Vector Spaces

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

As we mentioned in the introduction to the previous post, the vector space is the central object of study in linear algebra, generalizing the coordinate spaces learned in high school. To prepare for this, we defined abelian groups and fields in the last post.

Many linear algebra textbooks avoid these definitions and consider only $\mathbb{R}$-vector spaces or $\mathbb{C}$-vector spaces, but the more general case is not at all more complicated, so there is no need to restrict our attention to special cases.

Definition of Vector Spaces

Definition 1 Let $\mathbb{K}$ be a field and $V$ an abelian group. We say that $V$ is a vector space over $\mathbb{K}$, or simply a $\mathbb{K}$-vector space, if there is an additional operation (scalar multiplication) $\cdot:\mathbb{K}\times V\rightarrow V$ satisfying

$\alpha\cdot(\beta\cdot u)=(\alpha\beta)\cdot u$ for any $\alpha,\beta\in\mathbb{K}$ and any $u\in V$.
$\alpha\cdot(u+_{\tiny V}v)=(\alpha\cdot u)+_{\tiny V}(\alpha\cdot v)$ for any $\alpha\in\mathbb{K}$ and any $u,v\in V$.
$(\alpha+_{\tiny \mathbb{K}}\beta)\cdot u=(\alpha\cdot u)+_{\tiny V}(\beta\cdot u)$ for any $\alpha,\beta\in\mathbb{K}$ and any $u\in V$.
$1\cdot u=u$ for any $u\in V$, where $1\in\mathbb{K}$ is the multiplicative identity.

The elements of $V$ are called vectors.

As in the definition above, to avoid confusion we will write elements of the field $\mathbb{K}$ as $\alpha,\beta,\ldots$ and elements of a $\mathbb{K}$-vector space as $u,v,\ldots$. In the definition we distinguished $+_{\tiny V}$ and $+_{\tiny \mathbb{K}}$, but with this notation the elements around $+$ make it clear whether they belong to $\mathbb{K}$ or $V$, so we may simply write $+$ without risk of confusion.

Similarly, we will write $\alpha u$ instead of $\alpha\cdot u$. The only concern is that $\alpha\beta u$ could be read as either $(\alpha\beta)u$ or $\alpha(\beta u)$, but by the first condition of the definition both choices yield the same value, so this is not a problem.

A vector space is an abelian group $V$ equipped with the additional structure of $\mathbb{K}$-scalar multiplication. Therefore $V$ possesses all the properties of an abelian group. (§Abelian Groups and Fields, ⁋Proposition 2 and §Abelian Groups and Fields, ⁋Corollary 3)

The following properties are additional ones determined by the $\mathbb{K}$-scalar multiplication.

Proposition 2 Let $V$ be a $\mathbb{K}$-vector space. Then

$\alpha 0=0$ for any $\alpha\in\mathbb{K}$, and
$0v=0$ for any $v\in V$.

Conversely, if $\alpha v=0$, then either $\alpha=0$ or $v=0$.

Proof

The first two claims proceed similarly to §Abelian Groups and Fields, ⁋Proposition 6. For example,

\[\alpha0+\alpha0=\alpha(0+0)=\alpha0\]

so $\alpha0=0$, and similarly

\[0v+0v=(0+0)v=0v\]

so $0v=0$. Finally, suppose $\alpha v=0$ and $\alpha\neq 0$. Then $\alpha^{-1}\in\mathbb{K}$ exists with $\alpha\alpha^{-1}=1$, and thus

\[v=1v=(\alpha^{-1}\alpha)v=\alpha^{-1}(\alpha v)=\alpha^{-1}0=0\]

so $v=0$, and the proposition follows.

The $0$ appearing in part 1 of the proposition and on the right-hand side of part 2 are elements of $V$, while the $0$ on the left-hand side of part 2 is an element of $\mathbb{K}$. Strictly speaking these should be distinguished as $0_{\tiny V}$ and $0_{\tiny \mathbb{K}}$, but context makes the distinction clear, so we write them all as $0$.

Corollary 3 For any element $v$ of a $\mathbb{K}$-vector space $V$, $(-1)v=-v$ always holds.

Proof

This follows immediately from the equation

\[(-1)v+v=(-1)v+1v=((-1)+1)v=0v=0\]

and the uniqueness of additive inverses in $V$.

Examples of Vector Spaces

Let us now look at some examples of vector spaces.

