Vector Spaces and Subspaces

Why Vector Spaces for Wireless Communications?

Every modern wireless system transmits and receives vectors. When a base station equipped with NtN_t antennas sends a symbol vector x∈CNt\mathbf{x} \in \mathbb{C}^{N_t}, a receiver with NrN_r antennas observes

y=Hx+n,\mathbf{y} = \mathbf{H}\mathbf{x} + \mathbf{n},

where H∈CNr×Nt\mathbf{H} \in \mathbb{C}^{N_r \times N_t} is the channel matrix and n∈CNr\mathbf{n} \in \mathbb{C}^{N_r} is additive noise. The entirety of MIMO theory --- beamforming, spatial multiplexing, interference alignment --- reduces to geometric operations inside complex vector spaces: projections, subspace decompositions, and changes of basis.

A rigorous command of vector-space fundamentals is therefore not mathematical overhead; it is the language in which capacity-achieving schemes are conceived and analysed. This section builds that language from the axioms up, with Cn\mathbb{C}^n always in view as the workhorse space of communications engineering.

Definition:

Field

A field is a set F\mathbb{F} together with two binary operations, addition (++) and multiplication (⋅\cdot), satisfying the usual axioms of commutativity, associativity, distributivity, existence of additive identity 00, multiplicative identity 1≠01 \neq 0, additive inverses, and multiplicative inverses for every nonzero element.

Throughout this text the relevant fields are:

  • R\mathbb{R}, the real numbers, and
  • C\mathbb{C}, the complex numbers (where j2=−1j^{2} = -1).

Engineering convention uses jj for the imaginary unit (reserving ii for current). We follow this convention throughout.

Definition:

Vector Space

Let F\mathbb{F} be a field. A vector space over F\mathbb{F} is a non-empty set V\mathcal{V} equipped with two operations --- vector addition + ⁣:V×V→V+ \colon \mathcal{V} \times \mathcal{V} \to \mathcal{V} and scalar multiplication ⋅ ⁣:F×V→V\cdot \colon \mathbb{F} \times \mathcal{V} \to \mathcal{V} --- satisfying, for all u,v,w∈V\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathcal{V} and all α,β∈F\alpha, \beta \in \mathbb{F}:

# Axiom Statement
A1 Additive closure u+v∈V\mathbf{u} + \mathbf{v} \in \mathcal{V}
A2 Additive commutativity u+v=v+u\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}
A3 Additive associativity (u+v)+w=u+(v+w)(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})
A4 Existence of zero vector There exists 0∈V\mathbf{0} \in \mathcal{V} such that v+0=v\mathbf{v} + \mathbf{0} = \mathbf{v}
A5 Existence of additive inverse For each v\mathbf{v} there exists −v∈V-\mathbf{v} \in \mathcal{V} with v+(−v)=0\mathbf{v} + (-\mathbf{v}) = \mathbf{0}
S1 Scalar-multiplication closure αv∈V\alpha \mathbf{v} \in \mathcal{V}
S2 Compatibility of scalar multiplication α(βv)=(αβ)v\alpha(\beta \mathbf{v}) = (\alpha\beta)\mathbf{v}
S3 Multiplicative identity 1 v=v1\,\mathbf{v} = \mathbf{v}
D1 Distributivity over vector addition α(u+v)=αu+αv\alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v}
D2 Distributivity over scalar addition (α+β)v=αv+βv(\alpha + \beta)\mathbf{v} = \alpha\mathbf{v} + \beta\mathbf{v}

The axioms are not independent (e.g., additive closure is sometimes folded into the definition of the addition map), but listing all eight properties plus the two closure conditions makes verification of concrete examples systematic.

Definition:

The Space Cn\mathbb{C}^n

For a positive integer nn, define

Cn={ x=(x1,x2,…,xn) ⁣T:xk∈C,  k=1,…,n }\mathbb{C}^{n} = \bigl\{\,\mathbf{x} = (x_1, x_2, \dots, x_n)^{\!\mathsf{T}} : x_k \in \mathbb{C},\; k=1,\dots,n\,\bigr\}

with component-wise addition and scalar multiplication:

x+y=(x1+y1,  …,  xn+yn) ⁣T,α x=(αx1,  …,  αxn) ⁣T.\mathbf{x} + \mathbf{y} = (x_1+y_1,\;\dots,\;x_n+y_n)^{\!\mathsf{T}}, \qquad \alpha\,\mathbf{x} = (\alpha x_1,\;\dots,\;\alpha x_n)^{\!\mathsf{T}}.

