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 $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$. (Actually the following works for many filds whjich includes the rational numbers, integers modulo $p$, and so forth). A vector space over $\mathbb{F}$ also called “real” and “complex” vector spaces for $F = \mathbb{R},\mathbb{C}$ respectively, is a set $V$ of elements $\ket{v}$ with the following properties:

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

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

(141)#\[(\ket{v} + \ket{w}) + \ket{y} = \ket{v} + (\ket{w} + \ket{y})\]

a zero vector $\ket{0}\in V$ exists such that $\ket{v} + \ket{0} = \ket{v}$.

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

Multiplication is associative: For all $b \in \mathbb{F}$, $a(b\ket{v}) = (ab)\ket{v}$, where $ab$ is the standard mjultiplication of numbers in $\mathbb{F}$.
$1\ket{v} = \ket{v}$.
$0 \ket{v} = \ket{0}$.

Distributive properties

For all $a\in \mathbb{F}$, $\ket{v},\ket{w} \in V$,

(142)#\[a\left(\ket{v} + \ket{w}\right) = a \ket{v} + b \ket{w}\]

for all $a,b \in \mathbb{F}$, $\ket{v} \in V$,

(143)#\[(a + b)\ket{v} = a\ket{v} + b \ket{v}\]

From these rules we can also deduce the existence of an additive inverse: for enery $\ket{v} \in V$, there exists a vector $\ket{-v} \in V$ such that $\ket{v} + \ket{-v} = \ket{0}$. This can be seen bu construction: set $\ket{-v} = (-1) \ket{v}$. Then

(144)#\[\ket{v} + \ket{-v} = \ket{v} + (-1) \ket{v} = (1 + (-1) \ket{v} = 0 \ket{v} = \ket{0}\]

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.

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

\[\begin{split}\ket{v} = \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix}\end{split}\]

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

(145)#\[\begin{split}\begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} + \begin{pmatrix} d_1 \\ d_2 \\ \vdots \\ d_n \end{pmatrix} = \begin{pmatrix} c_1+ d_1 \\ c_2+d_2 \\ \vdots \\ c_n+d_n \end{pmatrix}\end{split}\]

(146)#\[\begin{split}a \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} = \begin{pmatrix} a c_1 \\ a c_2 \\ \vdots \\ a c_n \end{pmatrix}\end{split}\]

for any $a \in \mathbb{C}$.

Note we can do ths same with $c_k, d_k \in \mathbb{R}$: 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 $M_n(\mathbb{C})$. Addition and scalar multiplication are just matrix addition and scalar miltiplication (for $M \in M_n$, $aM$ is elementwise multiplication by $a$.
Degree-$n$ polynomials over $\mathbb{C}$:

(147)#\[\ket{a_0,\ldots a_n} = a_0 + a_1 x + a_2 x^2 + \ldots a_n x^n\]

with addition and scalar multiplication working in the standard way. Note that this is clearly equivalent to $\mathbb{C}^n$. 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 $\psi(x)$ 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 $\psi(0) = \psi(1) = 0$.
Neumann $\psi'(0) = \psi'(1) = 0$
Periodic $\psi(0) = \psi(1)$ (so $\psi$ is a function on a circle). However, the boundary condition $\psi(0) = a$, $\psi(1) = b$ for nonzero $a,b \in \mathbb{C}$ 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 $\mathbb{R}$: that is, functions $\psi(x)$ for $x\in \mathbb{R}$ such that

(148)#\[\int_{-\infty}^{\infty} dx |\psi(x)|^2 < \infty\]

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 $V = \mathbb{C}^3$,

(149)#\[\begin{split}M = \left\{ \begin{pmatrix} c_1 \\ c_2 \\ 0 \end{pmatrix} \forall c_i \in \mathbb{C} \right\}\end{split}\]

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

(150)#\[\begin{split}a \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix}\end{split}\]

for fixed $c_k\in \mathbb{C}^n$ and all $a \in \mathbb{C}^n$.

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

(151)#\[\begin{split}a \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} + \begin{pmatrix} d_1 \\ d_2 \\ \vdots \\ d_n \end{pmatrix} \end{split}\]

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 $\ket{v_1},\ldots,\ket{v_m} \in V$ is linearly independent if

(152)#\[\sum_{i = 1}^m a_k \ket{v_k} = 0 \Leftrightarrow a_k = 0 \forall\ $k = 1,\ldots,n\]

Let us give some examples.

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

\[\begin{split}\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\end{split}\]

Similarly these ar elinearly independent:

\[\begin{split}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\end{split}\]

However, these three are not:

\[\begin{split}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\ ; \ \begin{pmatrix} 1/2 \\ 1/2 \\ 1 \end{pmatrix}\end{split}\]

as we can see because

\[\begin{split}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} - 2 \begin{pmatrix} 1/2 \\ 1/2 \\ 1 \end{pmatrix} = 0\end{split}\]

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

\[\ket{n} = \sin\left(\frac{n\pi x}{L}\right)\]

are linearly independent. Similarly, for $n$th order polynomials, the monomials $\ket{k} = x^k$ 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 $\ket{k}$, $k = 1,\ldots,n$, then for any vector $\ket{v}$ there is a unique set of complex numbers $a_{k - 1,\ldots,n$ such that

(153)#\[\sum_{k = 1}^n a_k \ket{k} \]

Proof. Assume the contrary, that

(154)#\[\ket{v} = \sum_{k = 1}^n a_k \ket{k} = \sum_{k = 1}^n b_k \ket{k}\]

for $a_k,b_k \in \mathbb{C}$ different numbers. The point is that if this is true,

(155)#\[\ket{0} = \ket{v} - \ket{v} = \sum_{k = 1}^n a_k \ket{k} - \sum_{k = 1}^n b_k \ket{k} = \sum_{k = 1}^n (a_k + b_k)\ket{k}\]

but this cannot be zero of $a_k,b_k$ differ in any way, so we have a contradiction to our supposition. Thus $a_k = b_k$.