Dimension – "Math for Non-Geeks"

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 $ℝ,{ℝ}^2,{ℝ}^3,\mathrm{...}$ have the "obvious dimensions" $1,2,3,\mathrm{...}$ . The vector space ${ℝ}^3$ has a basis with $3$ elements, for example the standard basis $B_3=\{e_1,e_2,e_3\}$.

Similarly, for any field $K$, we want the vector space $K^n$ to have dimension ${\mathrm{d}\mathrm{i}\mathrm{m}}_K(K^n)=n$. Again, with the standard basis $B_n=\{e_1,\mathrm{...},e_n\}$ we find a basis with exactly $n$ vectors.

This suggests defining 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 ${\mathrm{d}\mathrm{i}\mathrm{m}}_K(V):=\mathrm{d}\mathrm{i}\mathrm{m}(V):=|B|$. (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 $\mathrm{d}\mathrm{i}\mathrm{m}(V)=\mathrm{\infty }$.

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 $B,B^{\prime }$ two bases of $V$. If $B$ is finite, then $B^{\prime }$ is also finite, and we have that $|B|=|B^{\prime }|$.

Proof (Uniqueness of the dimension)

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

$B=\{\underset{n\text{ components}}{\underbrace{(1,0,\dots ,0{)}^T}},\underset{n\text{ components}}{\underbrace{(0,1,\dots ,0{)}^T}},\cdots ,\underset{n\text{ components}}{\underbrace{(0,0,\dots ,1{)}^T}}\}$

This basis is finite, and we have that $|B|=n$. We get ${\mathrm{d}\mathrm{i}\mathrm{m}}_K(K^n)=n$.

Dimension of a polynomial space

Example

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

$K[x]:=\left\{\sum _{i=0}^n{a}_i{x}^i\right|n\in \mathbb{N},a_i\in K\mathrm{\forall }i\in \{0,\dots ,n\}\},$

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

Dimension of $ℂ$ as an $ℝ$-vector space

Example

We would like to determine the dimension of the complex numbers, understood as an $ℝ$-vector space. Each complex number $z\in ℂ$ can be written uniquely as $z=a+b\cdot i$, with $a,b\in ℝ$. From this we see that $\{1,i\}$ is a basis of $ℂ$ over $ℝ$. So ${\mathrm{d}\mathrm{i}\mathrm{m}}_{ℝ}(ℂ)=2$, i.e., as a real vector space, $ℂ$ is 2-dimensions. (Note that as a complex vector space, $ℂ$ 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, $\mathrm{\emptyset }$ is linearly independent by definition and therefore a basis of the null space. Thus, we have ${\mathrm{d}\mathrm{i}\mathrm{m}}_K(\{0\})=|\mathrm{\emptyset }|=0$.

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. $\mathrm{d}\mathrm{i}\mathrm{m}(U)\le \mathrm{d}\mathrm{i}\mathrm{m}(V)$.
2. If $\mathrm{d}\mathrm{i}\mathrm{m}(U)=\mathrm{d}\mathrm{i}\mathrm{m}(V)$, 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 $|B_U|\le |B_V|$, so $\mathrm{d}\mathrm{i}\mathrm{m}(U)\le \mathrm{d}\mathrm{i}\mathrm{m}(V)$. Now, if in addition $\mathrm{d}\mathrm{i}\mathrm{m}(U)=\mathrm{d}\mathrm{i}\mathrm{m}(V)$ 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 $U=\mathrm{span}(B_U)=\mathrm{span}(B_V)=V$.

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 $U:=\left\{\sum _{i=1}^n{a}_i{x}^i\right|n\in \mathbb{N},a_i\in K\mathrm{\forall }i\in \{0,\dots ,n\}\}$ the subspace of polynomials without a constant term. One can easily show (as above) that $B=\left\{x^i\right|i\in {\mathbb{N}}_{>}0\}$ is a basis of $U$. Thus, we see $\mathrm{d}\mathrm{i}\mathrm{m}(U)=\mathrm{\infty }=\mathrm{d}\mathrm{i}\mathrm{m}(V)$. But $U⊊V$, 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: $\dim (U+W)=\dim (U)+\dim (W)-\dim (U\cap W)$

Proof (Dimension formula)

Since $U,V$ are finite dimensional, the spaces $U+V,U\cap V$ are also finite dimensional. Set $n:=\dim (U\cap W)$. Then $\dim (U),\dim (W)\ge n$. So there are some $k,m\in \mathbb{N}$, such that $\dim (U)=n+k$ and $\dim (W)=n+m$. Furthermore, let $\{b_1,\dots ,b_n\}$ 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 $u_1,\dots ,u_k\in U$ and vectors $w_1,\dots ,w_m\in W$ , such that $\{b_1,\dots ,b_n,u_1,\dots ,u_k\}$ is a basis of $U$ and $\{b_1,\dots ,b_n,w_1,\dots w_m\}$ is a basis of $W$.

