Vector spaces

We have defined addition and scalar multiplication of vectors. These operations satisfy a set of laws that characterize vector algebra. Conversely, any set equipped with ``addition'' and ``scalar multiplication'' that satisfy the laws of vector algebra is considered a vector space. In this view, vectors are elements of a vector space.

We usually assume that vectors are elements of ${\mathbb{R}}^n$ for some $n\in \mathbb{N}$. This ${\mathbb{R}}^n$ is an example of a vector space.

Remark. In mathematics, in general, we tend to use the word ``space'' to denote a set with some additional ``structures''. e.g., vector space, probability space, topological space, Banach space, inner product space, Hilbert space, etc. □

The following results apply to any vector spaces (e.g. ${\mathbb{C}}^n$). Therefore we use $V$ to represent an arbitrary vector space. We use $K$ to represent the field underlying the vector space. We say $V$ is a vector space over the field $K$.

Example. 

• ${\mathbb{R}}^n$ is a vector space over the field $\mathbb{R}$.
• ${\mathbb{C}}^n$ is a vector space over the field $\mathbb{C}$.

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The following lemma is a list of algebraic laws for vectors. We use boldface letters such as $\mathbf{u},\mathbf{v}$ to represent vectors and Greek letters such as $\lambda ,\mu ,\cdots$ to represent scalars. 

Lemma 

1. For all $\mathbf{u},\mathbf{v}\in V$, $\mathbf{u}+\mathbf{v}\in V$.  (Vector addition is closed in $V$.)
2. For all $\mathbf{u},\mathbf{v}\in V$, $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$. (Vector addition is commutative.)
3. For any $\mathbf{u},\mathbf{v},\mathbf{w}\in V$, $(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})$. (Vector addition is associative.)
4. There exists an element $\mathbf{0}\in V$ such that for all $\mathbf{u}\in V$, $\mathbf{0}+\mathbf{u}=\mathbf{u}$. (The additive identity element exists and is called $\mathbf{0}$ (zero).)
5. For each $\mathbf{u}\in V$, there exists $\mathbf{-}\mathbf{u}$ such that $\mathbf{u}+(\mathbf{-}\mathbf{u})=\mathbf{0}$. (The additive inverse ($-\mathbf{u}$) always exists.)
6. For each $\lambda \in K$, and for each $\mathbf{u}\in V$, $\lambda \mathbf{u}\in V$. (Multiplication by a scalar is closed in $V$.)
7. For each $\mathbf{u}\in V$, for each $\lambda ,\mu \in K$, $(\lambda +\mu )\mathbf{u}=\lambda \mathbf{u}+\mu \mathbf{u}$. (The distributive law: multiplication by a vector distributes over scalar addition. [Note: Scalars are elements of the field $K$ so that addition and multiplication are defined.])
8. For all $\mathbf{u},\mathbf{v}\in V$, for all $\lambda \in K$, $\lambda (\mathbf{u}+\mathbf{v})=\lambda \mathbf{u}+\lambda \mathbf{v}$.(The distributive law: multiplication by a scalar distributes over vector addition.)
9. For all $\mathbf{u}\in V$, $1\cdot \mathbf{u}=\mathbf{u}$. (The multiplicative identity of $K$ acts as the multiplicative identity for the scalar multiplication in $V$.)
10. For all $\lambda ,\mu \in K$, for all $\mathbf{u}\in V$, $(\lambda \mu )\mathbf{u}=\lambda (\mu \mathbf{u})$. (Scalar multiplication is associative.)

Example (${\mathbb{R}}^2$). Let's prove some of the above laws for ${\mathbb{R}}^2$. Let $\mathbf{u}=(u_1,u_2)$ and $\mathbf{v}=(v_1,v_2)$ and $\lambda \in \mathbb{R}$. 

(Law 1) Recall that vector addition is defined by

$$\mathbf{u}+\mathbf{v}=(u_1,u_2)+(v_1,v_2)=(u_1+v_1,u_2+v_2).$$

Clearly, 

$$(u_1+v_1,u_2+v_2)\in {\mathbb{R}}^2.$$

Therefore, ${\mathbb{R}}^2$ is closed under vector addition.

(Law 6) Scalar multiplication is defined by

$$\lambda \mathbf{u}=\lambda (u_1,u_2)=(\lambda u_1,\lambda u_2),$$

and clearly,

$$(\lambda u_1,\lambda u_2)\in {\mathbb{R}}^2.$$

Thus, ${\mathbb{R}}^2$ is closed under scalar multiplication.

(Law 4) The zero element $\mathbf{0}$ is given by $\mathbf{0}=(0,0)$ where $0$ is the real number zero.

In fact,

$$\mathbf{0}+\mathbf{u}=(0,0)+(u_1,u_2)=(0+u_1,0+u_2)=(u_1,u_2).$$

(Law 5) The additive inverse of $\mathbf{u}=(u_1,u_2)$ is $-\mathbf{u}=(-u_1,-u_2)$. In fact,

$$\mathbf{u}+(-\mathbf{u})=(u_1,u_2)+(-u_1,-u_2)=(u_1-u_1,u_2-u_2)=(0,0).$$

(Law 9) The multiplicative identity for the scalar multiplication is $1\in \mathbb{R}$. In fact,

$$1\cdot (u_1,u_2)=(1\cdot u_1,1\cdot u_2)=(u_1,u_2).$$

You should check other laws, too. □

Definition (Vector space)

Let $V$ be a set equipped with addition and scalar multiplication, with scalars being elements of the field $K$. If all of the above laws (1-10) are satisfied, we say that $V$ is a vector space.

Although it may look similar, a vector space is not a field. For example, we don't have multiplication between vectors but only between scalars and vectors. A vector space is actually an algebraic structure of its own kind.

See also: Fields