MathWithNaziaa : Linear Algebra/Definition and Examples of Vector Spaces

 A vector space (over ) consists of a set along with two operations "" and "" subject to these conditions.

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

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

The best way to go through the examples below is to check all ten conditions in the definition. That check is written out at length in the first example. Use it as a model for the others. Especially important are the first condition "${\displaystyle {\vec {v}}+{\vec {w}}}$ is in ${\displaystyle V}$" and the sixth condition "${\displaystyle r\cdot {\vec {v}}}$ is in ${\displaystyle V}$". These are the closure conditions. They specify that the addition and scalar multiplication operations are always sensible — they are defined for every pair of vectors, and every scalar and vector, and the result of the operation is a member of the set

Example :

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

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

We shall check all of the conditions.

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

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

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

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

(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.

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

For the fourth condition we must produce a zero element — 

the vector of zeroes is it.

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

For 5, to produce an additive inverse, note that for 

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

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

any ${\displaystyle v_{1},v_{2}\in \mathbb {R} }$ we have

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

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

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

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

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 ${\displaystyle r,v_{1},v_{2}\in \mathbb {R} }$ ,

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

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