# Complex Vector Spaces > A quantum state is a vector. Not a metaphor — literally a unit vector in a complex vector space. > Before we can say what a qubit *is*, we need to be precise about the space it lives in: $\mathbb{C}^n$, > the complex vector space. You know real vector spaces; the upgrade to $\mathbb{C}$ is small in > definition but consequential in physics, because complex phases are where quantum interference lives. ## Learning Objectives After this lesson you will be able to: 1. State the axioms of a vector space over a field and verify them for $\mathbb{C}^n$. 2. Explain why quantum mechanics requires $\mathbb{C}$, not $\mathbb{R}$, as its scalar field. 3. Determine whether a set of complex vectors is **linearly independent** and compute the **dimension** of its span. 4. Identify **subspaces** and produce a **basis** for one. 5. Represent and manipulate complex vectors and bases in NumPy. --- ## Intuition A vector space is a setting where two operations make sense and play nicely together: you can **add** two vectors, and you can **scale** a vector by a number (a *scalar*). That's it. Everything else — bases, dimension, linear maps — is built from those two operations. In high-school physics the scalars were real numbers and vectors were arrows in space. In quantum mechanics the scalars are **complex numbers** and the "vectors" are abstract lists of complex amplitudes. Why complex? Because a quantum amplitude carries a **phase** as well as a magnitude, and relative phases between components are physically observable — they produce interference, the engine of quantum speedups. A real-amplitude theory cannot reproduce the observed physics (you'll prove a sharp version of this when we hit the Bloch sphere and interference in [Term 1](../../term-1-quantum-mechanics/c02-qubits-bloch/01-the-qubit.md)). For now: think "lists of complex numbers you can add and scale." --- ## Theory ### Fields The scalars of a vector space form a **field**: a set $\mathbb{F}$ with addition and multiplication that are associative, commutative, distributive, have identities $0$ and $1$, and admit additive inverses and (for nonzero elements) multiplicative inverses. We care about two fields: - $\mathbb{R}$, the real numbers. - $\mathbb{C} = \{\,a + bi : a,b \in \mathbb{R}\,\}$ with $i^2 = -1$, the **complex numbers**. We review $\mathbb{C}$ thoroughly in [0.3.1 Complex Numbers & Functions](../c03-fourier-methods/01-complex-numbers-functions.md); here we need only that it is a field and that every complex number $z = a+bi$ has a **conjugate** $z^* = a - bi$ and a **modulus** $|z| = \sqrt{a^2+b^2} = \sqrt{z^* z}$. ### Definition: vector space over a field A **vector space** over a field $\mathbb{F}$ is a set $V$ with two operations — vector addition $+ : V \times V \to V$ and scalar multiplication $\cdot : \mathbb{F} \times V \to V$ — satisfying, for all $u, v, w \in V$ and $a, b \in \mathbb{F}$: $$ \begin{aligned} &\textbf{(A1) associativity:} && (u + v) + w = u + (v + w) \\ &\textbf{(A2) commutativity:} && u + v = v + u \\ &\textbf{(A3) zero vector:} && \exists\, 0 \in V \text{ s.t. } v + 0 = v \\ &\textbf{(A4) additive inverse:} && \exists\, (-v) \text{ s.t. } v + (-v) = 0 \\ &\textbf{(M1) compatibility:} && a(bv) = (ab)v \\ &\textbf{(M2) scalar identity:} && 1\,v = v \\ &\textbf{(D1) distributivity over }V: && a(u+v) = au + av \\ &\textbf{(D2) distributivity over }\mathbb{F}: && (a+b)v = av + bv \end{aligned} $$ Elements of $V$ are **vectors**; elements of $\mathbb{F}$ are **scalars**. When $\mathbb{F} = \mathbb{C}$ we call $V$ a **complex vector space**. ### The space $\mathbb{C}^n$ The central example for us is $$ \mathbb{C}^n = \left\{\, \begin{pmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{pmatrix} : z_k \in \mathbb{C} \,\right\}, $$ with componentwise addition and scalar multiplication: $$ \begin{pmatrix} z_1 \\ \vdots \\ z_n \end{pmatrix} + \begin{pmatrix} w_1 \\ \vdots \\ w_n \end{pmatrix} = \begin{pmatrix} z_1 + w_1 \\ \vdots \\ z_n + w_n \end{pmatrix}, \qquad a \begin{pmatrix} z_1 \\ \vdots \\ z_n \end{pmatrix} = \begin{pmatrix} a z_1 \\ \vdots \\ a z_n \end{pmatrix}. $$ The axioms (A1)–(D2) follow directly from the field axioms of $\mathbb{C}$ applied componentwise. A single qubit's state space is $\mathbb{C}^2$; $n$ qubits live in $\mathbb{C}^{2^n}$ (we'll see *why* the dimension is $2^n$, not $2n$, in [0.1.7 Tensor Products](07-tensor-products.md)). > **Note on dimension counting.** As a *real* vector space, $\mathbb{C}^n$ has dimension $2n$ (each > complex component contributes a real and imaginary part). As a *complex* vector space it has > dimension $n$. Unless stated otherwise, "dimension" means the complex dimension. ### Linear combinations, span, independence Let $v_1, \dots, v_m \in V$. - A **linear combination** is any $\sum_{k=1}^m a_k v_k$ with $a_k \in \mathbb{F}$. - Their **span** $\operatorname{span}\{v_1,\dots,v_m\}$ is the set of *all* such linear combinations. It is always a subspace (see below). - The set is **linearly independent** if the only solution to $$ \sum_{k=1}^m a_k v_k = 0 \quad\text{is}\quad a_1 = a_2 = \cdots = a_m = 0. $$ Otherwise it is **linearly dependent**: some vector is a linear combination of the others. ### Basis and dimension A **basis** of $V$ is a linearly independent set that spans $V$. Two foundational facts (proved in any linear algebra text, e.g. [Axl]): - **Every basis of a finite-dimensional space has the same number of elements.** That number is the **dimension** $\dim V$. - **Unique representation:** if $\{e_1, \dots, e_n\}$ is a basis, every $v \in V$ has a *unique* expansion $v = \sum_k c_k e_k$. The $c_k$ are the **coordinates** of $v$ in that basis. The **standard (computational) basis** of $\mathbb{C}^n$ is $$ e_1 = \begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix},\ \ e_2 = \begin{pmatrix}0\\1\\\vdots\\0\end{pmatrix},\ \dots,\ e_n = \begin{pmatrix}0\\0\\\vdots\\1\end{pmatrix}. $$ For $\mathbb{C}^2$ we rename these $|0\rangle = e_1$ and $|1\rangle = e_2$ — the computational basis of a qubit. (That bra–ket notation is the subject of [0.1.3 Dirac Notation](03-dirac-notation.md); we adopt it as soon as inner products are in hand.) ### Subspaces A subset $W \subseteq V$ is a **subspace** if it is itself a vector space under the inherited operations. Equivalently (the **subspace test**), $W$ is a subspace iff: 1. $0 \in W$; 2. $W$ is closed under addition: $u, w \in W \Rightarrow u + w \in W$; 3. $W$ is closed under scalar multiplication: $a \in \mathbb{F},\, w \in W \Rightarrow a w \in W$. Spans are the prototypical subspaces. Subspaces matter physically: a measurement outcome corresponds to projecting a state onto a subspace (the eigenspace of an observable), which we develop in [Term 1.3](../../term-1-quantum-mechanics/c03-measurement/01-projective-measurement.md). --- ## Worked Examples ### Example 1 — Verifying linear independence in $\mathbb{C}^2$ Are $v_1 = \begin{pmatrix} 1 \\ i \end{pmatrix}$ and $v_2 = \begin{pmatrix} i \\ -1 \end{pmatrix}$ linearly independent over $\mathbb{C}$? Set $a v_1 + b v_2 = 0$: $$ a\begin{pmatrix}1\\i\end{pmatrix} + b\begin{pmatrix}i\\-1\end{pmatrix} = \begin{pmatrix} a + bi \\ ai - b \end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}. $$ From the top: $a = -bi$. Substitute into the bottom: $(-bi)i - b = -b i^2 - b = b - b = 0$, which holds for *every* $b$. So a nonzero solution exists (e.g. $b = 1,\ a = -i$): the vectors are **linearly dependent**. Indeed $v_2 = i\,v_1$ — multiplying by $i$ is a legal complex scaling, so over $\mathbb{C}$ these are the "same direction." (Over $\mathbb{R}$, treating $i$ as a forbidden scalar, they would look independent — a concrete reminder that the scalar field matters.) ### Example 2 — A basis for a subspace Let $W = \{ (z_1, z_2, z_3) \in \mathbb{C}^3 : z_1 + z_2 + z_3 = 0 \}$. Is $W$ a subspace, and what is $\dim W$? **Subspace check.** $0$ satisfies the constraint. If $u, w \in W$ then $(u+w)$'s components sum to $0 + 0 = 0$, and $a w$'s components sum to $a \cdot 0 = 0$. All three conditions hold, so $W$ is a subspace. **Basis.** The constraint $z_3 = -z_1 - z_2$ leaves $z_1, z_2$ free. Write a general element as $$ \begin{pmatrix} z_1 \\ z_2 \\ -z_1 - z_2 \end{pmatrix} = z_1 \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + z_2 \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}. $$ These two vectors span $W$ and are independent (neither is a scalar multiple of the other), so they form a basis and $\dim W = 2$. One linear constraint dropped the dimension from $3$ to $2$. --- ## Hands-on (Python) We make these abstractions concrete with NumPy. Recall from [Appendix B](../../appendix/B-python-numpy-refresher.md): **always use `dtype=complex`** for states. ```python import numpy as np # The computational basis of C^2 (a single qubit). ket0 = np.array([1, 0], dtype=complex) # |0> ket1 = np.array([0, 1], dtype=complex) # |1> # A linear combination (a superposition): (3 + 4i)|0> + (1 - 2i)|1> v = (3 + 4j) * ket0 + (1 - 2j) * ket1 print(v) # [3.+4.j 1.-2.j] def is_independent(*vectors): """A set of vectors is independent iff the matrix with them as columns has full column rank.""" M = np.column_stack(vectors) # shape (n, m) return np.linalg.matrix_rank(M) == M.shape[1] v1 = np.array([1, 1j], dtype=complex) v2 = np.array([1j, -1], dtype=complex) print(is_independent(v1, v2)) # False -> v2 = i*v1 (Example 1) # A genuinely independent pair: print(is_independent(ket0, ket1)) # True ``` ```python # Coordinates in the standard basis are just the components themselves. # Expansion v = sum_k c_k e_k is recovered trivially for the standard basis: coords = v # [3+4j, 1-2j] are the coordinates of v reconstructed = coords[0] * ket0 + coords[1] * ket1 print(np.allclose(reconstructed, v)) # True # Dimension of a span = rank of the matrix of spanning vectors. W_basis = [np.array([1, 0, -1], dtype=complex), np.array([0, 1, -1], dtype=complex)] print(np.linalg.matrix_rank(np.column_stack(W_basis))) # 2 (Example 2) ``` > Computing coordinates in a *non-standard* basis requires solving a linear system or inverting the > basis matrix — we'll do that systematically once we have the inner product in > [0.1.2](02-inner-products-norms.md), where for orthonormal bases it collapses to a simple projection. --- ## Exercises > Attempt before expanding. Solutions use the conventions of this lesson. **E1 (easy).** Verify axiom (D2), $(a+b)v = av + bv$, explicitly for $\mathbb{C}^2$ with $a = 1+i$, $b = 2$, $v = (i, 1)^T$.