Under these operations Cn\mathbb{C}^n is a vector space over C\mathbb{C}. The zero vector is 0=(0,0,…,0) ⁣T\mathbf{0} = (0,0,\dots,0)^{\!\mathsf{T}}.

In wireless communications nn usually equals the number of antenna elements at one end of the link. A transmit vector x∈CNt\mathbf{x} \in \mathbb{C}^{N_t} assigns a complex baseband signal to each of NtN_t antennas.

Definition:

Subspace

Let V\mathcal{V} be a vector space over F\mathbb{F}. A non-empty subset W⊆V\mathcal{W} \subseteq \mathcal{V} is a subspace of V\mathcal{V} if and only if the following three conditions hold:

  1. Contains the zero vector: 0∈W\mathbf{0} \in \mathcal{W}.
  2. Closed under addition: u,v∈W  ⟹  u+v∈W\mathbf{u}, \mathbf{v} \in \mathcal{W} \;\Longrightarrow\; \mathbf{u} + \mathbf{v} \in \mathcal{W}.
  3. Closed under scalar multiplication: α∈F,  v∈W  ⟹  αv∈W\alpha \in \mathbb{F},\; \mathbf{v} \in \mathcal{W} \;\Longrightarrow\; \alpha\mathbf{v} \in \mathcal{W}.

Equivalently, W\mathcal{W} is a subspace if and only if αu+βv∈W\alpha\mathbf{u} + \beta\mathbf{v} \in \mathcal{W} for all u,v∈W\mathbf{u},\mathbf{v} \in \mathcal{W} and all α,β∈F\alpha,\beta \in \mathbb{F} (closure under linear combinations).

Condition 1 can be replaced by the requirement that W\mathcal{W} is non-empty, since taking α=0\alpha = 0 in condition 3 then gives 0∈W\mathbf{0} \in \mathcal{W}. We list it explicitly for clarity.

Definition:

Linear Combination

Let V\mathcal{V} be a vector space over F\mathbb{F} and let v1,…,vk∈V\mathbf{v}_1, \dots, \mathbf{v}_k \in \mathcal{V}. A vector v∈V\mathbf{v} \in \mathcal{V} is a linear combination of v1,…,vk\mathbf{v}_1, \dots, \mathbf{v}_k if there exist scalars α1,…,αk∈F\alpha_1, \dots, \alpha_k \in \mathbb{F} such that

v=α1v1+α2v2+⋯+αkvk=∑i=1kαivi.\mathbf{v} = \alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \cdots + \alpha_k \mathbf{v}_k = \sum_{i=1}^{k} \alpha_i \mathbf{v}_i.

Definition:

Span

Let S={v1,…,vk}⊆V\mathcal{S} = \{\mathbf{v}_1, \dots, \mathbf{v}_k\} \subseteq \mathcal{V}. The span of S\mathcal{S} is the set of all linear combinations of vectors in S\mathcal{S}:

span⁡(S)={∑i=1kαivi:αi∈F}.\operatorname{span}(\mathcal{S}) = \Bigl\{\sum_{i=1}^{k}\alpha_i \mathbf{v}_i : \alpha_i \in \mathbb{F}\Bigr\}.

By convention span⁡(∅)={0}\operatorname{span}(\varnothing) = \{\mathbf{0}\}.

span⁡(S)\operatorname{span}(\mathcal{S}) is always a subspace of V\mathcal{V} --- in fact it is the smallest subspace containing S\mathcal{S}. If span⁡(S)=V\operatorname{span}(\mathcal{S}) = \mathcal{V} we say S\mathcal{S} spans (or generates) V\mathcal{V}.

