Introduction to Vector Spaces

A vector space (over $\mathbb {R}$) consists of a set $V$ along with two operations "$+$" and "$\cdot$" subject to these conditions.

1. For any ${\vec {v}},{\vec {w}}\in V:{\vec {v}}+{\vec {w}}\in V$.
2. For any ${\vec {v}},{\vec {w}}\in V:{\vec {v}}+{\vec {w}}={\vec {w}}+{\vec {v}}$.
3. For any ${\vec {u}},{\vec {v}},{\vec {w}}\in V:({\vec {v}}+{\vec {w}})+{\vec {u}}={\vec {v}}+({\vec {w}}+{\vec {u}})$.
4. There is a zero vector ${\vec {0}}\in V$ such that ${\vec {v}}+{\vec {0}}={\vec {v}}$ for all ${\vec {v}}\in V$.
5. Each ${\vec {v}}\in V$ has an additive inverse ${\vec {w}}\in V$ such that ${\vec {w}}+{\vec {v}}={\vec {0}}$.
6. If $r$ is a scalar, that is, a member of $\mathbb {R}$ and ${\vec {v}}\in V$ then the scalar multiple $r\cdot {\vec {v}}$ is in $V$.
7. If $r,s\in \mathbb {R}$ and ${\vec {v}}\in V$ then $(r+s)\cdot {\vec {v}}=r\cdot {\vec {v}}+s\cdot {\vec {v}}$.
8. If $r\in \mathbb {R}$ and ${\vec {v}},{\vec {w}}\in V$ , then $r\cdot ({\vec {v}}+{\vec {w}})=r\cdot {\vec {v}}+r\cdot {\vec {w}}$.
9. If $r,s\in \mathbb {R}$ and ${\vec {v}}\in V$ , then $(rs)\cdot {\vec {v}}=r\cdot (s\cdot {\vec {v}})$.
10. For any ${\vec {v}}\in V$ , $1\cdot {\vec {v}}={\vec {v}}$.

Remark: Because it involves two kinds of addition and two kinds of multiplication, that definition may seem confused. For instance, in condition 7 "$(r+s)\cdot {\vec {v}}=r\cdot {\vec {v}}+s\cdot {\vec {v}}$", the first "$+$" is the real number addition operator while the "$+$" to the right of the equals sign represents vector addition in the structure $V$ . These expressions aren't ambiguous because, e.g., $r$ and $s$ are real numbers so "$r+s$" can only mean real number addition.

Lemma 1.17: In any vector space $V$, for any ${\vec {v}}\in V$ and $r\in \mathbb {R}$, we have

1. $0\cdot {\vec {v}}={\vec {0}}$, and
2. $(-1\cdot {\vec {v}})+{\vec {v}}={\vec {0}}$, and
3. $r\cdot {\vec {0}}={\vec {0}}$.

Proof: For 1, note that ${\vec {v}}=(1+0)\cdot {\vec {v}}={\vec {v}}+(0\cdot {\vec {v}})$. Add to both sides the additive inverse of ${\vec {v}}$, the vector ${\vec {w}}$ such that ${\vec {w}}+{\vec {v}}={\vec {0}}$.

${\begin{array}{rl}{\vec {w}}+{\vec {v}}&={\vec {w}}+{\vec {v}}+0\cdot {\vec {v}}\\{\vec {0}}&={\vec {0}}+0\cdot {\vec {v}}\\{\vec {0}}&=0\cdot {\vec {v}}\end{array}}$

The second item is easy: $(-1\cdot {\vec {v}})+{\vec {v}}=(-1+1)\cdot {\vec {v}}=0\cdot {\vec {v}}={\vec {0}}$ shows that we can write "$-{\vec {v}}\,$" for the additive inverse of ${\vec {v}}$ without worrying about possible confusion with $(-1)\cdot {\vec {v}}$.

For 3, this $r\cdot {\vec {0}}=r\cdot (0\cdot {\vec {0}})=(r\cdot 0)\cdot {\vec {0}}={\vec {0}}$ will do.

Example 1

The set $\mathbb {R} ^{2}$ is a vector space if the operations "$+$" and "$\cdot$" have their usual meaning.

${\binom {x_{1}}{x_{2}}}+{\binom {y_{1}}{y_{2}}}={\binom {x_{1}+y_{1}}{x_{2}+y_{2}}}\qquad r\cdot {\binom {x_{1}}{x_{2}}}={\binom {rx_{1}}{rx_{2}}}$

We shall check all of the conditions.

