Help understanding Replacement theorem Friedberg

**gordo151091** · January 26th 2013, 09:47 AM

Hello everyone, im having trouble understanding the Replacement Theorem proof outline in Friedberg Linear Algebra 4th pg 45.

Theorem 1.10 (Replacement Theorem)

Let $V$ be a vector space that is generated by a set $G$ containing exactly $n$ vectors, and let $L$ be a linearly independent subset of $V$ containing exactly vectors. Then $m\leq n$ and there exists a subset $H$ of $G$ containing exactly $n-m$ vectors such that $L \cup H$ generates $V$

The proof is by induction on $m$ , I skipped the base case $m=0$ .

Inductive step

Let $L = \{v_{1},...,v_{m+1}\}$ , be a linearly independent subset of $V$ consisting of $m+1$ vectors. Then $\{v_{1},...,v_{m}\}$ is linearly independent, and so we may apply the induction hypothesis to conclude that $m\leq n$ and that theres is a subset $\{u_{1},...,v_{m-n}\}$ of $G$ such that $\{v_{1},...,v_{m}\} \cup \{u_{1},...,v_{m-n}\}$ generates $V$ . Thus there exist scalars $a_{1},..,a_{m},b_{1},...,b_{n-m}$ such that $a_{1}v_{1}+...+a_{m}v_{m}+b_{1}u_{1}+...+b_{n-m}u_{n-m}=v_{m+1}$ .

Note that $n-m>0$ , lest $v_{m+1}$ be a linear combination of $\{v_{1},...,v_{m}\}$ which contradicts the assumption that L is linearly independent.

I am having trouble understanding why does $n-m>0$ .

Could someone give me a hint so I can understand. Thanks!

Edit: Is it beacause the induction hypothesis gives us $m\leq n$ and and since $L$ has an extra vector means that they cannot be equal?

**emakarov** · January 26th 2013, 10:34 AM

Since m <= n, the only choices are m < n and m = n. If m = n, then from

$a_{1}v_{1}+...+a_{m}v_{m}+b_{1}u_{1}+...+b_{n-m}u_{n-m}=v_{m+1}$

we have

$a_{1}v_{1}+...+a_{m}v_{m}=v_{m+1}$

i.e., L is not linearly independent, contrary to the assumption.

**gordo151091** · January 26th 2013, 11:04 AM

Thanks, I over looked n-m=0

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