Infinite-Dimensional Vector Space Theorem

Infinite-Dimensional Vector Space Theorem

Recall that a vector space $V$ is said to be infinite-dimensional if there does not exist a finite set of vectors $\{ v_1, v_2, ..., v_n \}$ from $V$ such that $V = \mathrm{span} (v_1, v_2, ..., v_n)$.

Also recall that a set of vectors $\{ v_1, v_2, ..., v_n \}$ is linearly independent if for $a_1, a_2, ..., a_n \in \mathbb{F}$ we have that the equation $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$ implies that $a_1 = a_2 = ... = a_n = 0$.

We will now look at an important theorem which says that if we can construct a linearly independent set of vectors $\{ v_1, v_2, ..., v_n \}$ from $V$ for any positive integer $n$, then $V$ is infinite-dimensional.

 Theorem 1: Let $V$ be a vector space over the field $\mathbb{F}$. For each natural number $n$ there exists a set of $n$ linearly independent vectors $\{v_1, v_2, ..., v_n \}$ from $V$ if and only if $V$ is infinite-dimensional.
• Proof: $\Rightarrow$ Suppose that for each positive integer $n$ there exists a set of $n$ linearly independent vectors $\{ v_1, v_2, ..., v_n \}$ from $V$. Let $U_n = \mathrm{span} (v_1, v_2, ..., v_n)$. We note that $U_n$ is a subspace of $V$ spanned by a finite set of vectors.
• So for each $n \in \mathbb{N}$ we have a subspace $U_n$ of $V$ spanned by the linearly independent set of $n$ vectors $\{ v_1, v_2, ..., v_n \}$.
• If $V$ were finite-dimensional and spanned by $n$ vectors, then the subspace $U_{n+1}$ spanned by the $n+1$ linearly independent vectors $\{ v_1, v_2, ..., v_n, v_{n+1} \}$ would contradict the assumption that $\{ v_1, v_2, ..., v_n, v_{n+1} \}$ is linearly independent since every linearly independent set of vectors in $V$ has length less or equal to every spanning set of vectors in $V$ and clearly $n+1 \not ≤ n$. Thus $V$ is infinite-dimensional.
• $\Leftrightarrow$ Suppose that $V$ is infinite-dimensional. Then no list of vectors from $V$ spans $V$ which implies there is no upper bound for the length of a linearly independent set of vectors in $V$, so for each natural number $n$ there exists a set of linearly independent vectors $\{ v_1, v_2, ..., v_n \}$. $\blacksquare$