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Vectors and Vector Spaces | Mathematics

---:::note[Historical Context] The vector space was introduced by Giuseppe Peano (1888), building on Grassmann’s Ausdehnungslehre (1844). The modern axiomatic approach was established by Banach (1920). Every vector space has a basis (proved via Zorn’s Lemma, 1935). Vector spaces connect to representation theory, where groups act on vector spaces.

Section titled “---:::note[Historical Context] The vector space was introduced by Giuseppe Peano (1888), building on Grassmann’s Ausdehnungslehre (1844). The modern axiomatic approach was established by Banach (1920). Every vector space has a basis (proved via Zorn’s Lemma, 1935). Vector spaces connect to representation theory, where groups act on vector spaces.”

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

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

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

  1. Commutativity: u+v=v+u\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}
  2. Associativity of addition: (u+v)+w=u+(v+w)(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})
  3. Additive identity: There exists 0V\mathbf{0} \in V such that v+0=v\mathbf{v} + \mathbf{0} = \mathbf{v}
  4. Additive inverse: For each v\mathbf{v}There exists v-\mathbf{v} such that v+(v)=0\mathbf{v} + (-\mathbf{v}) = \mathbf{0}
  5. Compatibility of scalar multiplication: α(βv)=(αβ)v\alpha(\beta \mathbf{v}) = (\alpha\beta)\mathbf{v}
  6. Identity element of scalar multiplication: 1v=v1 \cdot \mathbf{v} = \mathbf{v}
  7. Distributivity over vector addition: α(u+v)=αu+αv\alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v}
  8. Distributivity over scalar addition: (α+β)v=αv+βv(\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. Rn\mathbb{R}^n with component-wise addition and scalar multiplication is a vector space Over R\mathbb{R}.

Example 2. The set Pn\mathcal{P}_n of all polynomials of degree at most nn with real coefficients, With the usual polynomial addition and scalar multiplication, is a vector space over R\mathbb{R}. Its dimension is n+1n + 1With standard basis {1,x,x2,,xn}\{1, x, x^2, \ldots, x^n\}.

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

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

Example 5 (Function spaces). The set F(R,R)\mathcal{F}(\mathbb{R}, \mathbb{R}) of all functions f:RRf : \mathbb{R} \to \mathbb{R} is a vector space over R\mathbb{R} under point-wise addition (f+g)(x)=f(x)+g(x)(f + g)(x) = f(x) + g(x) and scalar multiplication (αf)(x)=αf(x)(\alpha f)(x) = \alpha \cdot f(x). The spaces Ck(R)C^k(\mathbb{R}) of kk-times continuously differentiable functions and L2[a,b]L^2[a,b] of square-integrable functions are important subspaces of F(R,R)\mathcal{F}(\mathbb{R}, \mathbb{R}).

Example 6 (Sequence spaces). The set 2\ell^2 of all real sequences (a1,a2,a3,)(a_1, a_2, a_3, \ldots) With n=1an2<\sum_{n=1}^{\infty} a_n^2 \lt \infty is a vector space over R\mathbb{R}. This is the Infinite-dimensional analogue of Rn\mathbb{R}^n and is fundamental in functional analysis.

A subspace WW of a vector space VV is a subset WVW \subseteq V that is itself a vector space Under the same operations.

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

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

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

Proposition 1.2 (Closure under Linear Combinations). If WW is a subspace of VV Then WW is Closed under all finite linear combinations: for all v1,,vkW\mathbf{v}_1, \ldots, \mathbf{v}_k \in W and All α1,,αkF\alpha_1, \ldots, \alpha_k \in F

α1v1+α2v2++αkvkW\alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \cdots + \alpha_k \mathbf{v}_k \in W

Proof. We proceed by induction on kk. For k=1k = 1, α1v1W\alpha_1 \mathbf{v}_1 \in W by closure under Scalar multiplication. Assume the result holds for k1k - 1 vectors. Then

α1v1++αkvk=(α1v1++αk1vk1)+αkvk\alpha_1 \mathbf{v}_1 + \cdots + \alpha_k \mathbf{v}_k = (\alpha_1 \mathbf{v}_1 + \cdots + \alpha_{k-1} \mathbf{v}_{k-1}) + \alpha_k \mathbf{v}_k

By the inductive hypothesis, α1v1++αk1vk1W\alpha_1 \mathbf{v}_1 + \cdots + \alpha_{k-1} \mathbf{v}_{k-1} \in W And αkvkW\alpha_k \mathbf{v}_k \in W by closure under scalar multiplication. Their sum is in WW by Closure under addition. \blacksquare

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

1.4 Worked Example: Verifying Subspace Criteria

Section titled “1.4 Worked Example: Verifying Subspace Criteria”

Problem. Determine whether each of the following subsets of R3\mathbb{R}^3 is a subspace.

