# 12-240/Proofs in Vector Spaces

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...generating set as $\beta$, so $k = |L|\leq |\beta| = dimV$ ***(explanation needed, why? [or your question])*** since $L$ is a some linearly independent...
Theorem: Let $W$ be a subspace of a finite dimensional vector space $V$. Then $W$ is finite dimensional and $dimW \leq dimV$
Proof: Let $\beta$ be a basis for $V$. Then we know that $\beta$ is a finite set since $V$ is a finite dimensional. Then, for given a subspace $W$, let us construct a linearly independent set $L$ by adding vectors from $W$ such that $L=\{w_1,w_2, ... w_k\}$ is maximally linearly independent. In other words, adding any other vector from $W$ would make $L$ linearly dependent. Here, L has to be a finite set by the Replacement Theorem, if we choose the generating set as $\beta$, so $k = |L|\leq |\beta| = dimV$ since $L$ is a some linearly independent subset of $V$. Now we want to show that $L$ is a basis for $W$. Since $L$ is linearly independent, it suffices to show that $span(L)=W$. Suppose not:$span(L)\neq W$. (We know that $L \subseteq span(L) \subseteq W$ since $L$ is made of vectors from $W$.) Then $\exists w_a \in W : w_a \notin span(L)$ But this means $span(L)\cup \{w_a\}$ is linearly independent, which contradicts with maximally linearly independence of $L$. Therefore $span(L)=W$ and hence, $L$ is a basis for $W$