There are five conditions in item 1. For 1, closure of addition, note that for any $v_{1},v_{2},w_{1},w_{2}\in \mathbb {R}$ the result of the sum

${\binom {v_{1}}{v_{2}}}+{\binom {w_{1}}{w_{2}}}={\binom {v_{1}+w_{1}}{v_{2}+w_{2}}}$

is a column array with two real entries, and so is in $\mathbb {R} ^{2}$ . For 2, that addition of vectors commutes, take all entries to be real numbers and compute

${\binom {v_{1}}{v_{2}}}+{\binom {w_{1}}{w_{2}}}={\binom {v_{1}+w_{1}}{v_{2}+w_{2}}}={\binom {w_{1}+v_{1}}{w_{2}+v_{2}}}={\binom {w_{1}}{w_{2}}}+{\binom {v_{1}}{v_{2}}}$

(the second equality follows from the fact that the components of the vectors are real numbers, and the addition of real numbers is commutative). Condition 3, associativity of vector addition, is similar.

${\begin{aligned}{\Bigg (}{\binom {v_{1}}{v_{2}}}+{\binom {w_{1}}{w_{2}}}{\Bigg )}+{\binom {u_{1}}{u_{2}}}&={\binom {(v_{1}+w_{1})+u_{1}}{(v_{2}+w_{2})+u_{2}}}\\&={\binom {v_{1}+(w_{1}+u_{1})}{v_{2}+(w_{2}+u_{2})}}\\&={\binom {v_{1}}{v_{2}}}+{\Bigg (}{\binom {w_{1}}{w_{2}}}+{\binom {u_{1}}{u_{2}}}{\Bigg )}\end{aligned}}$

For the fourth condition we must produce a zero element — the vector of zeroes is it.

${\binom {v_{1}}{v_{2}}}+{\binom {0}{0}}={\binom {v_{1}}{v_{2}}}$

For 5, to produce an additive inverse, note that for any $v_{1},v_{2}\in \mathbb {R}$ we have

${\binom {-v_{1}}{-v_{2}}}+{\binom {v_{1}}{v_{2}}}={\binom {0}{0}}$

so the first vector is the desired additive inverse of the second.

The checks for the five conditions having to do with scalar multiplication are just as routine. For 6, closure under scalar multiplication, where $r,v_{1},v_{2}\in \mathbb {R}$ ,

$r\cdot {\binom {v_{1}}{v_{2}}}={\binom {rv_{1}}{rv_{2}}}$

is a column array with two real entries, and so is in $\mathbb {R} ^{2}$ . Next, this checks 7.

$(r+s)\cdot {\binom {v_{1}}{v_{2}}}={\binom {(r+s)v_{1}}{(r+s)v_{2}}}={\binom {rv_{1}+sv_{1}}{rv_{2}+sv_{2}}}=r\cdot {\binom {v_{1}}{v_{2}}}+s\cdot {\binom {v_{1}}{v_{2}}}$

For 8, that scalar multiplication distributes from the left over vector addition, we have this.

$r\cdot {\Bigg (}{\binom {v_{1}}{v_{2}}}+{\binom {w_{1}}{w_{2}}}{\Bigg )}={\binom {r(v_{1}+w_{1})}{r(v_{2}+w_{2})}}={\binom {rv_{1}+rw_{1}}{rv_{2}+rw_{2}}}=r\cdot {\binom {v_{1}}{v_{2}}}+r\cdot {\binom {w_{1}}{w_{2}}}$

The ninth

$(rs)\cdot {\binom {v_{1}}{v_{2}}}={\binom {(rs)v_{1}}{(rs)v_{2}}}={\binom {r(sv_{1})}{r(sv_{2})}}=r\cdot \left(s\cdot {\binom {v_{1}}{v_{2}}}\right)$

and tenth conditions are also straightforward.

$1\cdot {\binom {v_{1}}{v_{2}}}={\binom {1v_{1}}{1v_{2}}}={\binom {v_{1}}{v_{2}}}$

In a similar way, each $\mathbb {R} ^{n}$ is a vector space with the usual operations of vector addition and scalar multiplication. (In $\mathbb {R} ^{1}$, we usually do not write the members as column vectors, i.e., we usually do not write "$(\pi )$". Instead we just write "$\pi$".)

Example 2

The set $F=\{a\cos \theta +b\sin \theta \,{\big |}\,a,b\in \mathbb {R} \}$ of real-valued functions of the real variable $\theta$ is a vector space under the operations

$(a_{1}\cos \theta +b_{1}\sin \theta )+(a_{2}\cos \theta +b_{2}\sin \theta )=(a_{1}+a_{2})\cos \theta +(b_{1}+b_{2})\sin \theta$

and

$r\cdot (a\cos \theta +b\sin \theta )=(ra)\cos \theta +(rb)\sin \theta$

inherited from the space in the prior example. (We can think of $F$ as "the same" as $\mathbb {R} ^{2}$ in that $a\cos \theta +b\sin \theta$ corresponds to the vector with components $a$ and $b$.)