(a) W1={(x,y,z)R3:x+2yz=0}W_1 = \{(x, y, z) \in \mathbb{R}^3 : x + 2y - z = 0\}

(b) W2={(x,y,z)R3:x2+y2=1}W_2 = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 = 1\}

(c) W3={(x,y,z)R3:x=0 and y=z}W_3 = \{(x, y, z) \in \mathbb{R}^3 : x = 0 \mathrm{~and~} y = z\}

Solution

(a) Let u=(x1,y1,z1)\mathbf{u} = (x_1, y_1, z_1) and v=(x2,y2,z2)\mathbf{v} = (x_2, y_2, z_2) be in W1W_1 So x1+2y1z1=0x_1 + 2y_1 - z_1 = 0 and x2+2y2z2=0x_2 + 2y_2 - z_2 = 0. Then

(x1+x2)+2(y1+y2)(z1+z2)=(x1+2y1z1)+(x2+2y2z2)=0+0=0(x_1 + x_2) + 2(y_1 + y_2) - (z_1 + z_2) = (x_1 + 2y_1 - z_1) + (x_2 + 2y_2 - z_2) = 0 + 0 = 0

So u+vW1\mathbf{u} + \mathbf{v} \in W_1. For αR\alpha \in \mathbb{R}

(αx1)+2(αy1)(αz1)=α(x1+2y1z1)=α0=0(\alpha x_1) + 2(\alpha y_1) - (\alpha z_1) = \alpha(x_1 + 2y_1 - z_1) = \alpha \cdot 0 = 0

So αuW1\alpha \mathbf{u} \in W_1. Since W1W_1 is non-empty (e.g., 0W1\mathbf{0} \in W_1), it is a subspace.

(b) W2W_2 is not a subspace. For instance, (1,0,0)W2(1, 0, 0) \in W_2 since 12+02=11^2 + 0^2 = 1 But 2(1,0,0)=(2,0,0)W22 \cdot (1, 0, 0) = (2, 0, 0) \notin W_2 since 22+02=412^2 + 0^2 = 4 \neq 1. So W2W_2 is not closed Under scalar multiplication.

(c) Let u=(0,a,a)\mathbf{u} = (0, a, a) and v=(0,b,b)\mathbf{v} = (0, b, b) be in W3W_3. Then u+v=(0,a+b,a+b)W3\mathbf{u} + \mathbf{v} = (0, a + b, a + b) \in W_3 and αu=(0,αa,αa)W3\alpha \mathbf{u} = (0, \alpha a, \alpha a) \in W_3. Since (0,0,0)W3(0, 0, 0) \in W_3It is a non-empty subspace.

\blacksquare

1.5 Worked Example: Sum and Intersection of Subspaces

Section titled “1.5 Worked Example: Sum and Intersection of Subspaces”

Problem. Let U={(x,y,z)R3:z=0}U = \{(x, y, z) \in \mathbb{R}^3 : z = 0\} (the xyxy-plane) and W={(x,y,z)R3:x=0}W = \{(x, y, z) \in \mathbb{R}^3 : x = 0\} (the yzyz-plane). Find U+WU + W and UWU \cap W And verify the dimension formula.

Solution

UU has basis {(1,0,0),(0,1,0)}\{(1, 0, 0), (0, 1, 0)\} and dim(U)=2\dim(U) = 2. WW has basis {(0,1,0),(0,0,1)}\{(0, 1, 0), (0, 0, 1)\} and dim(W)=2\dim(W) = 2.

UW={(x,y,z):z=0 and x=0}={(0,y,0):yR}U \cap W = \{(x, y, z) : z = 0 \mathrm{~and~} x = 0\} = \{(0, y, 0) : y \in \mathbb{R}\} Which has basis {(0,1,0)}\{(0, 1, 0)\} and dimension 1.

U+W=span{(1,0,0),(0,1,0),(0,1,0),(0,0,1)}=span{(1,0,0),(0,1,0),(0,0,1)}=R3U + W = \mathrm{span}\{(1,0,0), (0,1,0), (0,1,0), (0,0,1)\} = \mathrm{span}\{(1,0,0), (0,1,0), (0,0,1)\} = \mathbb{R}^3 So dim(U+W)=3\dim(U + W) = 3.

Verify: dim(U+W)=dim(U)+dim(W)dim(UW)=2+21=3\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W) = 2 + 2 - 1 = 3. \checkmark \blacksquare

1.6 Intuition: What Does a Vector Space Capture Geometrically?

Section titled “1.6 Intuition: What Does a Vector Space Capture Geometrically?”

