Coordinate spaces – "Math for Non-Geeks"

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The coordinate space is the vector space of $n$-tuples $(x_{1},...,x_{n})$ with entries $x_{k}\in K$ in a field $K$, equipped with componentwise addition and scalar multiplication. An example is the vector space $\mathbb {R} ^{3}$ known from school, with vectors $(x_{1},x_{2},x_{3})$ and $K=\mathbb {R}$.

Derivation

In mathematics, one often uses existing structures to define new and more general ones. As we have already seen in introduction to vector space, we can extend from the vector spaces $\mathbb {R} ^{3}$ and $\mathbb {R} ^{2}$ over the real numbers $\mathbb {R}$ to more general vector spaces $\mathbb {R} ^{n}$ for every natural number $n$. For this we recall how the addition of two vectors and the scalar multiplication between a vector and a scalar works in $\mathbb {R} ^{3}$ and $\mathbb {R} ^{2}$: We have

${\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}+{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}={\begin{pmatrix}x_{1}+y_{1}\\x_{2}+y_{2}\\x_{3}+y_{3}\end{pmatrix}}\in \mathbb {R} ^{3}\quad {\text{und}}\quad {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}+{\begin{pmatrix}y_{1}\\y_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+y_{1}\\x_{2}+y_{2}\end{pmatrix}}\in \mathbb {R} ^{2}$

In other words, the addition and scalar multiplication are defined component-wise. That is, we perform the addition and the scalar multiplication in $\mathbb {R} ^{2}$ and $\mathbb {R} ^{3}$ by adding in each component and multiplying in each component with the scalar, respectively. In the same way we can define an addition and a scalar multiplication if our vectors do not consist of two or three but of $n$ real numbers. That is, on the set

$\mathbb {R} ^{n}=\{(x_{1},\dots ,x_{n})\mid x_{i}\in \mathbb {R} {\text{ for all }}1\leq i\leq n\}$

we define a component-wise vector addition and a scalar multiplication:

For the vector addition we use the real number addition and for the scalar multiplication the real number multiplication. Let $x=(x_{1},\dots ,x_{n})$ and $y=(y_{1},\dots ,y_{n})$ with $x_{i},y_{i}\in \mathbb {R}$, then the vector addition is defined by

$x+y=(x_{1}+y_{1},\dots ,x_{n}+y_{n})$

Let $x\in \mathbb {R} ^{n}$ and $\alpha \in \mathbb {R}$, then the scalar multiplication is defined by

$\alpha \cdot x=(\alpha \cdot x_{1},\dots ,\alpha \cdot x_{n})$

We can now easily verify that $\mathbb {R} ^{n}$ with this vector addition and scalar multiplication is a vector space over the field $\mathbb {R}$.

Thus we have transferred the known structure of the real numbers $\mathbb {R}$ and the vector spaces $\mathbb {R} ^{2}$ and $\mathbb {R} ^{3}$ to the vector space $\mathbb {R} ^{n}$. We also refer to $\mathbb {R} ^{n}$ as coordinate space of dimension $n$ over $\mathbb {R}$.

Simple generalization

If we look again at the definition of the vector space structure on $\mathbb {R} ^{n}$, we have used only multiplication and addition on $\mathbb {R}$. But now, any field $K$ admits a multiplication and addition. Thus the above construction also provides us with a way to define a coordinate space over arbitrary fields. This coordinate space is defined by taking the set

$K^{n}=\{(x_{1},\dots ,x_{n})|x_{i}\in K{\text{ for all }}1\leq i\leq n\}$

and equipping it with an addition and a scalar multiplication. For this we copy the definition of above and define it component-wise. That means we use in every component the addition and multiplication of $K$ to define the addition and scalar multiplication on $K^{n}$.

