# Dimension – Serlo

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In this article we define the dimension of a vector space and show some elementary properties like the dimension formula.

## Motivation

In this article we define the notion of a dimension of a vector space. It would be nice to do it in a way that common vector spaces like ${\displaystyle \mathbb {R} ,\mathbb {R} ^{2},\mathbb {R} ^{3},...}$ have the "obvious dimensions" ${\displaystyle 1,2,3,...}$ . The vector space ${\displaystyle \mathbb {R} ^{3}}$ has a basis with ${\displaystyle 3}$ elements, for example the standard basis ${\displaystyle B_{3}=\lbrace e_{1},e_{2},e_{3}\rbrace }$.

Similarly, for any field ${\displaystyle K}$, we want the vector space ${\displaystyle K^{n}}$ to have dimension ${\displaystyle \mathrm {dim} _{K}(K^{n})=n}$. Again, with the standard basis ${\displaystyle B_{n}=\lbrace e_{1},...,e_{n}\rbrace }$ we find a basis with exactly ${\displaystyle n}$ vectors.

This suggests to define the dimension of a vector space ${\displaystyle V}$ as the number of vectors of a basis of ${\displaystyle V}$. At this point, it is not yet clear that every basis has the same cardinality. We will prove this in the following.

## Definition of the dimension

Definition (Dimension of a vector space)

Let ${\displaystyle V}$ be a ${\displaystyle K}$-vector space and let ${\displaystyle B}$ be a basis of ${\displaystyle V}$. If ${\displaystyle B}$ is finite, we define the dimension of ${\displaystyle V}$ by ${\displaystyle \mathrm {dim} _{K}(V):=\mathrm {dim} (V):=|B|}$. (We usually omit the specification of the field if this is obvious from the context). In this case, we say that ${\displaystyle V}$ is finite dimensional. If ${\displaystyle B}$ is infinite instead, we say that ${\displaystyle V}$ is infinite-dimensional and write ${\displaystyle \mathrm {dim} (V)=\infty }$.

From this definition it is not clear that the dimension is independent of the choice of the basis of our vector space. For example, it could happen that a vector space has different bases with different numbers of elements. This does actually never happen, as the next theorem shows:

Theorem (Well-definedness of the dimension)

Let ${\displaystyle V}$ be a ${\displaystyle K}$-vector space and ${\displaystyle B,B'}$ two bases of ${\displaystyle V}$. If ${\displaystyle B}$ is finite, then ${\displaystyle B'}$ is also finite and we have that ${\displaystyle |B|=|B'|}$.

Proof (Uniqueness of the dimension)

Let ${\displaystyle B}$ be finite. Suppose that the basis ${\displaystyle B'}$ was infinite. Then, we could choose any ${\displaystyle |B|+1}$-elementary subset ${\displaystyle W}$ of ${\displaystyle B'}$. Note that ${\displaystyle W}$ is linearly independent, as it is a subset of the linearly independent set ${\displaystyle B'}$. This contradicts Steinitz's exchange theorem because of ${\displaystyle |B|<|W|}$. Thus ${\displaystyle B'}$ is finite. It remains to show the equality of cardinalities. Since ${\displaystyle B}$ is linearly independent and ${\displaystyle B'}$ is a basis of ${\displaystyle V}$, it follows from Steinitz's exchange theorem that ${\displaystyle |B|\leq |B'|}$. Analogously, we can prove ${\displaystyle |B'|\leq |B|}$. This establishes the theorem.

## Examples of dimensions

### Dimension of ${\displaystyle K^{n}}$

Example

In this example we want to verify that the dimension of ${\displaystyle K^{n}}$ is indeed ${\displaystyle n}$. A possible basis would be the canonical basis:

${\displaystyle B=\lbrace \underbrace {(1,0,\ldots ,0)^{T}} _{n{\text{ components}}},\,\underbrace {(0,1,\ldots ,0)^{T}} _{n{\text{ components}}},\,\cdots ,\,\underbrace {(0,0,\ldots ,1)^{T}} _{n{\text{ components}}}\rbrace }$

This basis is finite and we have that ${\displaystyle |B|=n}$. We get ${\displaystyle \mathrm {dim} _{K}(K^{n})=n}$.

