane). Find $U + W$ and $U \cap W$ And verify the dimension formula.

The eight axioms of a vector space are not arbitrary rules; they encode exactly the algebraic structure needed for the notion of “linear combination” to be meaningful. When you write $\alpha \mathbf{u} + \beta \mathbf{v}$, you are performing two geometric operations: stretching ($\alpha \mathbf{u}$) and then adding ($+ \beta \mathbf{v}$). The axioms guarantee that this operation behaves as your geometric intuition demands:

• Commutativity ($\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$) means that the order in which you place two arrows head-to-tail does not matter. Geometrically, the parallelogram rule gives the same diagonal regardless of which side you traverse first.
• Associativity ($(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$) means that when adding three vectors, the grouping is irrelevant. You can slide parentheses freely without changing the result.
• Additive identity ($\mathbf{0}$) means there exists a “do nothing” element: adding the zero vector leaves any vector unchanged, just as a displacement of zero does not move you.
• Additive inverse ($-\mathbf{v}$) means every displacement can be undone. If $\mathbf{v}$ moves you from $A$ to $B$, then $-\mathbf{v}$ moves you back from $B$ to $A$.
• Distributivity ($\alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v}$) means scaling distributes over addition: stretching the diagonal of a parallelogram is the same as stretching each side and then forming the diagonal.

The power of the abstract definition is that it applies to objects that have no obvious geometric meaning as arrows. The set of $2 \times 2$ matrices, the set of polynomials of degree $\leq 3$, and the set of continuous functions on $[0,1]$ all satisfy the same axioms. This means that any theorem proved using only the vector space axioms (such as the dimension formula or the rank-nullity theorem) automatically applies to all these seemingly unrelated objects.

Connection to other areas. Vector spaces are the stage on which linear algebra, functional analysis, and quantum mechanics are performed. In quantum mechanics, the state of a physical system is a vector in a complex Hilbert space (a complete inner product vector space), and observables are linear operators on that space. The superposition principle in physics is precisely the statement that linear combinations of state vectors are again valid states.

Problem. Let $V = C[0,1]$ be the vector space of continuous functions on $[0,1]$. Show that the subset $W = \{f \in C[0,1] : f(0) = 0\}$ is a subspace of $V$.

Problem. Let $A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{pmatrix}$. Find a basis for $\mathrm{null}(A)$ and verify it is a subspace of $\mathbb{R}^3$.

• The empty set is not a vector space. The subspace criterion requires the subset to be non-empty. The trivial subspace $\{\mathbf{0}\}$ is the smallest subspace of any vector space.
• Non-homogeneous conditions do not define subspaces. The set of solutions to $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} \neq \mathbf{0}$ is not a subspace (it is an affine subspace, or coset of the null space). For example, $\{(x, y) : x + y = 1\}$ is a line in $\mathbb{R}^2$ that does not pass through the origin.
• Closure must hold for all scalars. A set that is closed under addition and multiplication by positive scalars is not necessarily a subspace; it must also be closed under multiplication by $-1$. For example, the first quadrant $\{(x, y) : x \geq 0, y \geq 0\}$ is closed under addition and positive scalar multiplication, but $(-1) \cdot (1, 1) = (-1, -1)$ is not in the first quadrant.
• Do not confuse the field with the vector space. $\mathbb{R}$ is a field, but it is also a vector space over itself (of dimension 1). $\mathbb{R}^2$ is a vector space over $\mathbb{R}$ (of dimension 2), but it is not a field (you cannot multiply arbitrary pairs of vectors).
• Dimension is a property of the space, not the embedding. The polynomial space $\mathcal{P}_3(\mathbb{R})$ has dimension 4, even though its elements are “logically” embedded in $C(\mathbb{R})$, which is infinite-dimensional. The dimension depends on the vector space structure, not on how the space sits inside a larger space.

A vector space is any collection of objects that can be added together and scaled by numbers in the same way that arrows in space can. The power of the abstraction is that it applies far beyond geometry. Polynomials behave like vectors: you can add two polynomials and multiply one by a scalar. Functions behave like vectors: you can add two functions pointwise and multiply one by a constant. Even sequences of numbers form vector spaces. The axioms are directly the rules that make “linear combinations” meaningful — they guarantee that scaling and adding behave the way your geometric intuition expects.

The connection to function spaces is where the abstraction pays off most. In quantum mechanics, the state of a particle is a vector in an infinite-dimensional function space (a Hilbert space). The Schrodinger equation is a linear operator acting on these vectors, and solving it is structurally identical to solving a system of linear equations — except the “matrix” is replaced by a differential operator and the “vector” is a wavefunction. This is why linear algebra is the language of quantum theory: the superposition principle is literally the closure of a vector space under linear combinations.

The third row is the classic, and it is worth seeing why: the union of the $x$-axis and the $y$-axis contains $(1,0)$ and $(0,1)$ but not their sum. The sum of subspaces is a subspace; the union almost never is.

The seventh row is the one that matters for the rest of mathematics. A vector space has no notion of length, angle or order. Those come later — inner product spaces, normed spaces, ordered fields — and confusing an added structure with part of the definition is how people end up trying to take “intra-vector” dot products before an inner product has been defined.

• Infinite-dimensional spaces. Everything here carries over, but bases become unconstructible and dimension becomes a cardinal. That is the boundary between algebra and functional analysis, and it is crossed on the functional analysis track.
• Fields other than $\mathbb{R}$ and $\mathbb{C}$. Vector spaces over a finite field underpin error-correcting codes and cryptography, and the theory changes character when the field is not ordered.
• Modules. Drop the requirement that the scalars form a field and keep a ring, and bases can stop existing entirely. The structure theorem for finitely generated modules over a principal ideal domain is what survives, and it is what produces the Jordan form.
• Category theory. Vector spaces and linear maps form a category, and the whole of linear algebra can be restated as the statement that this category has certain properties. That restatement is what makes quantum information theory possible to write down.

Mix these with Linear Independence, Span, Basis and Dimension, Systems of Linear Equations, Newtonian Mechanics Review and Functional Analysis.

1. (Axioms.) Verify that the set of all $2\times2$ matrices is a vector space over $\mathbb{R}$, listing each axiom as you check it. Then show that the invertible ones are not, by exhibiting a failure of closure.
2. (Subspaces.) Show that the set of polynomials with $p(0) = 0$ is a subspace of the polynomial space, and find its dimension.
3. (Subspaces.) Give two subspaces of $\mathbb{R}^2$ whose union is not a subspace, and compute the smallest subspace containing both.
4. (Null space.) Show that the solution set of a homogeneous system $Ax = 0$ is a subspace, and identify exactly which step fails when $b \neq 0$.
5. (Functional analysis.) Show that the set of solutions of $y'' + y = 0$ is a two-dimensional vector space, and find a basis. Then explain why this makes “general solution” a precise statement.
6. (Research.) Show that the axiom “every vector space has a basis” implies the axiom of choice. This is why infinite-dimensional bases are unconstructible and why analysis avoids them.

• Linear Independence, Span, Basis, and Dimension: Basis and dimension are defined in terms of linear independence and span of vector space elements.
• Matrices: Matrices represent linear transformations between finite-dimensional vector spaces.
• Linear Transformations: Linear transformations are maps between vector spaces that preserve linear structure.
• Quantum Mechanics
• Graph Theory

• Eigenvalues and Eigenvectors
• Inner Product Spaces
• Linear Independence, Span, Basis, and Dimension
• Matrices