Definition: coordinate space

Definition (The set $K^{n}$)

Let $n\in \mathbb {N}$ and $K$ be a field. We define

$K^{n}:=\left\{\,(x_{1},\dots ,x_{n})\,{\bigg \vert }\,x_{i}\in K{\text{ for all }}1\leq i\leq n\,\right\}.$

We denote the elements of this set as $n$-tuples with entries in $K$.

Example (Examples for tuples)

As an example, $(5,7)$ is a $2$-tuple with entries in $\mathbb {Q}$ (we also just say tuple or pair instead of $2$-tuple).

$\left(\pi ,12,{\sqrt {31}}\right)$ is a $3$-tuple with entries in $\mathbb {R}$ (we also say triple instead of $3$-tuple).

Now, we equip the set $K^{n}$ with an addition and a scalar multiplication.

Definition (vector space operations on $K^{n}$)

The addition $\boxplus \colon K^{n}\times K^{n}\to K^{n}$ is defined by ${\begin{aligned}(x_{1},\dots ,x_{n})\boxplus (y_{1},\dots ,y_{n}):=(x_{1}+y_{1},\cdots ,x_{n}+y_{n}).\end{aligned}}$

Similarly, we define the scalar multiplication $\boxdot \colon K\times K^{n}\to K^{n}$ by

$\lambda \boxdot (x_{1},\dots ,x_{n}):=(\lambda \cdot x_{1},\dots ,\lambda \cdot x_{n}).$

We call $(K^{n},\boxplus ,\boxdot )$ coordinate space.

Hint

We have so far taken a tuple to be a row vector. That means, we have written $(x_{1},\dots ,x_{n})$ for an element in $K^{n}$. Just as well, instead of writing the elements as one row with $n$ columns, we could also write them as one column with $n$ entries. Then an element in $K^{n}$ would look like this:

${\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}$

This other representation does NOT change the properties of $K^{n}$ as vector space. If we take the vector $x\in K^{n}$ to be a row with $n$ columns, then $x$ is called a row vector. If we take $x$ in turn as a column with $n$ rows as a column vector.

It will prove useful for matrices (missing) to write the vectors in $K^{n}$ as column vectors. Therefore, from now on, we will work with column vectors. However, the notation of a column vector in a line is not very space-saving. Therefore, we introduce the following notation: Instead of

$x={\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\in K^{n}$

we write the vector as $x=(x_{1},\dots ,x_{n})^{T}\in K$. The symbol $^{T}$ means that this vector is transposed, i.e. the row vector is transformed into a column vector. This transposition is the same as for matrices (missing)

Coordinate spaces are vector spaces

In the article introduction to the vector space we used the above construction first over $\mathbb {R}$ and then over arbitrary fields to derive the vector space axioms. Moreover, field satisfies similar properties as vector space and we have used the former very directly to define addition and scalar multiplication on the coordinate space. Therefore, we can conjecture that the definition of $\boxplus$ and $\boxdot$ on $K^{n}$ defines a vector space structure, as well. And this is indeed true, as we will verify now.

Theorem ($K^{n}$ is a vector space)

$(K^{n},\boxplus ,\boxdot )$ is a $K$-vector space.

How to get to the proof? ($K^{n}$ is a vector space)

We proceed as in the article "Proofs for vector spaces", where the eight vector space axioms were checked, one by the other.

The definitions of $\boxplus ,\boxdot$ are chosen by copying the operations $+,\cdot$ in the field $K$ "in a natural (component-wise) way to the vector space $K^{n}$." We show that the vector space axioms follow directly from the corresponding field axioms. So we have to show in every step that the definitions of $\boxplus$ and $\boxdot$ induce the known properties of $+$ and $\cdot$ in the field $K$.