### Dimension of a polynomial space

Example

The polynomial space ${\displaystyle K[x]}$ over a field ${\displaystyle K}$ is defined as

${\displaystyle K[x]:=\left\lbrace \sum _{i=0}^{n}a_{i}x^{i}\,{\bigg |}\,n\in \mathbb {N} ,a_{i}\in K\,\,\forall i\in \lbrace 0,\ldots ,n\rbrace \right\rbrace ,}$

with coefficient-wise addition and scalar multiplication. From this we see that ${\displaystyle \lbrace x^{n}|n\in \mathbb {N} \rbrace }$ is a generator of ${\displaystyle K[x]}$. Since the vector space identifications operate coefficient-wise, this set is also linearly independent. From ${\displaystyle {\big |}\lbrace x^{n}|n\in \mathbb {N} \rbrace {\big |}=\infty }$ we finally get ${\displaystyle \mathrm {dim} _{K}(K[x])=\infty }$.

### Dimension of ${\displaystyle \mathbb {C} }$ as an ${\displaystyle \mathbb {R} }$-vector space

Example

We would like to determine the dimension of the complex numbers, understood as an ${\displaystyle \mathbb {R} }$-vector space. Each complex number ${\displaystyle z\in \mathbb {C} }$ can be written uniquely as ${\displaystyle z=a+b\cdot i}$, with ${\displaystyle a,b\in \mathbb {R} }$. From this we see that ${\displaystyle \lbrace 1,i\rbrace }$ is a basis of ${\displaystyle \mathbb {C} }$ over ${\displaystyle \mathbb {R} }$. So ${\displaystyle \mathrm {dim} _{\mathbb {R} }(\mathbb {C} )=2}$, i.e., as a real vector space, ${\displaystyle \mathbb {C} }$ is 2-dimensions. (Note that as a complex vector space, ${\displaystyle \mathbb {C} }$ is 1-dimensional).

### Dimension of the null space

Example

For every field ${\displaystyle K}$, the ("trivial") null space ${\displaystyle \{0\}}$ is a vector space. To determine its dimension, we need to find a basis. As we have already seen in the article on the null space, the null space is generated by the empty set. Furthermore, ${\displaystyle \emptyset }$ is linearly independent by definition and therefore a basis of the null space. Thus we have ${\displaystyle \mathrm {dim} _{K}(\{0\})=|\emptyset |=0}$.

## Properties of the dimension

We now prove some properties of the "dimension" notion:

Theorem

Let ${\displaystyle V}$ be a finite-dimensional ${\displaystyle K}$-vector space and ${\displaystyle U\subseteq V}$ a subspace. Then we have that:

1. ${\displaystyle \mathrm {dim} (U)\leq \mathrm {dim} (V)}$.
2. If ${\displaystyle \mathrm {dim} (U)=\mathrm {dim} (V)}$, then ${\displaystyle U=V}$ follows.

Proof

Let ${\displaystyle B_{U}}$ be a basis of ${\displaystyle U}$. Then ${\displaystyle B_{U}}$ is a linearly independent subset of ${\displaystyle V}$. Therefore, according to the basis completion theorem, there exists a basis ${\displaystyle B_{V}}$ of ${\displaystyle V}$, with ${\displaystyle B_{U}\subseteq B_{V}}$. It follows immediately that ${\displaystyle |B_{U}|\leq |B_{V}|}$, so ${\displaystyle \mathrm {dim} (U)\leq \mathrm {dim} (V)}$. Now, if in addition ${\displaystyle \mathrm {dim} (U)=\mathrm {dim} (V)}$ is assumed, then we have that ${\displaystyle |B_{U}|=|B_{V}|}$. Since ${\displaystyle B_{V}}$ is finite, we get ${\displaystyle B_{U}=B_{V}}$. So we can conclude ${\displaystyle U=\operatorname {span} (B_{U})=\operatorname {span} (B_{V})=V}$.

In order to show that it is important to assume ${\displaystyle V}$ to be finite-dimensional, consider an example of an infinite-dimensional vector space that has a proper infinite-dimensional subspace:

Example

Let ${\displaystyle V=K[x]}$ be the polynomial space over a field ${\displaystyle K}$ and ${\displaystyle U:=\left\lbrace \sum _{i=1}^{n}a_{i}x^{i}\,{\bigg |}\,n\in \mathbb {N} ,a_{i}\in K\,\,\forall i\in \lbrace 0,\ldots ,n\rbrace \right\rbrace }$ the subspace of polynomials without a constant term. One can easily show (as above) that ${\displaystyle B=\left\lbrace x^{i}\,{\bigg |}\,i\in \mathbb {N} _{>}0\right\rbrace }$ is a basis of ${\displaystyle U}$. Thus we see ${\displaystyle \mathrm {dim} (U)=\infty =\mathrm {dim} (V)}$. But ${\displaystyle U\subsetneq V}$, since the constant polynomial is ${\displaystyle 1\in V\setminus U}$.

