Dimension – Serlo

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 $\mathbb {R} ,\mathbb {R} ^{2},\mathbb {R} ^{3},...$ have the "obvious dimensions" $1,2,3,...$ . The vector space $\mathbb {R} ^{3}$ has a basis with $3$ elements, for example the standard basis $B_{3}=\lbrace e_{1},e_{2},e_{3}\rbrace$.

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

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

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

Proof (Uniqueness of the dimension)

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

Examples of dimensions

Dimension of $K^{n}$

Example

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

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

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

Dimension of a polynomial space

Example

The polynomial space $K[x]$ over a field $K$ is defined as

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

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

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

Example

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

Dimension of the null space

Example

For every field $K$, the ("trivial") null space $\{0\}$ 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, $\emptyset$ is linearly independent by definition and therefore a basis of the null space. Thus we have $\mathrm {dim} _{K}(\{0\})=|\emptyset |=0$.

Properties of the dimension

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

Theorem

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

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

Proof

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

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

Example

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

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 $U,\,W\subseteq V$ of a $K$-vector space $V$.

Theorem (Dimension formula)

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

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

Proof (Dimension formula)

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

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

Proof step: First, we show that $B$ is a generator.

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

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

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

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

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

Thus,

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

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

Proof step: Now we show the linear independence of $B$.

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

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

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

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

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

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

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

Now, we further have

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

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

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

So

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

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

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

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

Now, we have that

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

So:

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

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

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

Proof

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

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

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

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

Exercises: Dimension formula

Exercise (Dimension formula)

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

How to get to the proof? (Dimension formula)

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

Solution (Dimension formula)

We have the dimensional formula

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

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

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

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

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

From the dimensional formula we now conclude

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

In total we obtain:

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

Exercise (Dimension formula)

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

Solution (Dimension formula)

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

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

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

From the dimension formula we now obtain

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

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