Dimension of Subspace cannot be bigger than Parent Vector Space

Last updated Nov 1, 2022

Let $V$ be a Vector Space and let $W \subset V$ be a Vector Subspace of $V$. Then $\dim W \leq \dim V$.

First note that if $\dim V = \infty$, then this statement is trivially true since $\dim W \leq \infty$ is always true by definition of Dimension of a Vector Space.

Suppose $\dim V = n < \infty$ for some $n \in \mathbb{N}$. First observe that any Linearly Independent $S \subset W$ is also a Linearly Independent $S \subset V$, since linear independence is only defined in terms of the vectors and underlying Field. Therefore, because Dimension of a Vector Space upper bounds Linearly Independent Set size, we must have that $|S| \leq n$ for such $S$. Applying the process inremark (1) of Growing a Linearly Independent Set on $W$, we must terminate in at most $n$ steps. By that remark, we have that $R$ is a Vector Space Basis for $W$. Since $|R| \leq n$, $W$ is indeed a Finite-Dimensional Vector Space and $\dim W = |R| \leq n = \dim V$. $\blacksquare$

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