## Dimension formula

### Proof of the dimension formula

The following dimensional formula gives how to calculate the dimension of the sum of two finite dimensional subspaces ${\displaystyle U,\,W\subseteq V}$ of a ${\displaystyle K}$-vector space ${\displaystyle V}$.

Theorem (Dimension formula)

Let ${\displaystyle V}$ be a ${\displaystyle K}$-vector space and let ${\displaystyle U,W\subseteq V}$ be finite-dimensional subspaces. Then, we have:

${\displaystyle \operatorname {dim} (U+W)=\operatorname {dim} (U)+\operatorname {dim} (W)-\operatorname {dim} (U\cap W)}$

Proof (Dimension formula)

Since ${\displaystyle U,V}$ are finite dimensional, the spaces ${\displaystyle U+V,U\cap V}$ are also finite dimensional. set ${\displaystyle n:=\operatorname {dim} (U\cap W)}$. Then ${\displaystyle \operatorname {dim} (U),\operatorname {dim} (W)\geq n}$. So there are some ${\displaystyle k,m\in \mathbb {N} }$, such that ${\displaystyle \operatorname {dim} (U)=n+k}$ and ${\displaystyle \operatorname {dim} (W)=n+m}$. Furthermore, let ${\displaystyle \lbrace b_{1},\ldots ,b_{n}\rbrace }$ be a basis of ${\displaystyle U\cap W}$. Since ${\displaystyle U\cap W}$ is a subspace of ${\displaystyle U}$ and of ${\displaystyle W}$, according to the basis completion theorem there exist vectors ${\displaystyle u_{1},\ldots ,u_{k}\in U}$ and vectors ${\displaystyle w_{1},\,\ldots ,w_{m}\in W}$ , such that ${\displaystyle \lbrace b_{1},\,\ldots ,b_{n},\,u_{1},\ldots ,u_{k}\rbrace }$ is a basis of ${\displaystyle U}$ and ${\displaystyle \lbrace b_{1},\ldots ,b_{n},\,w_{1},\ldots w_{m}\rbrace }$ is a basis of ${\displaystyle W}$.

We now show that ${\displaystyle B:=\lbrace b_{1},\ldots ,b_{n},u_{1},\ldots u_{k},\,w_{1},\ldots w_{m}\rbrace }$ is a basis of ${\displaystyle U+W}$.

Proof step: First, we show that ${\displaystyle B}$ is a generator.

To do this, we show that any vector ${\displaystyle a\in U+W}$ can be represented as a linear combination of elements from ${\displaystyle B}$.

So let ${\displaystyle a\in U+W}$. Then there are some ${\displaystyle u\in U,\,w\in W}$ with ${\displaystyle a=u+w}$. Since ${\displaystyle u}$ is a linear combination of the basis ${\displaystyle \lbrace b_{1},\ldots ,b_{n},\,u_{1},\ldots ,u_{k}\rbrace }$ of ${\displaystyle U}$, we have

${\displaystyle u=\sum _{i=1}^{n}\lambda _{i}b_{i}+\sum _{j=1}^{k}\mu _{j}u_{j}{\text{ für geeignete }}\lambda _{i},\mu _{j}\in K}$

And ${\displaystyle v}$ is a linear combination of the basis ${\displaystyle \lbrace b_{1},\ldots ,b_{n},\,w_{1},\ldots ,w_{m}\rbrace }$ of ${\displaystyle W}$, so

${\displaystyle w=\sum _{i=1}^{n}\rho _{i}b_{i}+\sum _{j=1}^{m}\sigma _{j}w_{j}{\text{ für }}\rho _{i},\sigma _{j}\in K}$

Thus,

${\displaystyle a=\sum _{i=1}^{n}(\lambda _{i}+\rho _{i})b_{i}+\sum _{j=1}^{k}\mu _{j}u_{j}+\sum _{l=1}^{m}\sigma _{l}w_{l}.}$

So ${\displaystyle a}$ is linear combination of ${\displaystyle \lbrace b_{1},\ldots ,b_{n},\,u_{1},\ldots ,u_{k},\,w_{1},\ldots ,w_{m}\rbrace }$ and ${\displaystyle B}$ a generator of ${\displaystyle U+W}$.