Proof step: Associativity of addition

Let $x=(x_{1},\cdots ,x_{n})^{T},y=(y_{1},\cdots ,y_{n})^{T},z=(z_{1},\cdots ,z_{n})^{T}\in K^{n}$. Then, we have: ${\begin{aligned}(x\boxplus y)\boxplus z&=((x_{1},\cdots ,x_{n})^{T}\boxplus (y_{1},\cdots ,y_{n})^{T})\boxplus (z_{1},\cdots ,z_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{1}+y_{1},\cdots ,x_{n}+y_{n})^{T}\boxplus (z_{1},\cdots ,z_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=((x_{1}+y_{1})+z_{1},\cdots ,(x_{n}+y_{n})+z_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of addition in }}K\right.}\\[0.3em]&=(x_{1}+(y_{1}+z_{1}),\cdots ,x_{n}+(y_{n}+z_{n}))^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{1},\cdots ,x_{n})^{T}\boxplus (y_{1}+z_{1},\cdots ,y_{n}+z_{n})^{T}=x\boxplus (y\boxplus z).\end{aligned}}$ This shows the associativity of addition.

Proof step: Commutativity of addition

Let $x=(x_{1},\cdots ,x_{n})^{T},y=(y_{1},\cdots ,y_{n})^{T}\in K^{n}$. Then, we have: ${\begin{aligned}x\boxplus y&=(x_{1},\cdots ,x_{n})^{T}\boxplus (y_{1},\cdots ,y_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{1}+y_{1},\cdots ,x_{n}+y_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of addition in }}K\right.}\\[0.3em]&=(y_{1}+x_{1},\cdots ,y_{n}+x_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(y_{1},\cdots ,y_{n})^{T}\boxplus (x_{1},\cdots ,x_{n})^{T}=y\boxplus x.\end{aligned}}$ This shows the commutativity of addition.

Proof step: Neutral element of addition

We still need to show that there is a neutral element $0_{K^{n}}\in K^{n}$ for which we have that

$x\boxplus 0_{K^{n}}=x{\text{ for all }}x\in K^{n}$

Since we trace all properties back to the corresponding properties in $K$, here we use the neutral element of addition $0\in K$ to construct the neutral element of addition $0_{K^{n}}\in K^{n}$. That means, we set

$0_{K^{n}}=(\underbrace {0,0,\cdots ,0} _{n})^{T}.$

Is this neutral element of addition really neutral? For this, we have to check $x\boxplus 0_{K^{n}}=x$: Let $x=(x_{1},\cdots ,x_{n})^{T}\in K^{n}$. Then, we have:

$x\boxplus 0_{K^{n}}=(x_{1}+0,\cdots ,x_{n}+0)^{T}=(x_{1},\cdots ,x_{n})^{T}=x.$

Thus we have shown that $0_{K^{n}}\in K^{n}$ is the neutral element of addition.

Proof step: Inverse with respect to addition

Let $x=(x_{1},\cdots ,x_{n})^{T}\in K^{n}$.

We need to show that there exists a $y\in K^{n}$ such that $x\boxplus y=0_{K^{n}}$. Let us reduce this problem to the properties of arithmetic operations in $K$. In $K$ we have that if $a,b\in K$ and $a+b=0$, then $b=-a$. Therefore, for $y$ we choose the $n$ tuple $(-x_{1},\cdots ,-x_{n})^{T}$ as the potential inverse. Then, we have:

$x\boxplus y=(x_{1}+(-x_{1}),\cdots ,x_{n}+(-x_{n}))^{T}=(\underbrace {0,\cdots ,0} _{n})^{T}=0_{K^{n}}$

Thus we have shown that for any $x\in K^{n}$ there exists a $y\in K^{n}$ with $x\boxplus y=0_{K^{n}}$.