Definition:

Linear Independence

A set {v1,…,vk}⊆V\{\mathbf{v}_1, \dots, \mathbf{v}_k\} \subseteq \mathcal{V} is linearly independent if the equation

α1v1+α2v2+⋯+αkvk=0\alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \cdots + \alpha_k \mathbf{v}_k = \mathbf{0}

implies α1=α2=⋯=αk=0\alpha_1 = \alpha_2 = \cdots = \alpha_k = 0.

A set that is not linearly independent is called linearly dependent; equivalently, at least one vector in the set can be written as a linear combination of the others.

Definition:

Basis

A set B={b1,…,bn}⊆V\mathcal{B} = \{\mathbf{b}_1, \dots, \mathbf{b}_n\} \subseteq \mathcal{V} is a basis for V\mathcal{V} if:

  1. B\mathcal{B} is linearly independent, and
  2. span⁡(B)=V\operatorname{span}(\mathcal{B}) = \mathcal{V}.

Equivalently, B\mathcal{B} is a basis if and only if every vector v∈V\mathbf{v} \in \mathcal{V} can be written uniquely as

v=∑i=1nαibi,αi∈F.\mathbf{v} = \sum_{i=1}^{n} \alpha_i \mathbf{b}_i, \qquad \alpha_i \in \mathbb{F}.

The standard (canonical) basis for Cn\mathbb{C}^n is {e1,…,en}\{\mathbf{e}_1, \dots, \mathbf{e}_n\} where ek\mathbf{e}_k has a 11 in position kk and 00 elsewhere.

Definition:

Dimension

The dimension of a vector space V\mathcal{V}, denoted dim⁡(V)\dim(\mathcal{V}), is the number of vectors in any basis for V\mathcal{V}.

This is well-defined because all bases of V\mathcal{V} have the same cardinality (see TUniqueness of Dimension (Invariance of Basis Cardinality) below).

In particular, dim⁡(Cn)=n\dim(\mathbb{C}^n) = n (over C\mathbb{C}).

Theorem: Uniqueness of Dimension (Invariance of Basis Cardinality)

Let V\mathcal{V} be a vector space over a field F\mathbb{F}. If V\mathcal{V} has a finite basis, then every basis of V\mathcal{V} is finite and all bases have the same number of elements.

A basis is a "maximally efficient coordinate system" for the space. Because every vector has a unique expansion in a given basis, no basis can be "shorter" than another (it would miss some direction) or "longer" (it would contain a redundancy). The Steinitz exchange argument formalises this by showing that independent vectors can always be swapped into a spanning set one-by-one without losing the spanning property.

Show Hint

Use the Steinitz Exchange Lemma: if a set of mm vectors is linearly independent and another set of nn vectors spans the space, then m≤nm \le n.

Apply the lemma twice, once in each direction.

Proof
Steinitz Exchange Lemma

We first establish the following key lemma.

Lemma (Steinitz Exchange). Let L={u1,…,um}\mathcal{L} = \{\mathbf{u}_1, \dots, \mathbf{u}_m\} be a linearly independent set in V\mathcal{V} and let G={w1,…,wn}\mathcal{G} = \{\mathbf{w}_1, \dots, \mathbf{w}_n\} be a set that spans V\mathcal{V}. Then m≤nm \le n, and there exists a subset of G\mathcal{G} of size n−mn - m such that its union with L\mathcal{L} still spans V\mathcal{V}.

Proof of the lemma. We proceed by induction on mm.

Base case (m=0m = 0). The empty set is trivially linearly independent, 0≤n0 \le n is obvious, and G\mathcal{G} itself spans V\mathcal{V}.

Inductive step. Assume the result holds for m−1m-1. By the inductive hypothesis there exists a re-indexing of G\mathcal{G} such that

Sm−1={u1,…,um−1, wm,…,wn}\mathcal{S}_{m-1} = \{\mathbf{u}_1, \dots, \mathbf{u}_{m-1},\, \mathbf{w}_{m}, \dots, \mathbf{w}_{n}\}

spans V\mathcal{V} (and m−1≤nm - 1 \le n). Since Sm−1\mathcal{S}_{m-1} spans V\mathcal{V}, write

um=∑i=1m−1αiui+∑k=mnβkwk.\mathbf{u}_m = \sum_{i=1}^{m-1} \alpha_i \mathbf{u}_i + \sum_{k=m}^{n} \beta_k \mathbf{w}_k.