Proof step: Now we show the linear independence of ${\displaystyle B}$.

Let ${\displaystyle \lambda _{i},\mu _{j},\rho _{l}\in K}$, with

${\displaystyle \sum _{i=1}^{n}\lambda _{i}b_{i}+\sum _{j=1}^{k}\mu _{j}u_{j}+\sum _{l=1}^{m}\rho _{l}w_{l}=0}$

We have to show that ${\displaystyle \lambda _{i}=\mu _{j}=\rho _{l}=0}$ for all ${\displaystyle i,j,k}$. Let for this ${\displaystyle v=\sum _{i=1}^{n}\lambda _{i}b_{i}+\sum _{j=1}^{k}\mu _{j}u_{j}}$. Then, we have ${\displaystyle v\in U}$ and because of the above condition.

${\displaystyle v=-\sum _{l=1}^{m}\rho _{l}w_{l}.}$

So also ${\displaystyle v\in W}$ , i.e., ${\displaystyle v\in U\cap W}$ .

Thus ${\displaystyle v}$ can be represented as a linear combination of the basis ${\displaystyle \{b_{1},\ldots ,b_{n}\}}$ of ${\displaystyle U\cap W}$ and there exist ${\displaystyle \sigma _{1},\ldots ,\sigma _{n}\in K}$ such that

${\displaystyle v=\sum _{i=1}^{n}\sigma _{i}b_{i}.}$

Now, we further have

${\displaystyle 0=v-v=\sum _{i=1}^{n}(\lambda _{i}-\sigma _{i})b_{i}+\sum _{j=1}^{k}\mu _{j}u_{j}.}$

Since ${\displaystyle \{b_{1},\ldots ,b_{n},\,u_{1},\ldots ,u_{k}\}}$ is a Basis of ${\displaystyle U}$, it is linearly independent and

${\displaystyle \lambda _{1}=\sigma _{1},\ldots ,\lambda _{n}=\sigma _{n}{\text{ and }}\mu _{1}=\ldots =\mu _{k}=0.}$

So

${\displaystyle \sum _{i=1}^{n}\lambda _{i}b_{i}+\sum _{l=1}^{m}\rho _{l}v_{l}=0.}$

Since ${\displaystyle \{b_{1},\ldots ,b_{n},\,w_{1},\ldots ,w_{m}\}}$ is a Basis of ${\displaystyle W}$ and hence the vectors are linearly independent, we have that

${\displaystyle \lambda _{1}=\ldots \,=\lambda _{n}=0{\text{ and }}\rho _{1}=\ldots =\rho _{m}=0.}$

Thus all coefficients are zero and the vectors ${\displaystyle \{b_{1},\ldots ,b_{n},\,u_{1},\ldots ,u_{k},\,w_{1},\ldots ,w_{m}\}}$ are linearly independent. ${\displaystyle B}$ is therefore actually a basis.

Now, we have that

• ${\displaystyle \operatorname {dim} (U+W)=|B|=n+k+m}$
• ${\displaystyle \operatorname {dim} (U\cap W)=n}$
• ${\displaystyle \operatorname {dim} (U)=n+k}$
• ${\displaystyle \operatorname {dim} (W)=n+m}$

So:

${\displaystyle \operatorname {dim} (U+W)=n+k+m=(n+k)+(n+m)-n=\operatorname {dim} (U)+\operatorname {dim} (V)-\operatorname {dim} (U\cap W).}$

This is the claim to be proven.

Next, we consider a conclusion of the dimension formula that makes a statement about the sum of subspaces (missing). Visually, this means that the complement of a subspace in terms of dimension is the missing "remainder".