Proof step: Scalar distributive law

Let $\lambda ,\mu \in K$ and $x=(x_{1},\cdots ,x_{n})^{T}\in K^{n}$. Then, we have: ${\begin{aligned}(\lambda +\mu )\boxdot x&=(\lambda +\mu )\boxdot (x_{1},\cdots ,x_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=((\lambda +\mu )\cdot x_{1},\cdots ,(\lambda +\mu )\cdot x_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=(\lambda \cdot x_{1}+\mu \cdot x_{1},\cdots ,\lambda \cdot x_{n}+\mu \cdot x_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x_{1},\cdots ,\lambda \cdot x_{n})^{T}\boxplus (\mu \cdot x_{1},\cdots ,\mu \cdot x_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot (x_{1},\cdots ,x_{n})^{T})\boxplus (\mu \boxdot (x_{1},\cdots ,x_{n})^{T})\\[0.3em]&=(\lambda \boxdot x)\boxplus (\mu \boxdot x).\end{aligned}}$ Thus the scalar distributive law is also shown.

Proof step: Vectorial distributive law

Let $\lambda \in K$ and $x=(x_{1},\cdots ,x_{n})^{T},y=(y_{1},\cdots ,y_{n})^{T}\in K^{n}$. Then, we have: ${\begin{aligned}\lambda \boxdot (x\boxplus y)&=\lambda \boxdot ((x_{1},\cdots ,x_{n})^{T}\boxplus (y_{1},\cdots ,y_{n})^{T})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=\lambda \boxdot (x_{1}+y_{1},\cdots ,x_{n}+y_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \cdot (x_{1}+y_{1}),\cdots ,\lambda \cdot (x_{n}+y_{n}))^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=(\lambda \cdot x_{1}+\lambda \cdot y_{1},\cdots ,\lambda \cdot x_{n}+\lambda \cdot y_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x_{1},\cdots ,\lambda \cdot x_{n})^{T}\boxplus (\lambda \cdot y_{1},\cdots ,\lambda \cdot y_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot (x_{1},\cdots ,x_{n})^{T})\boxplus (\lambda \boxdot (y_{1},\cdots ,y_{n})^{T})\\[0.3em]&=(\lambda \boxdot x)\boxplus (\lambda \boxdot y).\end{aligned}}$ This establishes the vectorial distributive law.

Proof step: Associativity of multiplication

Let $\lambda ,\mu \in K$ and $x=(x_{1},\cdots ,x_{n})^{T}\in K^{n}$. Then, we have: ${\begin{aligned}(\lambda \cdot \mu )\boxdot x&=(\lambda \cdot \mu )\boxdot (x_{1},\cdots ,x_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=((\lambda \cdot \mu )\cdot x_{1},\cdots ,(\lambda \cdot \mu )\cdot x_{n})^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of multiplication in }}K\right.}\\[0.3em]&=(\lambda \cdot (\mu \cdot x_{1}),\cdots ,\lambda \cdot (\mu \cdot x_{n}))^{T}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \cdot x_{1},\cdots ,\mu \cdot x_{n})^{T})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \boxdot (x_{1},\cdots ,x_{n})^{T})\\[0.3em]&=\lambda \boxdot (\mu \boxdot x).\end{aligned}}$ This shows the associative law for multiplication.

Proof step: unitarity law

Let $x=(x_{1},\cdots ,x_{n})^{T}\in K^{n}$. Then, we have: $1_{K}\boxdot x=1_{K}\boxdot (x_{1},\cdots ,x_{n})^{T}=(1_{K}\cdot x_{1},\cdots ,1_{K}\cdot x_{n})^{T}=(x_{1},\cdots ,x_{n})^{T}=x.$

Thus we have also shown the unitary law.

Thus we have shown all eight vector space axioms and hence $(K^{n},\boxplus ,\boxdot )$ is indeed a $K$-vector space. }}

Relation to the field being a vector space

We have already seen that $K$ is a $K$-vector space. This is a special case of the coordinate spaces $K^{n}$, because it is $K^{1}=K$. Here we take the vectors $(x)\in K^{1}$ to be elements of the field. We then write instead of the $1$-tuple $(x)$ only $x$, instead of $(1)$ only $1$ and instead of $(0)$ only $0$.

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