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...

Theorems & Proofs

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

Replacement Theorem: Let V be a vector space generated by G (perhaps linearly dependent) where |G|=n and let L be a linearly independent subset of V such that |L|=m. Then m \leq n and there exists a subset H of G with |H| = n-m and span(H \cup L)=V.

Proof: We will prove by induction hypothesis on m=|L|:

For m = 0: L = \emptyset, 0 \leq n and H=G so, span(H \cup L) = span(H) = span(G) = V

Now, suppose true for m: