Dimension of a Vector Space

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Before we precisely define what the dimension of a vector space is, we will first look at a very important theorem regarding bases that will give intuition to the subsequent definition.

Theorem 1: Let

$V$

be a finite dimensional vector space. If

$B_1$

and

$B_2$

are bases of

$V$

, then the size of

$B_1$

and

$B_2$

are equal, that is

\rvert B_1 \rvert = \rvert B_2 \rvert

Proof: Let $$V$$ be a finite dimensional vector space, and let $$B_1$$ and $$B_2$$ be bases of $$V$$ . To prove this we will use the fact that a basis by definition is both a linearly independent set of vectors from $$V$$ and a spanning set of vectors from $$V$$ .

Since $$B_1$$ is a basis, $$B_1$$ by definition is also a linearly independent set of vectors. Also, since $$B_2$$ is a basis, $$B_2$$ by definition is a spanning set of $$V$$ . By the Finite Dimensional Linearly Independent Set of Vectors Theorem the size of the linearly independent set of vectors $$B_1$$ must be less than or equal to the spanning set of vectors $$B_2$$ , and so $\rvert B_1 \rvert ≤ \rvert B_2 \rvert$ .

Similarly, since $$B_2$$ is a basis, $$B_2$$ by definition is also a linearly independent set of vectors. Also, since $$B_1$$ is a basis, $$B_1$$ by definition is a spanning set of $$V$$ . Once again by the Finite Dimensional Linearly Independent Set of Vectors Theorem, the size of the linearly independent set of vectors $$B_2$$ must be less than or equal to the spanning set of vectors $$B_1$$ and so $\rvert B_2 \rvert ≤ \rvert B_1 \rvert$ .

Since $\rvert B_1 \rvert ≤ \rvert B_2 \rvert$ and $\rvert B_2 \rvert ≤ \rvert B_1 \rvert$ , it follows that $\rvert B_1 \rvert = \rvert B_2 \rvert$ , and so both bases $$B_1$$ and $$B_2$$ have the same size. $\blacksquare$

We will now formally define the dimension of a vector space.

Definition: Let

$V$

be a finite dimensional vector space over the field

\mathbb{F}

. The Dimension of $$V$$ denoted

\dim_{\mathbb{F}} V

is the number of vectors in any basis of

$V$

. if

$V$

is an infinite dimensional vector space over

\mathbb{F}

then we write

\dim_{\mathbb{F}} V = \infty

We note from the theorem above, the dimension of a vector space is unique since the size of any two bases is of $$V$$ is always the same. Let's now look at some examples.

Consider the vector space $\mathbb{R}^n$ over the field of real numbers. We note that any basis of $\mathbb{R}^n$ will have $$n$$ vectors since the standard basis vectors form a basis, that is if $$e_1 = (1, 0, 0, ..., 0)$$ , $$e_2 = (0, 1, 0, ..., 0)$$ , …, $$e_n = (0, 0, ..., 0, 1)$$ then $V = \mathrm{span} (e_1, e_2, ..., e_n)$ and $\{ e_1, e_2, ..., e_n \}$ is linearly independent. Therefore $\dim_{\mathbb{R}} \mathbb{R}^n = n$ .

For another example, consider the set of all polynomials of degree less than or equal to $$n$$ , $\wp_n (\mathbb{F})$ over the real numbers. This is a finite dimensional vector space, and we know one such basis of this vector space to be $\{ 1, x, x^2, ..., x^n \}$ which contains $$n + 1$$ elements, so $\dim_{\mathbb{R}} \wp_n(\mathbb{F}) = n + 1$ .

Yet another example is the vector space of complex numbers $\mathbb{C}$ over the set of real numbers. Once such basis for $\mathbb{C}$ is the set $\{ 1, i \}$ which spans $\mathbb{C}$ and is linearly independent. So $\dim_{\mathbb{R}} \mathbb{C} = 2$ , which we should be familiar with since we can graph a complex number along a $$2$$ -dimensional real-axis and imaginary-axis.

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However, it is important to note that the underlying field is important when determining the dimension of a vector space. The vector space of complex numbers over the set of complex numbers has dimension $$1$$ , that is $\dim_{\mathbb{C}} \mathbb{C} = 1$ as you should verify.

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Dimension of a Vector Space