We now show that $B:=\{b_1,\dots ,b_n,u_1,\dots u_k,w_1,\dots w_m\}$ 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 $\{b_1,\dots ,b_n,u_1,\dots ,u_k\}$ of $U$, we have

$u=\sum _{i=1}^n{\lambda }_i{b}_i+\sum _{j=1}^k{\mu }_j{u}_j\text{ for suitable }{\lambda }_i,{\mu }_j\in K$

And $v$ is a linear combination of the basis $\{b_1,\dots ,b_n,w_1,\dots ,w_m\}$ of $W$, so

$w=\sum _{i=1}^n{\rho }_i{b}_i+\sum _{j=1}^m{\sigma }_j{w}_j\text{ for }{\rho }_i,{\sigma }_j\in K$

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 $\{b_1,\dots ,b_n,u_1,\dots ,u_k,w_1,\dots ,w_m\}$ 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 $\{b_1,\dots ,b_n\}$ of $U\cap W$ and there exist ${\sigma }_1,\dots ,{\sigma }_n\in K$ 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 $\{b_1,\dots ,b_n,u_1,\dots ,u_k\}$ is a Basis of $U$, it is linearly independent and

${\lambda }_1={\sigma }_1,\dots ,{\lambda }_n={\sigma }_n\text{ and }{\mu }_1=\dots ={\mu }_k=0.$

So

$\sum _{i=1}^n{\lambda }_i{b}_i+\sum _{l=1}^m{\rho }_l{v}_l=0.$

Since $\{b_1,\dots ,b_n,w_1,\dots ,w_m\}$ is a Basis of $W$ and hence the vectors are linearly independent, we have that

${\lambda }_1=\dots ={\lambda }_n=0\text{ and }{\rho }_1=\dots ={\rho }_m=0.$

Thus all coefficients are zero and the vectors $\{b_1,\dots ,b_n,u_1,\dots ,u_k,w_1,\dots ,w_m\}$ are linearly independent. $B$ is therefore actually a basis.

Now, we have that

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

So: $\dim (U+W)=n+k+m=(n+k)+(n+m)-n=\dim (U)+\dim (V)-\dim (U\cap W).$

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:

$\dim (V)=\dim (U)+\dim (W)$

Proof

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

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

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

$\dim (V)=\dim (U)+\dim (W)$

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 $\mathrm{d}\mathrm{i}\mathrm{m}(U)=3$, $\mathrm{d}\mathrm{i}\mathrm{m}(W)=5$, $\mathrm{d}\mathrm{i}\mathrm{m}(V)=7$. What dimension can $U\cap W$ and $U+W$ have?

How to get to the proof? (Dimension formula)

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

Solution (Dimension formula)

We have the dimensional formula

$\mathrm{d}\mathrm{i}\mathrm{m}(U+W)=\mathrm{d}\mathrm{i}\mathrm{m}(U)+\mathrm{d}\mathrm{i}\mathrm{m}(W)-\mathrm{d}\mathrm{i}\mathrm{m}(U\cap W)$

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

$\mathrm{d}\mathrm{i}\mathrm{m}(U+W)=\mathrm{d}\mathrm{i}\mathrm{m}(U)+\mathrm{d}\mathrm{i}\mathrm{m}(W)-\mathrm{d}\mathrm{i}\mathrm{m}(U\cap W)=8-\mathrm{d}\mathrm{i}\mathrm{m}(U\cap W)\ge 8-3=5$

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

$\mathrm{d}\mathrm{i}\mathrm{m}(U+W)\in \{5,6,7\}$

From the dimensional formula we now conclude

$\mathrm{d}\mathrm{i}\mathrm{m}(U\cap W)=\mathrm{d}\mathrm{i}\mathrm{m}(U)+\mathrm{d}\mathrm{i}\mathrm{m}(W)-\mathrm{d}\mathrm{i}\mathrm{m}(U+W)=3+5-\mathrm{d}\mathrm{i}\mathrm{m}(U+W)\in \{1,2,3\}$

In total, we obtain:

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

Exercise (Dimension formula)

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

$v=\left(\begin{array}{c}2\alpha +\beta \\ \alpha -\beta \\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}\lambda \\ \mu \\ \lambda \\ \mu \end{array}\right).$

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

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

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