Solution

$(a+b) = 3 + i$, so $(a+b)v = (3+i)(i,\,1)^T = ((3+i)i,\ 3+i)^T = (-1+3i,\ 3+i)^T$. Separately, $av = (1+i)(i,1)^T = (i + i^2,\ 1+i)^T = (-1+i,\ 1+i)^T$ and $bv = 2(i,1)^T = (2i, 2)^T$. Then $av + bv = (-1+i+2i,\ 1+i+2)^T = (-1+3i,\ 3+i)^T$. They match. ∎

**E2 (easy).** Show that $\{(1,1)^T, (1,-1)^T\}$ is a basis of $\mathbb{C}^2$ and find the coordinates of $(0,1)^T$ in it.

Solution

Independence: $a(1,1)^T + b(1,-1)^T = (a+b,\ a-b)^T = 0$ forces $a+b=0$ and $a-b=0$, hence $a=b=0$. Two independent vectors in a 2-dimensional space form a basis. For coordinates solve $a+b = 0,\ a-b = 1 \Rightarrow a = \tfrac12,\ b = -\tfrac12$. So $(0,1)^T = \tfrac12(1,1)^T - \tfrac12(1,-1)^T$. (Up to normalization these are $|+\rangle, |-\rangle$.)

**E3 (medium).** Let $W = \{ (z_1, z_2) \in \mathbb{C}^2 : z_1 = \overline{z_2} \}$ (bar = complex conjugate). Is $W$ a subspace of the *complex* vector space $\mathbb{C}^2$? Is it a subspace of $\mathbb{C}^2$ viewed as a *real* vector space?

Solution

Not a complex subspace: take $w = (1,1) \in W$ (since $\overline{1}=1$). Scale by $i$: $iw = (i, i)$; but $\overline{i} = -i \neq i$, so $iw \notin W$ — closure under complex scaling fails. It *is* a real subspace: it contains $0$, and is closed under addition and **real** scaling (for real $a$, $\overline{a z_2} = a\overline{z_2}$). This is the recurring lesson that "subspace" depends on the scalar field.

**E4 (medium).** Prove that any list of vectors containing the zero vector is linearly dependent.

Solution

Suppose $0$ is among $v_1, \dots, v_m$, say $v_1 = 0$. Then $1\cdot v_1 + 0\cdot v_2 + \cdots + 0\cdot v_m = 0$ is a vanishing linear combination with a **nonzero** coefficient ($a_1 = 1$). By definition the list is linearly dependent. ∎

**E5 (hard).** Show that $\mathbb{C}$ is a vector space over $\mathbb{R}$ of dimension $2$, but over $\mathbb{C}$ of dimension $1$. Generalize to state $\dim_\mathbb{R} \mathbb{C}^n$ vs $\dim_\mathbb{C} \mathbb{C}^n$.

Solution

Over $\mathbb{R}$: $\{1, i\}$ is independent (no real $a,b$ with $a\cdot 1 + b\cdot i = 0$ except $a=b=0$) and spans (every $z = a + bi$), so $\dim_\mathbb{R}\mathbb{C} = 2$. Over $\mathbb{C}$: $\{1\}$ spans (every $z = z\cdot 1$) and is independent, so $\dim_\mathbb{C}\mathbb{C} = 1$. In general $\dim_\mathbb{C}\mathbb{C}^n = n$ while $\dim_\mathbb{R}\mathbb{C}^n = 2n$, because each complex coordinate contributes two real degrees of freedom. This is why a single qubit, living in $\mathbb{C}^2$, has $2\cdot 2 = 4$ real parameters before we impose normalization and discard global phase — leaving the $2$ real parameters of the Bloch sphere (Term 1.2).

--- ## Checkpoint 1. State the eight vector-space axioms from memory (group them: 4 additive, 2 multiplicative, 2 distributive). 2. Why does quantum mechanics use $\mathbb{C}$ rather than $\mathbb{R}$ as its scalar field? 3. What is the difference between a *spanning set*, a *linearly independent set*, and a *basis*? 4. Give the complex and real dimensions of $\mathbb{C}^3$. 5. State the three conditions of the subspace test.

Answers

1. (A1) associativity, (A2) commutativity, (A3) zero vector, (A4) additive inverses; (M1) scalar-mult compatibility, (M2) scalar identity $1v=v$; (D1) distributivity over vector addition, (D2) distributivity over scalar addition. 2. Quantum amplitudes carry a relative **phase**; complex scalars encode magnitude *and* phase, and relative phases produce interference, which experiment confirms. A real theory cannot reproduce it. 3. A spanning set reaches every vector by linear combinations; an independent set has no redundancy; a basis is both at once (minimal spanning set / maximal independent set). 4. $\dim_\mathbb{C}\mathbb{C}^3 = 3$; $\dim_\mathbb{R}\mathbb{C}^3 = 6$. 5. Contains $0$; closed under vector addition; closed under scalar multiplication.

--- ## Further Reading - **[Axl]** Axler, *Linear Algebra Done Right*, Ch. 1–2 — vector spaces, span, independence, bases. - **[NC]** Nielsen & Chuang, §2.1.1 — bases and linear independence in the QC context. - **[Pre]** Preskill, *Ph219*, Ch. 2 — the linear-algebra setup as physicists use it. --- ← Prev: [Term 0 Overview](../README.md) · Up: [Term 0](../README.md) · Next: [Inner Products & Norms](02-inner-products-norms.md) →