Background on vector spaces

First, some notation and background. If $S$ and $T$ are any two sets, let $S\times T$ denote the Cartesian product of $S$ and $T$ defined by

$$S\times T=\{(s,t)\ \vert\ s\in S,\ t\in T\}.$$

In other words, $S\times T$ is simply the set of ordered pair of elements in $S$ and $T$. For example, $\mathbb{R}\times \mathbb{R}$ (which is also written as $\mathbb{R}^2$) is the Cartesian plane, where the first coordinate usually corresponds to the ``$x$-axis'' and the second coordinate to the ``$y$-axis''. More generally, if $S_1$, $S_2$, ..., $S_n$ are $n$ sets, then $S_1\times S_2 \times ...\times S_n$ is defined by

$$S_1\times S_2 \times ...\times S_n =\{(s_1,s_2,...,s_n)\ \vert\ s_i\in S_i,\ i=1,2,...,n\}.$$

If all these sets $S_i$ are equal to each other, say $S=S_1=...=S_n$, then we write this as $S^n$. In other words, $S^n$ is simply the set of ordered $n$-tuples of elements in $S$. For example, $\mathbb{R}^3$ is Cartesian $3$-space, where the first coordinate usually corresponds to the ``$x$-axis'', the second coordinate to the ``$y$-axis'', and the third coordinate the ``$z$-axis''. Let $F$ be any field and let $V=F^n$ denote the set of $n$-tuples of elements of $F$. Define vector addition $+$ and scalar multiplication $\cdot$ as follows: for ${\bf v}=(v_1,...,v_n), {\bf w}=(w_1,w_2,...,w_n)\in V$ and $c\in F$, define vector addition and scalar multiplication ``component-wise''

$${\bf v}+{\bf w}=(v_1+w_1,v_2+w_2,...,v_n+w_n),$$

$$c\cdot {\bf v}=(cv_1,cv_2,...,cv_n).$$

It is easy to check that with these two operations, $V$ satisfies the following axioms for a vector space.

Definition 3.1.1   Let $F$ be a field and $V$ be a set with operations $+:V\times V\rightarrow V$, written $+:({\bf u},{\bf v})\longmapsto {\bf u}+{\bf v}$, and $\cdot :F\times V\rightarrow V$, $+:(a,{\bf v})\longmapsto a\cdot {\bf v}$. $V$ is called a vector space over $F$ (or an $F$-vector space) if the following properties hold. For all ${\bf u},{\bf v},{\bf w}\in V$ and $a,b\in F$,

• ${\bf u}+{\bf v}={\bf v}+{\bf u}$ (commutativity)
• $({\bf u}+{\bf v})+{\bf w}={\bf u}+({\bf v}+{\bf w})$ (associativity)
• the vector ${\bf0}=(0,0,...,0)\in V$ satisfies ${\bf u}+{\bf0}={\bf u}$ (the zero vector ${\bf0}$ is the additive identity),
• for each ${\bf v}\in V$ the element $(-1){\bf v}=-{\bf v}\in V$ satisfies ${\bf v}+(-{\bf v})={\bf0}$ (each element ${\bf v}$ has an additive inverse $-{\bf v}$)
• $(a+b){\bf v}=a{\bf v}+b{\bf v}$ and $a({\bf v}+{\bf w})=a{\bf v}+a{\bf w}$ (distributive laws)
• $(ab){\bf v}=a(b{\bf v})$
• $1\cdot {\bf v}={\bf v}$.

In particular, every vector space must contain at least one element (the zero vector). The elements of $V$ are called vectors and the elements of $F$ are called scalars. Any subset $W\subset V$ of a vector space which is closed under vector addition (restricting $+$ from $V$ to $W$) and scalar multiplication is called a subspace of $V$.

Example 3.1.2   If $V=F^3$ (which reduces to the usual Euclidean $3$-space when $F=\mathbb{R}$) then $W=\{(x,y,0)\ \vert\ x,y\in F\}$ is a vector subspace but $W=\{(x,y,1)\ \vert\ x,y\in F\}$ is not.

