Dimension of a vector space

The dimension¹ of a vector space is the cardinality of any basis for that space. #m/def/linalg Thus all possible bases for a vector space have the same cardinality.²

Proof

Let $𝑉$ be a vector space. First we show that if ${𝑣 𝑖} 𝑛 𝑖 = 1$ are linearly independent in $𝑉$ and $𝑉 = s p a n {𝑠 𝑖} 𝑚 𝑖 = 1$ , then $𝑛 \leq 𝑚$ .

Let $𝐴 = {𝑣 𝑖} 𝑛 𝑖 = 1$ and $𝐵 = {𝑠 𝑖} 𝑚 𝑖 = 1$ We iterate the following steps, starting with $𝑘 = 1$ and incrementing until exhaustion:

Move a vector $𝑣 𝑘$ out of $𝐴$ into $𝐵$ .
Since $s p a n (𝐵) = 𝑉$ , $𝑣 𝑘$ is a linear combination of other elements of $𝐵$ , so one of the $𝑠 𝑖$ can be removed from $𝐵$ and still $s p a n (𝐵) = 𝑉$ . Without loss of generality by reïndexing we remove $𝑠 𝑘$ from $𝐵$ . $𝐴$ remains linearly independent.

If $𝑚 < 𝑛$ , we eventually exhaust all $𝑠 𝑖$ and $𝐴$ and $𝐵$ will partition ${𝑣 𝑖} 𝑛 𝑖 = 1$ . But then $𝐴 \subseteq s p a n (𝐵)$ , i.e. some $𝑣 𝑖$ are linear combinations of other $𝑣 𝑖$ , which is a contradiction. Therefore, $𝑛 \leq 𝑚$ .

It follows immediately that if a vector space $𝑉$ has any finite spanning set, then any two bases of $𝑉$ have the same size.

Now consider $𝑉$ with no finite spanning set. Let $B = {𝑏 𝑖} 𝑖 \in 𝐼$ and $C = {𝑐 𝑗} 𝑗 \in 𝐽$ be two distinct bases for $𝑉$ . Then any $𝑐 𝑗$ can be written as a finite linear combination of $𝑏 𝑖$ with nonzero coëfficients, say

𝑐 𝑗 = \sum 𝑖 \in 𝑈 𝑗 𝜆 𝑖 𝑏 𝑖

but because $C$ is a basis, it follows

⋃ 𝑗 \in 𝐽 𝑈 𝑗 = 𝐼

for if the vectors in $C$ can be expressed as finite linear combinations of vectors in a proper subset $B' \subset B$ , then $𝑉 = s p a n B'$ , which is a contradiction. Since $| 𝑈 𝑗 | < ℵ 0$ for all $𝑗 \in 𝐽$ , it follows from Upper bound on the cardinality of an arbitrary union that

| B | = | 𝐼 | \leq ℵ 0 | C | = | C |

By the same token $| C | \leq | B |$ thus by the Schröder-Bernstein theorem $| B | = | C |$ .

The vector spaces of a given dimension form an Isomorphism class.³

Vector spaces with multiple dimensionalities

Unless a vector space is over a prime field, there are typically multiple dimensionalities assignable to a vector space, depending on which ground field is being considered. This is distinguished by a subscript, for example $d i m ℂ ℂ = 1$ but $d i m ℝ ℂ = 2$ . If no ground field is specified, assume the topical field $𝕂$ .

#state/tidy | #SemBr | #lang/en

sometimes dimensionality, especially in plural since dimensions is confusing ↩
2008. Advanced Linear Algebra, pp. 48ff. ↩
2008. Advanced Linear Algebra, pp. 63–64 ↩

Dimension of a vector space

Footnotes