| richard r Poster Post Number: 14 |
Hello, Could someone please help me with this topic. What is meant by a vector subspace ? What are linearly independent vectors, and what is a basis ? Also, what does it mean to say vectors "span" a subspace ? Some simple explanations would be really appreciated. Thanks Richard |
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| Tristan Marshall Regular poster Post Number: 61 |
Do you know what a vector space is? How would you define a vector? If you know what a vector space is, then a vector subspace is a subset (of a vector space) that is also a vector space. A simple example would be if we take our vector space to be R3 (the space consisting of three ordered co-ordinates). The set of elements with the last co-ordinate equal to zero forms a subspace. A set V of vectors spans a subspace W if any element W can be written as a sum of elements in V. Two vectors v1, v2 are said to be linearly independent if and only if the only solution to l1 v1+l2v2 = 0 is l1=l2=0. For thevector space R2 (i.e. 2-D Cartesian co-ordinate vectors) thistranslates as 'the vectors don't point in the same direction'. Similarly, a set of n vectors v1, ..., vnare said to be linearly independent if and only if the only solution to l1v1 +¼+lnvn=0 is l1 = ¼ = ln=0. A basis is a linearly independent spanning set. It can be proved that all bases contain the same number of elements, which wecall the dimension. |
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| James Veteran poster Post Number: 792 |
I know I'm just repeating whats been said, lol, but I've written it all out now. A vector subspace is just a vector space itself, but all the vectors are also in a larger vector space. Eg. a line is a subspace of a plane, a plane is a subspace of R3 . A set of vectors, {u1 ,u2 ...un } are linearly independent vectors if the only solution to a1 u1 + a2 u2 + ... an un = 0 is where all the ai = 0 Eg, in R2 , the vectors (1,0) and (0,1) are linearly independent, but (3,0) and (7,0) are not, as 7(3,0) - 3(7,0) = (0,0) = 0 A set of vectors {u1 ,u2 ...un } which span the space have the property that, for any vector, v in the space, v = a1 u1 + a2 u2 + ... an un has at least 1 solution. So every single vector can be made from some combination of the ui . A basis is a spanning set which is linearly independent. In R3 , {(0,0,1), (0,1,0), (1,0,0), (3,7,4)} spans the space (but isnt a basis), as (x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1) + 0(3,7,4) But you can see that we have a redundant vector in the spanning set, we dont actually need all 4. We can get rid of any 1 of them and still have a spanning set(I've just shown how you can get rid of (3,7,4) because its most obvious, but you can get rid of any of the other 3 instead). A basis is a spanning set where we have gotten rid of all the redundant vectors. It means that for any vector, v, v = a1 u1 + a2 u2 + ... an un has a unique solution. |