Definition of Vector Space
Definition1
When the elements of a non-empty set $V$ satisfy the following ten rules for two operations, addition and scalar multiplication, $V$ is called a vector space over field2 $\mathbb{F}$, and the elements of $V$ are called vectors.
For $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $k, l \in \mathbb{F}$,
(A1) If $\mathbf{u}, \mathbf{v}$ is an element of $V$, then $\mathbf{u}+\mathbf{v}$ is also an element of $V$.
(A2) $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$
(A3) $(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})$
(A4) For all $\mathbf{u}$ in $V$, there exists $\mathbf{0}$ in $V$ that satisfies $\mathbf{u} + \mathbf{0} = \mathbf{0} + \mathbf{u} = \mathbf{u}$. This $\mathbf{0}$ is called the zero vector.
(A5) For all $\mathbf{u}$ in $V$, there exists $\mathbf{v}$ in $V$ that satisfies $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} = \mathbf{0}$. This $\mathbf{v}$ is called the negative of $\mathbf{u}$ and is denoted by $\mathbf{v} = -\mathbf{u}$.
(M1) If $\mathbf{u}$ is an element of $V$, then $k \mathbf{u}$ is also an element of $V$.
(M2) $k(\mathbf{u} + \mathbf{v})=k\mathbf{u} + k\mathbf{v}$
(M3) $(k+l)\mathbf{u}=k\mathbf{u}+ l\mathbf{u}$
(M4) $k(l\mathbf{u})=(kl)(\mathbf{u})$
(M5) For $1\in \mathbb{F}$, $1\mathbf{u} = \mathbf{u}$
Explanation
- The term linear space is also used.
Naturally, the scalar (field) does not need to be real numbers. Specifically, when $\mathbb{F} = \mathbb{R}$, it is called a real vector space, and when $\mathbb{F} = \mathbb{C}$, it is called a complex vector space.
In undergraduate linear algebra, mainly $\mathbb{R}^{n}$ or $\mathbb{C}^{n}$ are discussed. $\mathbb{R}^{n}$ refers to the vector space containing ordered pairs of real numbers $n$. In other words, it signifies the $n$-dimensional Euclidean space, specifically $\mathbb{R}^{3}$ refers to the three-dimensional space extensively covered in high school mathematics and calculus.
There are various sets that can become vector spaces. A set of functions can also be a vector space, which is called a function space.
In physics, something with magnitude and direction is called a vector. This concept is generalized in linear algebra. For example, consider a set $M_{m\times n}(\mathbb{R})$ that collects real number matrices of size $m\times n$. Then, it can be seen that $M_{m\times n}(\mathbb{R})$ satisfies all ten rules mentioned above. Therefore, a set of matrices of the same size becomes a vector space, and each matrix within it becomes a vector. If you are encountering this abstract vector space for the first time, the fact that matrices can be vectors might be surprising, but it makes sense when thinking about how vectors in coordinate spaces have been denoted.
To determine whether a set is a vector space, one must examine whether it satisfies the definition above. It might seem intuitively to be a vector space when it is not, and vice versa. Since there can be cases entirely different from intuition, it is advisable to carefully examine each one when solving problems. Also, the zero vector $\mathbf{0}$ and the scalar $0$ are entirely different entities and should be distinguished. Typically, vectors are expressed in bold type in textbooks.
Theorem 1
Let $V$ be a vector space and $\mathbf{u}$ be an element of $V$.
(1a) The zero vector of $V$ is unique.
(1b) The negative of $\mathbf{u}$ is unique.
Proof
This is a proof using the definition of vector spaces.
(1a)
Let’s say $\mathbf{0},\mathbf{0}^{\prime}$ is the zero vector of $V$. Then, by the definition of vector spaces, the following holds.
$$ \begin{align*} \mathbf{0} &= \mathbf{0} + \mathbf{0}^{\prime} && \text{by (A4)} \\ &= \mathbf{0}^{\prime} + \mathbf{0} && \text{by (A2)} \\ &= \mathbf{0}^{\prime} && \text{by (A4)} \end{align*} $$
Therefore, the two zero vectors are the same.
