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Suppose {\(v_1, ..., v_m\)} spans a vector space V and suppose \(v_i\) is linear combination of the preceding vectors. Show that {\(v_1, ..., v_{i1}, v_{i+1}, ..., v_m\)}spans V.
How to start?
 one year ago
 one year ago
Suppose {\(v_1, ..., v_m\)} spans a vector space V and suppose \(v_i\) is linear combination of the preceding vectors. Show that {\(v_1, ..., v_{i1}, v_{i+1}, ..., v_m\)}spans V. How to start?
 one year ago
 one year ago

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amoodaryaBest ResponseYou've already chosen the best response.0
are {v1 ,..,vn} independent ?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Hmm... suppose i) they are independent ii) they are dependent It is not specified in the question. So, I would like to know how it works for these two cases.
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
In English, (sort of), if a set of vectors (let's call it S) which contains v[i] spans V, then if you take away v[i], the resulting set would still span V, yes? :)
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
Oh, and apparently, v[i] is a linear combination of the other vectors. My bad :D
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
Hey, @RolyPoly you know what linearly independent/dependent means right?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
But that wasn't really important, you're right... Anyway, to show that a set spans a vector space, take an arbitrary element of said vector space (V) and show that it can be written as a linear combination of the set which you want to show is a spanning set... (mouthful, that was)
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
If \(v_i\) is the linear combination, then \(v_i=a_1v_1+\ldots+a_{i1}v_{i1}+a_{i+1}v_{i+1}+\ldots+a_mv_m\). Where not all \(a_k\) are equal to zero. That means that any vector in \(V\) can be represented as \(a_1v_1+\ldots+a_{i1}v_{i1}+a_iv_i+a_{i+1}v_{i+1}+\ldots+a_mv_m\). But \(v_i\) is a linear combination of previous vectors. So any vector in \(V\) can be represented as \(a_1v_1+\ldots+a_{i1}v_{i1}+a_{i+1}v_{i+1}+\ldots+a_mv_m\). The End.
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
Let's take an arbitrary element of V, (let's call it w) Since we know the set S spans V, then all elements of V, w included can be written as \[w=a_{1}v_{1}+a_{2}v_{2}+ \ ... \ +a_{i}v_{i} + \ ... \ + a_{m}v_{m}\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Just a double check with my concept: linear dependent: c1v1, c2v2, ... , cnvn =0 for c1, c2, ..., cn not zero?!
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
They can be zeros, but not all at the same time.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
So, all zero = linear independence, right?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
If all of the scalar coefficients have to be zero to make the linear combination equal to the zero vector, then yes, the vectors are linearly independent.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Okay, thanks. Also, what is the purpose of stating that ''Where not all ak are equal to zero.''?? I don't understand this part..
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
Because, if every a[k] has to be zero for that particular linear combination to be equal to the zero vector, then that means that that set is linearly independent. You may or may not have taken this up yet, but if a set of vectors is linearly independent, then under no circumstances is any of them a linear combination of the others. (Proof on demand)
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
Sorry, I spoke too soon. That one over there was pointless ^ My bad anyway, actually, it doesn't matter if all the a[k]'s are zero, it wouldn't make a difference, although that's too trivial a case to tackle
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
So, if not all ak are equal to 0, then, it is linear dependent. So, it can be a linear combination of others. If all ak are equal to 0, then it is linear independent. So, it can't be a linear combination of the others. Is that right? (I would really love to learn more before I formally take this course. So, you are welcomed to introduce more relevant concepts :) Thanks!)
 one year ago
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