> [!NOTE] Lemma (Dimension of Subspaces) > Let $V$ be a FDVS and $W \subset V$ a [[Vector subspace|vector subspace]] of $V$. Then $W$ is also finite dimensional and $\dim W \leq\dim V$ with equality iff $W=V$. *Proof*. Suppose $W$ is not an FDVS. Then there is an [[Linear Independence#^d0db64|infinite set]] $S \subset W$ of linearly independent elements of $V$, which [[Linear Independence#^fbfe78|contradicts]] the fact that $V$ is finite dimensional. Now let $B \subset W$ be a basis of $W$. Note that they are also linearly independent elements of $V$ so by *sifting lemma*, there is a set $L \subset V$ so that $B \cup L$ is a basis of $V$. Thus $\dim W = |B| \leq |B \cup L| = \dim V $with equality iff $s=n$ which is to say $B$ is a basis of $V$ and so $W=V$. $\square$