Vector spaces

Vector spaces#

Definition#

We will consider complex vector spaces but let us start with a bit of generality so that we can compare them to more familiar real vector spaces. Consider $F = R$ or $C$ . (Actually the following works for many filds whjich includes the rational numbers, integers modulo $p$ , and so forth). A vector space over $F$ also called “real” and “complex” vector spaces for $F = R, C$ respectively, is a set $V$ of elements $| v ⟩$ with the following properties:

Vector addition. For all $| v ⟩, | w ⟩ \in V$ there is a notion of addition $| v ⟩ + | w ⟩ \in V$ wuth the following properties:

Addition is commutative: $| v ⟩ + | w ⟩ = | w ⟩ + | v ⟩$ .
Addition is *associative. If there is a third vector $| y ⟩ \in V$ ,

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(| v ⟩ + | w ⟩) + | y ⟩ = | v ⟩ + (| w ⟩ + | y ⟩)

a zero vector $| 0 ⟩ \in V$ exists such that $| v ⟩ + | 0 ⟩ = | v ⟩$ .

Scalar multiplication. For all $a \in F$ , $| v ⟩ \in V$ , there is a notion of scalar multiplication such that $a | v ⟩ \in V$ with the following properties:

Multiplication is associative: For all $b \in F$ , $a (b | v ⟩) = (a b) | v ⟩$ , where $a b$ is the standard mjultiplication of numbers in $F$ .
$1 | v ⟩ = | v ⟩$ .
$0 | v ⟩ = | 0 ⟩$ .

Distributive properties

For all $a \in F$ , $| v ⟩, | w ⟩ \in V$ ,

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a (| v ⟩ + | w ⟩) = a | v ⟩ + b | w ⟩

for all $a, b \in F$ , $| v ⟩ \in V$ ,

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(a + b) | v ⟩ = a | v ⟩ + b | v ⟩

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| v ⟩ + | - v ⟩ = | v ⟩ + (- 1) | v ⟩ = (1 + (- 1) | v ⟩ = 0 | v ⟩ = | 0 ⟩

Note that I have not yet introduced any notion ofthe length of a vector, whetehr two vectors are orthogonal, and so on. As we will see, this requires som eadditional structure.

Examples#

Theer are a number of more and less familiar examples.

$C^{n}$ , the space of $n$ -component column vectors

\begin{array}{r} | v ⟩ = (\begin{array}{c} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{array}) \end{array}

with $c_{k} \in C$ . We define vector addition and scalar multiplication in the usual way:

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\begin{array}{r} (\begin{array}{c} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{array}) + (\begin{array}{c} d_{1} \\ d_{2} \\ ⋮ \\ d_{n} \end{array}) = (\begin{array}{c} c_{1} + d_{1} \\ c_{2} + d_{2} \\ ⋮ \\ c_{n} + d_{n} \end{array}) \end{array}

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\begin{array}{r} a (\begin{array}{c} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{array}) = (\begin{array}{c} a c_{1} \\ a c_{2} \\ ⋮ \\ a c_{n} \end{array}) \end{array}

for any $a \in C$ .

Note we can do ths same with $c_{k}, d_{k} \in R$ : then we have a real vector space. Here the zero vector is defined by $c_{k} = 0$ .

The space of $n \times n$ complex-valued matrices $M_{n} (C)$ . Addition and scalar multiplication are just matrix addition and scalar miltiplication (for $M \in M_{n}$ , $a M$ is elementwise multiplication by $a$ .
Degree- $n$ polynomials over $C$ :

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| a_{0}, \dots a_{n} ⟩ = a_{0} + a_{1} x + a_{2} x^{2} + \dots a_{n} x^{n}

with addition and scalar multiplication working in the standard way. Note that this is clearly equivalent to $C^{n}$ . Note also that there is no reason for $n$ to be finite – we could work with the space of all polynomials.

Complex functions on an interval: let $x \in [0, 1]$ . The set of all functions $ψ (x)$ forms a vector space under the standard addition and scalar multiplication of functions if we choose the right boundary conditions. These boundary conditions yield vector spaces:

Dirichlet $ψ (0) = ψ (1) = 0$ .
Neumann $ψ^{'} (0) = ψ^{'} (1) = 0$
Periodic $ψ (0) = ψ (1)$ (so $ψ$ is a function on a circle). However, the boundary condition $ψ (0) = a$ , $ψ (1) = b$ for nonzero $a, b \in C$ is not a vector space under standard addition of functions: the sum of two such functions does not satisfy the required boundary conditions and so is not in $V$ .