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(1b)
Let’s say $\mathbf{v}, \mathbf{v}^{\prime}$ is the negative of $\mathbf{u}$. Then, by the definition of vector spaces, the following holds.
$$ \begin{align*} \mathbf{v} &= \mathbf{v} + \mathbf{0} && \text{by (A4)} \\ &= \mathbf{v} + \left( \mathbf{u} + \mathbf{v}^{\prime} \right) && \text{by (A5)} \\ &= \left( \mathbf{v} + \mathbf{u} \right) + \mathbf{v}^{\prime} && \text{by (A3)} \\ &= \mathbf{0} + \mathbf{v}^{\prime} && \text{by (A5)} \\ &= \mathbf{v}^{\prime} && \text{by (A4)} \end{align*} $$
Therefore, the two negatives of $\mathbf{u}$ are the same.
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Theorem 2
Let $V$ be a vector space, $\mathbf{u}$ be an element of $V$, and $k$ be a scalar.
(2a) $0 \mathbf{u} = \mathbf{0}$
(2b) $k \mathbf{0} = \mathbf{0}$
(2c) $(-1) \mathbf{u} = -\mathbf{u}$
(2d) If $k \mathbf{u} = \mathbf{0}$, then $k = 0$ or $\mathbf{u} = \mathbf{0}$.
Proof
This is a proof using the definition of vector spaces.
(2a)
$$ \begin{align*} && 0\mathbf{u} &= (0 + 0)\mathbf{u} \\ && &= 0\mathbf{u} + 0\mathbf{u} &&\text{by (M3)} \\ & & \\ \implies && 0\mathbf{u}+(-0\mathbf{u}) &= 0\mathbf{u} + 0\mathbf{u} +(-0\mathbf{u}) \\ \implies && \mathbf{0} &= 0\mathbf{u} &&\text{by (A5)} \end{align*} $$
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(2b)
$$ \begin{align*} && k\mathbf{0} &= k(\mathbf{0} + \mathbf{0}) &&\text{by (A4)} \\ && &= k\mathbf{0} + k\mathbf{0} &&\text{by (M2)} \\ & & \\ \implies && k\mathbf{0}+(-k\mathbf{0}) &= k\mathbf{0} + k\mathbf{0} +(-k\mathbf{0}) \\ \implies && \mathbf{0} &= k\mathbf{0} &&\text{by (A5)} \end{align*} $$
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(2c)
$$ \begin{align*} \mathbf{u} + (-1)\mathbf{u} &= 1 \mathbf{u} + (-1)\mathbf{u} &&\text{by (M5)} \\ &= \big( 1 + (-1) \big) \mathbf{u} &&\text{by (M3)} \\ &= 0 \mathbf{u} \\ &= \mathbf{0} &&\text{by (a2)} \end{align*} $$
Thus, by (A5), $(-1)\mathbf{u}$ is the negative of $\mathbf{u}$, and since the negative of $\mathbf{u}$ is unique by (1b),
$$ (-1)\mathbf{u} = -\mathbf{u} $$
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(2d)
$k$ must either be $0$ or not $0$, so let’s consider both scenarios.
If $k=0$
This satisfies the conclusion.
If $k\ne 0$
Since $k$ is not $0$, it can be divided by $k$, hence
$$ \begin{align*} && k \mathbf{u} &= \mathbf{0} \\ \implies && \mathbf{u} &= \frac{1}{k}\mathbf{0} \\ && &= \mathbf{0} && \text{by (2b)} \end{align*} $$
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See Also
Abstract Algebra
The $F$-vector spaces discussed in the documents below are essentially no different from the vector spaces mentioned above, albeit from a slightly different perspective. Vector spaces in linear algebra are abstractions of intuitive Euclidean spaces, whereas in abstract
algebra, vector spaces are brought into the realm of ‘algebra’ in the true sense.
Conversely, $R$-modules generalize the scalar field $F$ of $F$-vector spaces into scalar rings $R$, thereby revealing their identity in a naming that is indifferent to the history and meaning of $F$-vector fields. From the perspective of group $G$, it’s about adding a new operation $\mu$ to ring $R$, thus also being a module.