Not all βk\beta_k can be zero, for otherwise um\mathbf{u}_m would be a linear combination of u1,…,um−1\mathbf{u}_1, \dots, \mathbf{u}_{m-1}, contradicting the linear independence of L\mathcal{L}. Hence m≤nm \le n (there is at least one wk\mathbf{w}_k term) and we may choose some index ℓ\ell with βℓ≠0\beta_\ell \neq 0. Solving for wℓ\mathbf{w}_\ell and substituting shows that

Sm={u1,…,um, wm+1,…,wn}\mathcal{S}_{m} = \{\mathbf{u}_1, \dots, \mathbf{u}_{m},\, \mathbf{w}_{m+1}, \dots, \mathbf{w}_{n}\}

(after re-indexing so that wℓ\mathbf{w}_\ell becomes wm\mathbf{w}_m) still spans V\mathcal{V}. This completes the induction and proves the lemma. □\square

Applying the lemma to two bases

Now let B1={b1,…,bp}\mathcal{B}_1 = \{\mathbf{b}_1, \dots, \mathbf{b}_p\} and B2={c1,…,cq}\mathcal{B}_2 = \{\mathbf{c}_1, \dots, \mathbf{c}_q\} be two bases of V\mathcal{V}.

  • B1\mathcal{B}_1 is linearly independent and B2\mathcal{B}_2 spans V\mathcal{V}, so the Steinitz Exchange Lemma gives p≤qp \le q.
  • B2\mathcal{B}_2 is linearly independent and B1\mathcal{B}_1 spans V\mathcal{V}, so the Steinitz Exchange Lemma gives q≤pq \le p.

Therefore p=qp = q. Every basis of V\mathcal{V} has the same cardinality, and the dimension dim⁡(V)\dim(\mathcal{V}) is well-defined. ■\blacksquare

Theorem: Extension of Linearly Independent Sets to a Basis

Let V\mathcal{V} be a finite-dimensional vector space with dim⁡(V)=n\dim(\mathcal{V}) = n. Every linearly independent set L⊆V\mathcal{L} \subseteq \mathcal{V} with ∣L∣≤n|\mathcal{L}| \le n can be extended to a basis for V\mathcal{V}. In particular, every linearly independent set in V\mathcal{V} contains at most nn vectors.

If a linearly independent set does not yet span the whole space, there is some direction it "misses." We can always adjoin a vector from that missing direction without destroying independence, and repeat until we have nn vectors --- at which point the Steinitz argument guarantees we span the space.

Show Hint

Iteratively adjoin vectors that lie outside the current span.

Use the Steinitz Exchange Lemma to show the process terminates.

Proof
Iterative extension procedure

Let L={v1,…,vm}\mathcal{L} = \{\mathbf{v}_1, \dots, \mathbf{v}_m\} be linearly independent with m≤nm \le n. Fix a basis B={b1,…,bn}\mathcal{B} = \{\mathbf{b}_1, \dots, \mathbf{b}_n\} for V\mathcal{V}.

If span⁡(L)=V\operatorname{span}(\mathcal{L}) = \mathcal{V} then L\mathcal{L} is already a basis and m=nm = n by TUniqueness of Dimension (Invariance of Basis Cardinality).

Otherwise, there exists bk1∈B\mathbf{b}_{k_1} \in \mathcal{B} with bk1∉span⁡(L)\mathbf{b}_{k_1} \notin \operatorname{span}(\mathcal{L}). Set L1=L∪{bk1}\mathcal{L}_1 = \mathcal{L} \cup \{\mathbf{b}_{k_1}\}.

Preservation of independence

We claim L1\mathcal{L}_1 is linearly independent. Suppose

∑i=1mαivi+γ bk1=0.\sum_{i=1}^{m} \alpha_i \mathbf{v}_i + \gamma\,\mathbf{b}_{k_1} = \mathbf{0}.

