12-240/Proofs in Vector Spaces

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...generating set as [math]\displaystyle{ \beta }[/math], so [math]\displaystyle{ k = |L|\leq |\beta| = dimV }[/math] ***(explanation needed, why? [or your question])*** since [math]\displaystyle{ L }[/math] is a some linearly independent...

Theorems & Proofs

Theorem: Let [math]\displaystyle{ W }[/math] be a subspace of a finite dimensional vector space [math]\displaystyle{ V }[/math]. Then [math]\displaystyle{ W }[/math] is finite dimensional and [math]\displaystyle{ dimW \leq dimV }[/math]

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

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

Proof: We will prove by induction hypothesis on [math]\displaystyle{ m=|L| }[/math]:

For [math]\displaystyle{ m = 0 }[/math]: [math]\displaystyle{ L = \emptyset }[/math], [math]\displaystyle{ 0 \leq n }[/math] and [math]\displaystyle{ H=G }[/math] so, [math]\displaystyle{ span(H \cup L) = span(H) = span(G) = V }[/math]

Now, suppose true for [math]\displaystyle{ m }[/math]: