Infinite / Finite-Dimensional Vector Space Comparison Theorem

# Infinite / Finite-Dimensional Vector Space Comparison Theorem

 Theorem 1: If $U$ is a subspace of a finite-dimensional vector space $V$, then $U$ is finite-dimensional.
• Proof: First consider the case where $U$ is the zero subspace. In such a case, clearly $\{ 0 \}$ is finite-dimensional since it can be spanned by any vector in $U$ when multiplied by zero.
• Now consider the case where $U \neq \{ 0 \}$. Let $u_1 \in U$ where $u_1 \neq 0$.
• Step 1: If $U = \mathrm{span} (u_1)$ then we are done, and $U$ is finite-dimensional. If not, then there exists another vector $u_2 \in U$ where $u_2 \not \in \mathrm{span} (u_1)$.
• Step 2: If $U = \mathrm{span} (u_1, u_2)$ then once again, we are done and so $U$ is finite-dimensional. If not, then there exists another vector $u_3 \in U$ where $u_3 \not \in \mathrm{span} (u_1, u_2)$.
• Step j: If $U = \mathrm{span} (u_1, u_2, ..., u_j)$ then $U$ is finite-dimensional, and if not, then there exists another vector $u_{j+1} \in U$ where $u_{j+1} \not \in \mathrm{span} (u_1, u_2, ..., u_j)$.
• Now since $V$ is finite-dimensional, $V$ is spanned by a finite set of vectors call them $\{ v_1, v_2, ... v_n \}$. Now notice that each $\{ u_1, u_2, ..., u_j \}$ is a linearly independent set of vectors in $U$, and since $U$ is a subspace of $V$, then $\{ u_1, u_2, ..., u_j \}$ is a linearly independent set of vectors in $V$ as well. Recall from the Finite-Dimensional Linearly Independent Set of Vectors Theorem theorem that any linearly independent set of vectors from a vector space $V$ has a size that is less than or equal to a spanning set of $V$, and so $j ≤ n$. Therefore, after at most $n$ steps, we will obtain a spanning set for $U$, and so $U$ is finite-dimensional. $\blacksquare$
 Corollary 1: If $U$ is a subspace of a vector space $V$, and $U$ is infinite-dimensional, then $V$ is also infinite-dimensional.
• Proof: This is just the contrapositive to the theorem proven above, that is if $U$ is not finite-dimensional, then $V$ is not finite-dimensional. $\blacksquare$

Let's now look at an example of applying the theorem from above.

## Example 1

Show that the vector space of infinite sequences $\mathbb{F}^{\infty}$ is infinite-dimensional.

Consider the subspaces $\mathbb{F}^n$, $n = 1, 2, ...$. Each of these subspaces can be spanned by the linearly independent set of lists $\{ (1, 0, 0, ..., 0), (0, 1, 0, ..., 0), ..., (0, 0, 0, ..., 0) \}$ of length $n$.

So as $n \to \infty$, we will always have a spanning set of linearly independent lists of vectors for the subspace $\mathbb{F}^n$ of $\mathbb{F}^{\infty}$ which implies that no set of sequences in $\mathbb{F}^{\infty}$ span $\mathbb{F}^{\infty}$, so $\mathbb{F}^{\infty}$ is infinite-dimensional.