If γ≠0\gamma \neq 0 we could solve for bk1\mathbf{b}_{k_1} as a linear combination of v1,…,vm\mathbf{v}_1, \dots, \mathbf{v}_m, contradicting bk1∉span⁡(L)\mathbf{b}_{k_1} \notin \operatorname{span}(\mathcal{L}). Hence γ=0\gamma = 0, and then all αi=0\alpha_i = 0 by the independence of L\mathcal{L}.

Termination

Repeat the extension: at each stage the cardinality increases by one and the set remains linearly independent. By the Steinitz Exchange Lemma, a linearly independent set in V\mathcal{V} has at most nn elements, so the process terminates after at most n−mn - m additions. The resulting set has nn linearly independent vectors in an nn-dimensional space, hence it spans V\mathcal{V} (otherwise one could adjoin yet another vector, producing n+1n+1 independent vectors, contradicting the Steinitz bound). Therefore it is a basis. ■\blacksquare

Example: Linear Independence and Basis Verification in C3\mathbb{C}^3

Consider the following three vectors in C3\mathbb{C}^3:

v1=(1j0),v2=(011+j),v3=(j01).\mathbf{v}_1 = \begin{pmatrix} 1 \\ j \\ 0 \end{pmatrix}, \qquad \mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 1+j \end{pmatrix}, \qquad \mathbf{v}_3 = \begin{pmatrix} j \\ 0 \\ 1 \end{pmatrix}.

(a) Determine whether {v1,v2,v3}\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\} is linearly independent over C\mathbb{C}.

(b) If it is, verify that it forms a basis for C3\mathbb{C}^3.

Solution
Set up the dependence equation

We require all solutions (α1,α2,α3)∈C3(\alpha_1, \alpha_2, \alpha_3) \in \mathbb{C}^3 of

α1v1+α2v2+α3v3=0.\alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \alpha_3 \mathbf{v}_3 = \mathbf{0}.

Writing this component-wise:

{α1+j α3=0,j α1+α2=0,(1+j) α2+α3=0.\begin{cases} \alpha_1 + j\,\alpha_3 = 0, \\ j\,\alpha_1 + \alpha_2 = 0, \\ (1+j)\,\alpha_2 + \alpha_3 = 0. \end{cases}

Row-reduce the coefficient matrix

Form the coefficient matrix and row-reduce:

A=(10jj1001+j1).\mathbf{A} = \begin{pmatrix} 1 & 0 & j \\ j & 1 & 0 \\ 0 & 1+j & 1 \end{pmatrix}.

Step 1. Eliminate the jj in position (2,1)(2,1) via R2←R2−j R1R_2 \leftarrow R_2 - j\,R_1:

(10j01−j201+j1)=(10j01101+j1),\begin{pmatrix} 1 & 0 & j \\ 0 & 1 & -j^2 \\ 0 & 1+j & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & j \\ 0 & 1 & 1 \\ 0 & 1+j & 1 \end{pmatrix},

since −j2=−(−1)=1-j^2 = -(-1) = 1.

Step 2. Eliminate position (3,2)(3,2) via R3←R3−(1+j) R2R_3 \leftarrow R_3 - (1+j)\,R_2:

(10j011001−(1+j))=(10j01100−j).\begin{pmatrix} 1 & 0 & j \\ 0 & 1 & 1 \\ 0 & 0 & 1-(1+j) \end{pmatrix} = \begin{pmatrix} 1 & 0 & j \\ 0 & 1 & 1 \\ 0 & 0 & -j \end{pmatrix}.

Since −j≠0-j \neq 0, the matrix has rank 33.

Conclude independence and basis

(a) The only solution is α1=α2=α3=0\alpha_1 = \alpha_2 = \alpha_3 = 0, so the set is linearly independent over C\mathbb{C}.

(b) Because dim⁡(C3)=3\dim(\mathbb{C}^3) = 3 and we have three linearly independent vectors in C3\mathbb{C}^3, the set {v1,v2,v3}\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\} is a basis for C3\mathbb{C}^3.

