Induction of linear independent vectors

**bmp05** · May 23rd 2009, 04:12 AM

This is perhaps an easy question- more of a question about mathematical induction than linear algebra, perhaps:

Let $V$ be a vector-space over a field $\mathbb{F}$ ; $v_1,...,v_n$ are linear independent vectors in $V$ . Prove by induction that the vectors $w_i = \sum_{k=1} ^{i} v_k, 1 \leq i \leq n$ , are linear independent.

So, the question is asking to prove that the sum of linearly independent vectors is a vector which is linearly independent.

The condition for linear independence is:
$\sum_{k=1} ^{i} a_k v_k = 0, 1 \leq i \leq n$ , when and only when $a_i = 0$ for all $1 \leq i \leq k$

So $v_1 = a_1v_1 = 0 = a_1w_1$ , but how do you make an induction step out of this? For $w_2$ , the proposition is that $v_1 + v_2$ is also linearly independent:

$w_2 = v_1 + v_2 = a_1v_1 + a_2v_2 ; a_1=a_2=0$ , so you can bracket the scalars out... hmm? But I still can't see how you can make an induction step. I'm sure it's kind of easy, but that's only making it more frustrating.

**bmp05** · May 23rd 2009, 04:36 AM

Could you use the unit-vectors to prove this, or perhaps the dimension of the vector space? dim(V) = n?

**NonCommAlg** · May 23rd 2009, 05:00 AM

Originally Posted by bmp05

This is perhaps an easy question- more of a question about mathematical induction than linear algebra, perhaps:

Let $V$ be a vector-space over a field $\mathbb{F}$ ; $v_1,...,v_n$ are linear independent vectors in $V$ . Prove by induction that the vectors $w_i = \sum_{k=1} ^{i} v_k, 1 \leq i \leq n$ , are linear independent.

So, the question is asking to prove that the sum of linearly independent vectors is a vector which is linearly independent.

The condition for linear independence is:
$\sum_{k=1} ^{i} a_k v_k = 0, 1 \leq i \leq n$ , when and only when $a_i = 0$ for all $1 \leq i \leq k$

So $v_1 = a_1v_1 = 0 = a_1w_1$ , but how do you make an induction step out of this? For $w_2$ , the proposition is that $v_1 + v_2$ is also linearly independent:

$w_2 = v_1 + v_2 = a_1v_1 + a_2v_2 ; a_1=a_2=0$ , so you can bracket the scalars out... hmm? But I still can't see how you can make an induction step. I'm sure it's kind of easy, but that's only making it more frustrating.

the induction is over $n,$ the number of vectors. for n = 1, there's nothing to prove. suppose the claim is true for n (induction hypothesis) and take any n + 1 linearly independent vectors

$v_1, \cdots , v_{n+1}.$ suppose $c_1v_1 + c_2(v_1+v_2) + \cdots + c_n(v_1 + \cdots + v_n) + c_{n+1}(v_1 + \cdots + v_{n+1})=0.$ if we show that $c_1=c_2 = \cdots = c_{n+1}=0,$ we're done. first note that if $c_{n+1}=0,$ then

$c_1=c_2 = \cdots = c_n = 0$ by induction hypothesis and so we're done. if $c_{n+1} \neq 0,$ then we'll have: $v_{n+1} = -c_{n+1}^{-1}c_1v_1 - c_{n+1}^{-1}c_2(v_1 + v_2) - \cdots - c_{n+1}^{-1}(c_n + c_{n+1})(v_1 + \cdots + v_n),$ which is not

possible because $v_1, \cdots , v_{n+1}$ are linearly independent. this completes our induction.

**bmp05** · May 23rd 2009, 10:40 AM

So, you're proving the proposition for all of n- per definition- and then you prove it for n+1? But once again, this relies on the definition of linear independence. It's a strange question, I think. But thanks for your help NCA- btw. can you recommend a good text book for a first course in linear algebra?

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