CoCalc -- 6-27.md

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title: 6/27

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Plan

Theorem (Replacement Theorem): Let $V$ be a vectir spaces spanned by a set $G$ of cardinality $n$ , and let $L$ be a linearly independent subset of $V$ containing exactly $m$ vectors. Then $m\leq n$ and there exists a subset $H$ of $G$ containing exactly $n-m$ vectors such that $L\cup H$ generates $V$ .
- Proof by induction on $m$ . Base case: $m=0$ ....
- Suppose the replacement theorem is true for some $m\geq 0$ . We prove the theorem is true for $m+1$ . Let $L=\{v_1,\ldots,v_{m+1}\}$ be a L.I. set.
- The set $\{v_1,\ldots,v_m\}$ is also linearly independent. So we can apply the induction hypothesis and deduce that there is a subset $\{u_1,\ldots,u_{n-m}\}$ of $G$ such that together they span $V$
- This means ParseError: KaTeX parse error: Expected group after '_' at position 2: v_̲_{m+1} is the span of the 2 sets. So we can deduced $n-m > 0$
- Also, we can deduce that some ParseError: KaTeX parse error: Expected group after '_' at position 2: u_̲_i is not needed.
Corollary: Suppose $V$ is a vector space having a finite basis. Then every basis for $V$ has the same number of vectors.
Corollary: A L.I. (or spanning) set of top cardinality is a basis. L.I. sets can be extended.
Subspaces have a lower dimension.