Coplex square-integrable functions on $R$ : that is, functions $ψ (x)$ for $x \in R$ such that

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\int_{- \infty}^{\infty} d x | ψ (x) |^{2} < \infty

Subspaces#

A set $M \subset V$ is a vector subspace if it is a vector space under the same laws for addition and svcalar multiplication. A standard example is any plane through the origin, such as $V = C^{3}$ ,

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\begin{array}{r} M = {(\begin{array}{c} c_{1} \\ c_{2} \\ 0 \end{array}) \forall c_{i} \in C} \end{array}

Similarly, any complex line through the origin, defined as the set of vectors

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\begin{array}{r} a (\begin{array}{c} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{array}) \end{array}

for fixed $c_{k} \in C^{n}$ and all $a \in C^{n}$ .

A counterexample is any complex line that does not run through the origin, defined as the set of all vectors

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\begin{array}{r} a (\begin{array}{c} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{array}) + (\begin{array}{c} d_{1} \\ d_{2} \\ ⋮ \\ d_{n} \end{array}) \end{array}

with $c_{k}, d_{k}$ fixed and the same for all vectors in this space, and $a$ any complex number. It is clear that the sum of two vectors is a different vector, if there is at least one $d_{k} \neq 0$ .

Linear independence#

Definition. A set of vectors $| v_{1} ⟩, \dots, | v_{m} ⟩ \in V$ is linearly independent if

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\sum_{i = 1}^{m} a_{k} | v_{k} ⟩ = 0 \Leftrightarrow a_{k} = 0 \forall $ k = 1, \dots, n

Let us give some examples.

$C^{3}$ . These vectors are linearly independent:

\begin{array}{r} (\begin{array}{c} 1 \\ 0 \\ 0 \end{array}); (\begin{array}{c} 0 \\ 1 \\ 0 \end{array}); (\begin{array}{c} 0 \\ 0 \\ 1 \end{array}) \end{array}

Similarly these ar elinearly independent:

\begin{array}{r} (\begin{array}{c} 1 \\ 0 \\ 1 \end{array}); (\begin{array}{c} 0 \\ 1 \\ 1 \end{array}); (\begin{array}{c} 0 \\ 0 \\ 1 \end{array}) \end{array}

However, these three are not:

\begin{array}{r} (\begin{array}{c} 1 \\ 0 \\ 1 \end{array}); (\begin{array}{c} 0 \\ 1 \\ 1 \end{array}); (\begin{array}{c} 1 / 2 \\ 1 / 2 \\ 1 \end{array}) \end{array}

as we can see because

\begin{array}{r} (\begin{array}{c} 1 \\ 0 \\ 1 \end{array}) + (\begin{array}{c} 0 \\ 1 \\ 1 \end{array}) - 2 (\begin{array}{c} 1 / 2 \\ 1 / 2 \\ 1 \end{array}) = 0 \end{array}

In the space of functions on the interval $[0, 1]$ satisfying periodic boundary conditions, the vectors

| n ⟩ = \sin (\frac{n π x}{L})

are linearly independent. Similarly, for $n$ th order polynomials, the monomials $| k ⟩ = x^{k}$ are a linearly independent set of $n$ th order polynomials.

Dimension of a vector space#

Definition: the dimension of a vector space $V$ is the maximum number of linearly independent vectors in $V$ . Any such maximal collection is called a basis.

Theorem: Given a basis $| k ⟩$ , $k = 1, \dots, n$ , then for any vector $| v ⟩$ there is a unique set of complex numbers $a_{k - 1,\ldots,n$ such that

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\sum_{k = 1}^{n} a_{k} | k ⟩

Proof. Assume the contrary, that

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| v ⟩ = \sum_{k = 1}^{n} a_{k} | k ⟩ = \sum_{k = 1}^{n} b_{k} | k ⟩

for $a_{k}, b_{k} \in C$ different numbers. The point is that if this is true,

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| 0 ⟩ = | v ⟩ - | v ⟩ = \sum_{k = 1}^{n} a_{k} | k ⟩ - \sum_{k = 1}^{n} b_{k} | k ⟩ = \sum_{k = 1}^{n} (a_{k} + b_{k}) | k ⟩

but this cannot be zero of $a_{k}, b_{k}$ differ in any way, so we have a contradiction to our supposition. Thus $a_{k} = b_{k}$ .