Dimension

We’ve been discussing finite-dimensional vector spaces, but we have not defined the dimension of such an object. A reasonable definition of dimension would have the dimension of ${\mathbb{F}}^n$ be equal to $n$.

Notice that the standard basis

$$(1,0,\dots ,0),(0,1,0,\dots ,0),\dots ,(0,\dots ,0,1)$$

of ${\mathbb{F}}^n$ has length of $n$. Thus, it’s tempting to define the dimension as the length of the basis. Fortunately, this works out because all bases in a given vector space have the same length.

Thus, we can define dimension as below.

Definition: Dimension, $\dim V$

• The dimension of a finite-dimensional vector space is the length of any basis of the vector space.
• The dimension of a finite-dimensional vector space $V$ is denoted by $\dim V$.

This relies on the following theorem:

Basis length does not depend on basis

Any two bases of a finite-dimensional vector space have the same length.

Proof. Suppose $V$ is finite-dimensional. Let $B_1$ and $B_2$ be two bases of $V$. Then $B_1$ is linearly independent in $V$ in $B_2$ spans $V$, so the length of $B_1$ is at most the length of $B_2$ (see here). Interchanging the roles of $B_1$ and $B_2$, we also see that the length of $B_2$ is at most the length of $B_1$. Thus, the length of $B_1$ equals the length of $B_2$, as desired.

Examples

• $\dim {\mathbb{F}}^n=n$ because the standard basis of ${\mathbb{F}}^n$ has length $n$.
• $\dim {\mathcal{P}}_m(\mathbb{F})=m+1$ because the standard basis $1,z,\dots ,z^m$ of ${\mathcal{P}}_m(\mathbb{F})$ has length $m+1$.
• If $U=\{(x,x,y)\in {\mathbb{F}}^3:x,y\in \mathbb{F}\}$, then $\dim U=2$ because $(1,1,0),(0,0,1)$ is a basis of $U$.
• If $U=\{(x,y,z)\in {\mathbb{F}}^3:x+y+z=0\}$, then $\dim U=2$ because the list $(1,-1,0),(1,0,-1)$ is a basis of $U$.