Alternatively, det⁡(A)=1⋅1⋅(−j)−j⋅1⋅(1+j)+⋯\det(\mathbf{A}) = 1 \cdot 1 \cdot (-j) - j \cdot 1 \cdot (1+j) + \cdots; computing in full gives det⁡(A)=−j−j(1+j)=−j−j+1=1−2j≠0\det(\mathbf{A}) = -j - j(1+j) = -j - j + 1 = 1 - 2j \neq 0, confirming invertibility and hence that the columns span C3\mathbb{C}^3.

Remark: In practice, checking det⁡(A)≠0\det(\mathbf{A}) \neq 0 is the fastest test, but row reduction reveals the structure more clearly and is numerically more stable for large systems.

Vector space

A set V\mathcal{V} over a field F\mathbb{F} with addition and scalar multiplication operations satisfying the ten axioms (A1--A5, S1--S3, D1--D2) listed in DVector Space.

Related: Vector Space, Subspace, Basis

Dimension

The cardinality of any basis for a vector space V\mathcal{V}. All bases share the same cardinality by TUniqueness of Dimension (Invariance of Basis Cardinality). Notation: dim⁡(V)\dim(\mathcal{V}).

Related: Dimension, Uniqueness of Dimension (Invariance of Basis Cardinality), Basis

Span

The set of all linear combinations of a given collection of vectors. span⁡(S)\operatorname{span}(\mathcal{S}) is the smallest subspace containing S\mathcal{S}.

Related: Span, Linear Combination, Subspace

Quick Check

What is dim⁡R(Cn)\dim_{\mathbb{R}}(\mathbb{C}^n), the dimension of Cn\mathbb{C}^n when it is viewed as a vector space over R\mathbb{R} (rather than over C\mathbb{C})?

nn

2n2n

n2n^2

n/2n/2

Correction:
2n2n

Over R\mathbb{R}, every complex component xk=ak+j bkx_k = a_k + j\,b_k requires two real coordinates. A real basis for Cn\mathbb{C}^n is {e1,…,en, je1,…,jen}\{\mathbf{e}_1, \dots, \mathbf{e}_n,\, j\mathbf{e}_1, \dots, j\mathbf{e}_n\}, which has 2n2n elements. Hence dim⁡R(Cn)=2n\dim_{\mathbb{R}}(\mathbb{C}^n) = 2n.

Quick Check

Which of the following subsets of C2\mathbb{C}^2 is a subspace (over C\mathbb{C})?

{(x1,x2)T:∣x1∣2+∣x2∣2≤1}\{(x_1, x_2)^\mathsf{T} : |x_1|^2 + |x_2|^2 \le 1\} (the closed unit ball)

{(x1,x2)T:x1+j x2=0}\{(x_1, x_2)^\mathsf{T} : x_1 + j\,x_2 = 0\}

{(x1,x2)T:x1∈R}\{(x_1, x_2)^\mathsf{T} : x_1 \in \mathbb{R}\}

{(1,0)T}\{(1, 0)^\mathsf{T}\} (a single nonzero vector)

Correction:
{(x1,x2)T:x1+j x2=0}\{(x_1, x_2)^\mathsf{T} : x_1 + j\,x_2 = 0\}

This set is the solution space of the homogeneous linear equation x1+j x2=0x_1 + j\,x_2 = 0, hence it is closed under addition and scalar multiplication and contains the zero vector. The unit ball fails closure under scalar multiplication (scaling by α\alpha with ∣α∣>1|\alpha| > 1 leaves the ball).

Why This Matters: Cn\mathbb{C}^n and Antenna Arrays

In a MIMO (Multiple-Input Multiple-Output) system, the baseband signal at a uniform linear array (ULA) of nn antenna elements is naturally represented as a vector x∈Cn\mathbf{x} \in \mathbb{C}^n. The kk-th component xkx_k is the complex baseband signal --- amplitude and phase --- fed to the kk-th element.

The array steering vector for a plane wave arriving at angle θ\theta takes the form

a(θ)=1n(1ej2πdλsin⁡θej2π2dλsin⁡θ⋮ej2π(n−1)dλsin⁡θ)∈Cn,\mathbf{a}(\theta) = \frac{1}{\sqrt{n}}\begin{pmatrix} 1 \\ e^{j 2\pi \frac{d}{\lambda}\sin\theta} \\ e^{j 2\pi \frac{2d}{\lambda}\sin\theta} \\ \vdots \\ e^{j 2\pi \frac{(n-1)d}{\lambda}\sin\theta} \end{pmatrix} \in \mathbb{C}^n,

where dd is the inter-element spacing and λ\lambda is the carrier wavelength.

