Dimension of a Vector Space

# Dimension of a Vector Space

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.

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.