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 αu+βv\alpha \mathbf{u} + \beta \mathbf{v}, you are performing two geometric operations: stretching (αu\alpha \mathbf{u}) and then adding (+βv+ \beta \mathbf{v}). The axioms guarantee that this operation behaves as your geometric intuition demands:

  • Commutativity (u+v=v+u\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 ((u+v)+w=u+(v+w)(\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 (0\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 (v-\mathbf{v}) means every displacement can be undone. If v\mathbf{v} moves you from AA to BB, then v-\mathbf{v} moves you back from BB to AA.
  • Distributivity (α(u+v)=αu+αv\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 2×22 \times 2 matrices, the set of polynomials of degree 3\leq 3, and the set of continuous functions on [0,1][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.

1.7 Worked Example: Infinite-Dimensional Vector Space

Section titled “1.7 Worked Example: Infinite-Dimensional Vector Space”

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

Solution

We verify the subspace criterion.

Non-empty: The zero function 0(x)=0\mathbf{0}(x) = 0 satisfies 0(0)=0\mathbf{0}(0) = 0, so 0W\mathbf{0} \in W.

Closure under addition: Let f,gWf, g \in W, so f(0)=0f(0) = 0 and g(0)=0g(0) = 0. Then (f+g)(0)=f(0)+g(0)=0+0=0(f + g)(0) = f(0) + g(0) = 0 + 0 = 0, so f+gWf + g \in W.

Closure under scalar multiplication: Let fWf \in W and αR\alpha \in \mathbb{R}. Then (αf)(0)=αf(0)=α0=0(\alpha f)(0) = \alpha \cdot f(0) = \alpha \cdot 0 = 0, so αfW\alpha f \in W.

By the subspace criterion, WW is a subspace of VV.

Geometric intuition: WW is the subspace of functions that “start at the origin.” In the infinite-dimensional space C[0,1]C[0,1], this is a closed hyperplane (codimension 1 subspace). The complementary subspace consists of constant functions: every fC[0,1]f \in C[0,1] can be written uniquely as f=(ff(0))+f(0)f = (f - f(0)) + f(0), where ff(0)Wf - f(0) \in W and f(0)f(0) is a constant function. \blacksquare

1.8 Worked Example: Null Space as a Subspace

Section titled “1.8 Worked Example: Null Space as a Subspace”

Problem. Let A=(121242)A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{pmatrix}. Find a basis for null(A)\mathrm{null}(A) and verify it is a subspace of R3\mathbb{R}^3.

Solution

Row-reduce AA:

(121242)R22R1(121000)\begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{pmatrix} \xrightarrow{R_2 - 2R_1} \begin{pmatrix} 1 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix}

The RREF has one pivot (column 1). The free variables are x2x_2 and x3x_3. Setting x2=sx_2 = s, x3=tx_3 = t: x1=2s+tx_1 = -2s + t.

null(A)={s(210)+t(101):s,tR}\mathrm{null}(A) = \left\{ s\begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} : s, t \in \mathbb{R} \right\}

A basis is {(2,1,0)T,(1,0,1)T}\{(-2, 1, 0)^T, (1, 0, 1)^T\} and dim(null(A))=2\dim(\mathrm{null}(A)) = 2.

Verification that null(A) is a subspace: This follows from the general theorem that the null space of any matrix is a subspace. Geometrically, null(A)\mathrm{null}(A) is the set of all vectors x\mathbf{x} such that Ax=0A\mathbf{x} = \mathbf{0}, i.e., the vectors that AA “collapses” to the origin. Since AA is a linear map, the preimage of the origin is always a subspace. This is a consequence of the fact that linear maps send subspaces to subspaces, and the inverse image of a subspace under a linear map is a subspace.

Dimension check: By rank-nullity, rank(A)+nullity(A)=3\mathrm{rank}(A) + \mathrm{nullity}(A) = 3. Since rank(A)=1\mathrm{rank}(A) = 1 (one pivot), nullity(A)=2\mathrm{nullity}(A) = 2, consistent with our basis. \blacksquare

  • The empty set is not a vector space. The subspace criterion requires the subset to be non-empty. The trivial subspace {0}\{\mathbf{0}\} is the smallest subspace of any vector space.
  • Non-homogeneous conditions do not define subspaces. The set of solutions to Ax=bA\mathbf{x} = \mathbf{b} with b0\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}\{(x, y) : x + y = 1\} is a line in R2\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 1-1. For example, the first quadrant {(x,y):x0,y0}\{(x, y) : x \geq 0, y \geq 0\} is closed under addition and positive scalar multiplication, but (1)(1,1)=(1,1)(-1) \cdot (1, 1) = (-1, -1) is not in the first quadrant.
  • Do not confuse the field with the vector space. R\mathbb{R} is a field, but it is also a vector space over itself (of dimension 1). R2\mathbb{R}^2 is a vector space over R\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 P3(R)\mathcal{P}_3(\mathbb{R}) has dimension 4, even though its elements are “logically” embedded in C(R)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.


flowchart TD
A[1_Vectors And Vector Spaces] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]