Example 4 The simplest example of a vector space is $\{0\}$. There is only one way to give this set an addition structure (namely $0+0=0$), and with this structure the set is an abelian group. Moreover, for any field $\mathbb{K}$ there is also only one way to define scalar multiplication on this set (namely $\alpha 0=0$), and this scalar multiplication makes $\{0\}$ a $\mathbb{K}$-vector space. We call this the trivial space.

A slightly less trivial example is the field itself. For any field $\mathbb{K}$, $\mathbb{K}$ is a $\mathbb{K}$-vector space. Since $\mathbb{K}$ is a field, it is trivially an abelian group under addition. We only need to give it a scalar multiplication structure, which is simply multiplication in $\mathbb{K}$, i.e. $\mathbb{K}\times \mathbb{K}\rightarrow \mathbb{K}$. One can verify that this scalar multiplication satisfies all the conditions of Definition 1, and thus $\mathbb{K}$ is a $\mathbb{K}$-vector space in its own right.

More generally, let $\mathbb{K}$ be a field and suppose there is another field $\mathbb{K}'$ with $\mathbb{K}'\supseteq \mathbb{K}$. Then $\mathbb{K}'$ is a $\mathbb{K}$-vector space. Since $\mathbb{K}'$ is a field, it is an abelian group under addition as before, and scalar multiplication by an element $\alpha\in\mathbb{K}$ is given by viewing $\alpha$ as an element of $\mathbb{K}'$ and using the multiplication in $\mathbb{K}'$. For example, $\mathbb{C}$ is an $\mathbb{R}$-vector space, and $\mathbb{R}$ is a $\mathbb{Q}$-vector space.

Example 5 Now suppose a field $\mathbb{K}$ is given. The Euclidean $n$-dimensional space is the $\mathbb{K}$-vector space consisting of $n$-tuples

\[\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix},\qquad a_i\in\mathbb{K}\text{ for all $i$}\]

Addition and scalar multiplication are defined by

\[\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix}+\begin{pmatrix}b_1\\b_2\\\vdots\\b_n\end{pmatrix}=\begin{pmatrix}a_1+b_1\\a_2+b_2\\\vdots\\a_n+b_n\end{pmatrix},\qquad \alpha\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix}=\begin{pmatrix}\alpha a_1\\\alpha a_2\\\vdots\\\alpha a_n\end{pmatrix}\]

When $\mathbb{K}=\mathbb{R}$ and $n=2,3$, this gives the familiar coordinate plane and coordinate space.

Euclidean spaces will be especially important objects for us. In the example above we used column notation instead of the ordered pair $(a_1, a_2, \ldots, a_n)$, which is closely related to the fundamental theorem of linear algebra.

However, no matter how compelling the reason, insisting on this $\begin{pmatrix}a_1\\a_2\\ \vdots\\a_n\end{pmatrix}$ notation throughout the text would be foolish. Therefore, in the main text we will use notation such as $(a_1\;a_2\;\cdots\;a_n)^t$, or following high school convention, simply $(a_1,a_2,\ldots, a_n)$.

The two vector spaces examined above are fairly concrete examples. As the next example shows, vector spaces are not always visualizable like the coordinate plane or coordinate space.

Example 6 Let $I$ be an interval and consider the set $\Fun(I,\mathbb{R})$ of functions from $I$ to $\mathbb{R}$. If we define addition and scalar multiplication on this set by

\[f+g:t\mapsto f(t)+g(t),\qquad \alpha f:t\mapsto \alpha f(t)\]

then one can verify that $\Fun(I,\mathbb{R})$ has a vector space structure. That is, $f+g$ is the function sending each $t\in I$ to $f(t)+g(t)$, and $\alpha f$ is the function sending each $t\in I$ to $\alpha f(t)$.

Moreover, various subsets of $\Fun(I,\mathbb{R})$ are also $\mathbb{R}$-vector spaces. For example, the set $C(I)$ of continuous functions from $I$ to $\mathbb{R}$ is an $\mathbb{R}$-vector space, and more generally one can verify that the set $C^k(I)$ of functions whose $k$th derivative is continuous is also an $\mathbb{R}$-vector space.

If we think of $\Fun(I,\mathbb{R})$ as the product set $\mathbb{R}^I$, then Example 6 can be viewed as a natural generalization of Example 5. (§Product of Sets, ⁋Definition 1)

References

[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.

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Definition of Vector Spaces

Examples of Vector Spaces

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