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2.A Vector spaces

At first, one might define a vector as a directed line-segment; in other words, an arrow of a given length pointing in some direction. We say that two vectors are equal if they have the same length and the same direction (though not necessarily the same position in space). In these notes we will usually denote vectors by small letters in boldface or with small arrows over them, such as $\vec{v}$ . When handwriting, it is also customary to underline vectors: $\underline{v}$ .

Usually, we will think of vectors as being elements of ${\mathbb{R}}^{n}$ ; in other words, an ordered list of real numbers, $\vec{x}=(\alpha_{1},\alpha_{2},\cdots,\alpha_{n})$ , where $\alpha_{i}\in{\mathbb{R}}$ . Here $\alpha_{i}$ are called the coordinates of $\vec{x}$ (see also 2.E). Instead of picturing this as a point in space, I recommend visualizing this element $\vec{x}\in{\mathbb{R}}^{n}$ as a directed line-segment (i.e. an arrow) starting at the origin $\vec{0}:=(0,\cdots,0)$ , and ending at the point $(\alpha_{1},\cdots,\alpha_{n})$ . If $V={\mathbb{R}}^{n}$ and $F={\mathbb{R}}$ , then the following statements should be familiar to you:

V1.

$V$ has a binary operation called vector addition: If $\vec{x},\vec{y}\in V$ , then $\vec{x}+\vec{y}\in V$ .
V2.

Addition is commutative: If $\vec{x},\vec{y}\in V$ , then $\vec{x}+\vec{y}=\vec{y}+\vec{x}$ .
V3.

Addition is associative: If $\vec{x},\vec{y},\vec{z}\in V$ , then $\vec{x}+(\vec{y}+\vec{z})=(\vec{x}+\vec{y})+\vec{z}$ .
V4.

There is an additive identity: $\exists\vec{0}\in V$ , called a zero vector, such that $\forall\vec{x}\in V$ we have $\vec{x}+\vec{0}=\vec{x}$ .
V5.

There are additive inverses: If $\vec{x}\in V$ , then $\exists\vec{y}\in V$ such that $\vec{x}+\vec{y}=\vec{0}$ , where $\vec{0}$ is a zero vector from V4.
V6.

$F$ is a field, and there is a scalar multiplication operation: If $\vec{x}\in V$ and $\alpha\in F$ then $\alpha\vec{x}\in V$ .
V7.

The multiplicative identity in $F$ operates as follows: If $\vec{x}\in V$ then $1\vec{x}=\vec{x}$ .
V8.

Scalar multiplication is compatible with multiplication in $F$ : If $\alpha,\beta\in F$ then $\alpha(\beta\vec{x})=(\alpha\beta)\vec{x}$ .
V9.

Distributivity of vector addition: If $\alpha\in F$ , $\vec{x},\vec{y}\in V$ then $\alpha(\vec{x}+\vec{y})=\alpha\vec{x}+\alpha\vec{y}$ .
V10.

Distributivity of field addition: If $\alpha,\beta\in F$ , $\vec{x}\in V$ then $(\alpha+\beta)\vec{x}=\alpha\vec{x}+\beta\vec{x}$ .

If $V$ is a set and $F$ is a field, and there is an addition $+$ (as in V1), and a scalar multiplication $\cdot$ (as in V6), which obey all of the above axioms, then the quadruple $(V,F,+,\cdot)$ is called a vector space. In that case, elements of the set $V$ are called vectors, and $F$ is called its field of scalars; we will also say that $V$ is a vector space over the field $F$ . The main example of a vector space that we will consider in this module is ${\mathbb{R}}^{n}$ , but there are other important ones as well. This is another abstract definition (see the discussion in Section 1.A). The above rules are called the vector space axioms.

Example 2.1.

i.
The set ${\mathbb{R}}^{n}$ for any positive integer $n\geq 1$ is a vector space over ${\mathbb{R}}$ . If $\vec{x}=(x_{1},x_{2},\cdots,x_{n})$ , $\vec{y}=(y_{1},y_{2},\cdots,y_{n})$ , and $\alpha\in{\mathbb{R}}$ , then the addition and scalar multiplication operations are defined as follows:
1. $\vec{x}+\vec{y}:=(x_{1}+y_{1},x_{2}+y_{2},\cdots,x_{n}+y_{n})$
2. $\alpha\vec{x}:=(\alpha x_{1},\alpha x_{2},\cdots,\alpha x_{n})$
ii.

The set $F^{n}:=\{(x_{1},\cdots,x_{n})\;|\;x_{i}\in F\}$ , for any field $F$ , with addition and scalar multiplication defined as in the $F={\mathbb{R}}$ case, is a vector space over $F$ . For example, axiom V3 follows from the field axiom F5; and V8 follows from F6; and so on.
iii.

The set $\operatorname{M}_{{n}}({{\mathbb{C}}})$ of square $n\times n$ complex matrices is a vector space over ${\mathbb{C}}$ , with the usual matrix addition and scalar multiplication.
iv.

As a generalization of the previous example, matrices $\operatorname{M}_{{n\times m}}({F})$ with $n$ rows and $m$ columns and whose coefficients are in $F$ , with usual addition and scalar multiplication, is a vector space over the field $F$ .
v.