Key link to this section: The set {a(θ1),…,a(θK)}\{\mathbf{a}(\theta_1), \dots, \mathbf{a}(\theta_K)\} for KK distinct directions of arrival is linearly independent (for almost all choices of θk\theta_k) whenever K≤nK \le n. This is precisely the linear-independence concept of DLinear Independence applied to Cn\mathbb{C}^n. A receiver with nn antennas can therefore resolve up to nn spatial paths --- a fact that underpins spatial multiplexing and beamforming gain.

See full treatment in Chapter 3، Section 2

Why This Matters: Signal and Noise Subspaces

After receiving TT snapshots from nn antennas, the sample covariance matrix R^∈Cn×n\hat{\mathbf{R}} \in \mathbb{C}^{n \times n} can be eigen-decomposed. Its column space splits into a signal subspace (spanned by eigenvectors corresponding to the KK largest eigenvalues) and a noise subspace (spanned by the remaining n−Kn - K eigenvectors). These are complementary subspaces of Cn\mathbb{C}^n in the sense of DSubspace. Algorithms such as MUSIC and ESPRIT exploit this decomposition for high-resolution direction-of-arrival estimation.

See full treatment in Chapter 7، Section 3

Common Mistake: Confusing R2n\mathbb{R}^{2n} with Cn\mathbb{C}^n

Mistake:

Treating Cn\mathbb{C}^n and R2n\mathbb{R}^{2n} as interchangeable. A common error is to claim that a set of 2n2n vectors in Cn\mathbb{C}^n can be linearly independent (over C\mathbb{C}), reasoning that Cn≅R2n\mathbb{C}^n \cong \mathbb{R}^{2n} "has 2n2n real degrees of freedom."

Correction:

Cn\mathbb{C}^n is nn-dimensional over C\mathbb{C} and 2n2n-dimensional over R\mathbb{R}. The dimension depends on the scalar field.

  • Over C\mathbb{C}: at most nn vectors can be linearly independent.
  • Over R\mathbb{R}: the same set Cn\mathbb{C}^n has dimension 2n2n, and up to 2n2n vectors can be R\mathbb{R}-linearly independent.

In wireless communications the scalar field is almost always C\mathbb{C} (complex baseband representation), so the relevant dimension is nn.

Example: In C1=C\mathbb{C}^1 = \mathbb{C}, the vectors 11 and jj are R\mathbb{R}-linearly independent but C\mathbb{C}-linearly dependent (since j=j⋅1j = j \cdot 1).

Common Mistake: Forgetting that span⁡(∅)={0}\operatorname{span}(\varnothing) = \{\mathbf{0}\}

Mistake:

Assuming span⁡(∅)\operatorname{span}(\varnothing) is undefined or equals the empty set. This sometimes leads to incorrect conclusions about the dimension of the trivial subspace.

Correction:

By convention span⁡(∅)={0}\operatorname{span}(\varnothing) = \{\mathbf{0}\}, which is the trivial subspace of any vector space. Its dimension is 00, and the empty set serves as its (unique) basis. This convention ensures that "every subspace has a basis" holds without exception.

Key Takeaway

The core message of this section in three bullets:

  1. Cn\mathbb{C}^n is the stage. Virtually every signal, channel, and noise vector in wireless communications lives in Cn\mathbb{C}^n for some nn. Mastering this space is non-negotiable.

  2. Dimension is the fundamental invariant. It tells you how many independent directions a (sub)space has --- equivalently, how many spatial streams a MIMO system can support or how many parameters a signal model requires.

  3. Bases are coordinate systems. Choosing a good basis (eigenvectors, steering vectors, DFT columns) is the single most recurring design step in communications theory. All bases for a given subspace have the same size, but their structure profoundly affects algorithm performance.

Prerequisites & NotationInner Products and Norms