Suppose $V$ is a vector space over a field $F$. If there are non-zero vectors ${\bf f}_1,{\bf f}_2,...,{\bf f}_k$ such that every ${\bf v}\in V$ can be written as a ``linear combination'' of these ${\bf v}_i$'s, i.e.,

$${\bf v}=c_1{\bf f}_1+c_2{\bf f}_2+...+c_k{\bf f}_k,$$

for some scalar coefficients $c_i\in F$ ( $i=1,2,...,k$) then we say that $V$ is spanned by the elements ${\bf f}_1,{\bf f}_2,...,{\bf f}_k$. We also say that elements ${\bf f}_1,{\bf f}_2,...,{\bf f}_k$ form a spanning set for $V$.

Example 3.1.3   If $V=F^3$ is then $V$ is spanned by ${\bf e}_1=(1,0,0),{\bf e}_2=(0,1,0),{\bf e}_3 =(0,0,1)$. In fact, if ${\bf f}_1=(1,1,0),{\bf f}_2=(1,0,0), {\bf f}_3=(0,0,1),{\bf f}_4=(0,1,0)$ then ${\bf f}_1,...,{\bf f}_4$ forms a spanning set for $V$ as well (since it contains the set ${\bf e}_1,{\bf e}_2,{\bf e}_3$).

Definition 3.1.4   If a vector space $V$ has a finite spanning set then we say that $V$ is finite dimensional. The dimension of $V$ over $F$ is the smallest number of vectors in a spanning set of $V$. The dimension of $V$ is written $dim(V)=k$. If $V$ is of dimension $k$ then there are $k$ vectors which span $V$ and any such set of $k$ spanning vectors of $V$ is called a basis of $V$.

Lemma 3.1.5   Let $V$ be a $k$-dimensional vector space over $F$ and let ${\bf f}_1,{\bf f}_2,...,{\bf f}_k$ be a basis of $V$. For each ${\bf v}\in V$ there is a unique set of coefficients $c_1,...,c_k$ in $F$ such that

$${\bf v}=c_1{\bf f}_1+c_2{\bf f}_2+...+c_k{\bf f}_k.$$

proof: Since ${\bf f}_1,{\bf f}_2,...,{\bf f}_k$ is a spanning set, for each ${\bf v}\in V$ there is a set of coefficients $c_1,...,c_k$ such that the above equality holds. To see that this is unique, suppose

$${\bf v}=c_1{\bf f}_1+c_2{\bf f}_2+...+c_k{\bf f}_k =c'_1{\bf f}_1+c'_2{\bf f}_2+...+c'_k{\bf f}_k.$$

We must show $c_1=c'_1$, ..., $c_k=c'_k$. Suppose not, to get a contradiction. Suppose $c_i\not= c'_i$, for some $i$. For typographical simplicity we will assume $i=1$. Subtracting, we get ${\bf0}=(c_1-c'_1){\bf f}_1+(c_2-c'_2){\bf f}_2+...+ (c_k-c'_k){\bf f}_k$. By our assumption,

$${\bf f}_1={c_2-c'_2\over c'_1-c_1}{\bf f}_2+...+ {c_k-c'_k\over c'_1-c_1}{\bf f}_k.$$

Let ${\bf w}\in V$ be arbitrary. Substitutinig this expression into a basis expansion

$${\bf w}=b_1{\bf f}_1+b_2{\bf f}_2+...+b_k{\bf f}_k , \ \ \ \ \ {\bf w}\in V,$$

yields a shorter expansion of the form

$${\bf w}=b'_2{\bf f}_2+...+b'_k{\bf f}_k .$$

Since ${\bf w}$ was arbitrary, this implies $V$ is $k-1$ dimensional, a contradiction. $\Box$ This lemma is important. It says that we may identify any vector in a finite-dimensional vector space with its list of coefficients with respect to a fixed basis.

Lemma 3.1.6   The vectors ${\bf e}_1=(1,0,0,...,0)$, ${\bf e}_2=(0,1,0,...,0)$,..., ${\bf e}_n=(0,0,...,0,1)$, form a basis for $F^n$.

The basis in the above lemma is called the standard basis. proof: If ${\bf v}=(v_1,...,v_n)\in F^n$ then ${\bf v}=v_1{\bf e}_1+...v_n{\bf e}_n$, so the vectors ${\bf e}_1,...,{\bf e}_n$ span $F^n$. If any one of these vectors were omitted, say ${\bf e}_i$, then it would be impossible to span all of $V$ (${\bf e}_i$ itself would not be in the span of the others, for example). Thus no proper subset of $\{{\bf e}_1,...,{\bf e}_n\}$ spans $F^n$, so they form a basis. $\Box$