Theorem

Let ${\displaystyle V}$ be a finite-dimensional ${\displaystyle K}$-vector space and let ${\displaystyle U,W\subseteq V}$ be subspaces, with ${\displaystyle U\oplus W=V}$. Then, we have:

${\displaystyle \operatorname {dim} (V)=\operatorname {dim} (U)+\operatorname {dim} (W)}$

Proof

First, because of ${\displaystyle U,W\subseteq V}$, both subspaces are finite-dimensional. Using the dimensional formula, we conclude

${\displaystyle \operatorname {dim} (V)=\operatorname {dim} (U+W)=\operatorname {dim} (U)+\operatorname {dim} (W)-\operatorname {dim} (U\cap W)=\operatorname {dim} (U)+\operatorname {dim} (W)-\operatorname {dim} (\{0\}).}$

As shown in the example above, we have that ${\displaystyle \mathrm {dim} (\{0\})=0}$ and get

${\displaystyle \operatorname {dim} (V)=\operatorname {dim} (U)+\operatorname {dim} (W)}$

### Exercises: Dimension formula

Exercise (Dimension formula)

Let ${\displaystyle V}$ be a ${\displaystyle K}$-vector space and let ${\displaystyle U,W\subseteq V}$ be subspaces of ${\displaystyle V}$. Further, let ${\displaystyle \mathrm {dim} (U)=3}$, ${\displaystyle \mathrm {dim} (W)=5}$, ${\displaystyle \mathrm {dim} (V)=7}$. What dimension can ${\displaystyle U\cap W}$ and ${\displaystyle U+W}$ have?

How to get to the proof? (Dimension formula)

Bound ${\displaystyle \mathrm {dim} (U\cap W)}$ from above and apply the dimension formula.

Solution (Dimension formula)

We have the dimensional formula

${\displaystyle \mathrm {dim} (U+W)=\mathrm {dim} (U)+\mathrm {dim} (W)-\mathrm {dim} (U\cap W)}$

Further, we have that ${\displaystyle \mathrm {dim} (U\cap W)\leq \mathrm {dim} (U)=3}$, since ${\displaystyle U\cap W}$ is a subspace of ${\displaystyle U}$. So

${\displaystyle \mathrm {dim} (U+W)=\mathrm {dim} (U)+\mathrm {dim} (W)-\mathrm {dim} (U\cap W)=8-\mathrm {dim} (U\cap W)\geq 8-3=5}$

Since ${\displaystyle U+W\subseteq V}$ holds, we also have that ${\displaystyle \mathrm {dim} (U+W)\leq 7}$. Both results together yield

${\displaystyle \mathrm {dim} (U+W)\in \lbrace 5,6,7\rbrace }$

From the dimensional formula we now conclude

${\displaystyle \mathrm {dim} (U\cap W)=\mathrm {dim} (U)+\mathrm {dim} (W)-\mathrm {dim} (U+W)=3+5-\mathrm {dim} (U+W)\in \lbrace 1,2,3\rbrace }$

In total we obtain:

${\displaystyle \mathrm {dim} (U\cap W)\in \{1,2,3\}\,{\text{und}}\,\mathrm {dim} (U+W)\in \{5,6,7\}.}$

Exercise (Dimension formula)

Consider the ${\displaystyle \mathbb {Q} }$-vector space ${\displaystyle V:=\mathbb {Q} ^{4}}$ and the subspaces ${\displaystyle U:=\operatorname {span} (\{(2,1,0,0)^{T},(1,-1,0,0)^{T}\})}$, ${\displaystyle W:=\operatorname {span} (\{(1,0,1,0)^{T},(0,1,0,1)^{T}\})}$. Show that ${\displaystyle V=U\oplus W}$ holds.

Solution (Dimension formula)

We first show ${\displaystyle U\cap W=\{0\}}$. Let for this be ${\displaystyle v\in U\cap W}$. Then there exist ${\displaystyle \alpha ,\beta ,\lambda ,\mu \in \mathbb {Q} }$ with

${\displaystyle v={\begin{pmatrix}2\alpha +\beta \\\alpha -\beta \\0\\0\end{pmatrix}}={\begin{pmatrix}\lambda \\\mu \\\lambda \\\mu \end{pmatrix}}.}$

It follows that ${\displaystyle \lambda =\mu =0}$ and hence ${\displaystyle v=0}$. So we have that ${\displaystyle U\cap W=\{0\}}$.

From the dimension formula we now obtain

${\displaystyle \mathrm {dim} (U+W)=\mathrm {dim} (U)+\mathrm {dim} (W)-\mathrm {dim} (U\cap W)=2+2-0=4=\mathrm {dim} (V).}$

Using the theorem above about properties of the dimension, it follows that ${\displaystyle V=U+W}$. Together we get ${\displaystyle V=U\oplus W}$.