The set $\mathcal{P}_{n}(F)$ of polynomials of degree less than or equal to $n$ , with coefficients in the field $F$ . In other words,

$\mathcal{P}_{n}(F):=\{c_{0}+c_{1}x+c_{2}x^{2}+\cdots+c_{n}x^{n}\;|\;c_{0},c_{1% },c_{2},\cdots,c_{n}\in F\}.$

You add polynomials by adding their coefficients. The zero vector in $\mathcal{P}_{n}(F)$ is $\vec{0}=0+0x+\cdots+0x^{n}$ . In fact, all of the vector space axioms are satisfied.
vi.
The set of all functions from ${\mathbb{R}}$ to ${\mathbb{R}}$ is a vector space over ${\mathbb{R}}$ . Addition and scalar multiplication are defined as follows:
1. $(f+g)(x):=f(x)+g(x)$
2. $(\alpha f)(x):=\alpha(f(x))$
where $\alpha\in{\mathbb{R}}$ and $f,g\colon{\mathbb{R}}\to{\mathbb{R}}$ .

Exercise 2.2:

Let $V=\operatorname{M}_{{2}}({{\mathbb{R}}})$ be the set of real $2\times 2$ matrices, with the usual matrix addition and scalar multiplication.

$V$ is a vector space over $\mathbb{R}$ . Why is $V$ not a vector space over $\mathbb{C}$ ?

[End of Exercise]

A key property of vector spaces is that they contain all linear combinations of all of their vectors.

Definition 2.3:

A linear combination of the vectors $\vec{v_{1}},\cdots,\vec{v_{r}}\in V$ is an vector of the following form:

\alpha_{1}\vec{v_{1}}+\alpha_{2}\vec{v_{2}}+\cdots+\alpha_{r}\vec{v_{r}}=\sum_% {i=1}^{r}\alpha_{i}\vec{v_{i}},

For some scalars $\alpha_{1},\cdots,\alpha_{r}\in F$ .

The above sum is written without brackets, which is only possible without risk of ambiguity due to the axiom V3.

At first, it is easiest to think about linear combinations within the vector space $V={\mathbb{R}}^{n}$ . But you will also need to think about linear combinations when $V$ is a vector space of matrices (so $+$ is matrix addition), or when $V$ is a vector space of functions (so $+$ refers to the addition of functions).

Exercise 2.4:

In the vector space $\mathbb{R}^{4}$ , determine whether or not $(1,2,3,4)$ is a linear combination of the two vectors $(1,-2,-1,4)$ and $(-1,4,3,-4)$ .

[Hint: You may need to solve a system of linear equations.]

[End of Exercise]

Here are some other elementary properties that follow from the axioms. These might seem “obvious” to you, but once you attempt to deduce them using only the axioms, you will realise how tricky and unintuitive the proofs can be.

Lemma 2.5.

Let $V$ be a vector space over a field $F$ .

i.

There is a unique zero vector; in other words, any two zero vectors are equal to each other, so we may call it “the zero vector”.
ii.

For each $\vec{v}\in V$ , there is a unique additive inverse; in other words, there is only one vector $\vec{w}\in V$ that obeys $\vec{v}+\vec{w}=\vec{0}$ . We write that vector “ $\vec{-v}$ ”.
iii.

$0\vec{v}=\vec{0}$ , for any $\vec{v}\in V$ .
iv.

$(-\alpha)\vec{v}=-(\alpha\vec{v})$ , for any $\alpha\in F$ and $\vec{v}\in V$ (notation as in part (ii)).
v.

For any $\alpha\in F$ , $\alpha\vec{0}=\vec{0}.$

Proof.

(i) Assume $\vec{0_{1}},\vec{0_{2}}\in V$ are two zero vectors; that means we have the following two equations for any $\vec{v}\in V$ :

$\vec{v}+\vec{0_{1}}=\vec{v}$ ,
$\vec{v}+\vec{0_{2}}=\vec{v}$ .

That is how a zero vector was defined, in axiom V4. Since these equations are true for any $\vec{v}\in V$ , they must be true for $\vec{v}=\vec{0_{1}}$ , in which case the second equation becomes $\vec{0_{1}}+\vec{0_{2}}=\vec{0_{1}}$ . But they must also be true for $\vec{v}=\vec{0_{2}}$ , in which case the first equation becomes $\vec{0_{2}}+\vec{0_{1}}=\vec{0_{2}}$ . Finally, by V2, we have the following equalities:

\vec{0_{1}}=\vec{0_{1}}+\vec{0_{2}}=\vec{0_{2}}+\vec{0_{1}}=\vec{0_{2}}.

So there can be only one zero vector.

(iii) We carefully proceed as follows. By F7 we have that $0+0=0$ in $F$ , and so

0\vec{v}=(0+0)\vec{v}

Now by V10 the above is equal to

0\vec{v}+0\vec{v}

Since $0\vec{v}\in V$ , by V5, there is an additive inverse $(-0\vec{v})$ which is also in $V$ . By V1, we can add this element to both sides of $0\vec{v}=0\vec{v}+0\vec{v}$ , which we have already established, to obtain

0\vec{v}+(-0\vec{v})=(0\vec{v}+0\vec{v})+(-0\vec{v})

Now by V3 the right hand side is equal to

0\vec{v}+(0\vec{v}+(-0\vec{v})).

By V5 the above is equal to

0\vec{v}+\vec{0}.

By V4, the above is eqaul to $0\vec{v}$ . But by V5, the left hand side of the above equation equals $\vec{0}$ . Therefore $0\vec{v}=\vec{0}$ .

For parts (ii) and (iv), see Exercise 2.6. For part (v) see Exercise 2.54. ∎

Exercise 2.6:

Try to carefully construct your own proofs, similar to the ones above, by proving Lemma 2.5(ii) and (iv), labelling the axioms used at each step.

[End of Exercise]