# Determine whether or not a vector belongs to a Sp(V) and so on

1. Mar 3, 2009

### kesun

Given: V={(1,1,3,2),(-1,2,1,-4),(0,3,4,-2)}; w=(-1,5,12,-9)

1) determine whether the given vector w belongs to Sp(V);
2) use the row reduction from 1) to explain whether or not V is a basis for Sp(V);
3) if V is a basis for Sp(V), and w is in Sp(V), determine, (w)[v], where [v] denotes a subscript.

For 1) I am not very sure about the Sp(V)..Am I supposed to make an augmented matrix of V and w and reduce it RREF, and then go from there? Say, if the result is consistent, then it belongs to Sp(V)?
By the definition of a basis, I need to show that V is linearly independent and is a spanning set for Sp(V) (?). How to show V is a spanning set of itself? Since it's Sp(V), is V already a spanning set of Sp(V)?
Suppose that V is a basis for Sp(V) and w is in SP(V), do I write V and w as an augmented matrix and solve straight away for (w)[v]?

2. Mar 3, 2009

### Staff: Mentor

You might be getting ahead of yourself by talking about augmented matrices and such, but not mentioning what it means for a vector to be in the span of a set of vectors. The augmented matrices are the techniques used to find out these things.

For the sake of brevity, let's call the vectors in V v1, v2, and v3. w is in span V if there are constants c1, c2, and c3 such that w = c1*v1 + c2*v2 + c3*v3. That's the fundamental idea. You can set up an augmented matrix to determine the constants.

3. Mar 3, 2009

### kesun

OH! This is exactly the idea that I need to clarify! Thank you very much! This span thingy is making a lot more sense to me now! :D

4. Mar 3, 2009

### kesun

This part is still getting my head spinning: From the row reduction in i), explain whether or not V is a basis for Sp(V).

V is a set of vectors. Isn't a basis just a single vector? How am I supposed to get going with this part? Please help me again..

Thanks!

5. Mar 3, 2009

### Staff: Mentor

A basis could consist of a single vector, in which the subspace would be a single line through the origin in whatever vector space you're working in. More often a basis is a set of vectors that 1) is a linearly independent set, and 2) spans the subspace in question.

For the three vectors in set V to be a basis for Span(V), they must be linearly independent. When you row reduced your augmented matrix in part i, if you got a unique solution for the constants c1, c2, and c3, that guarantees that the three vectors are linearly independent. To understand this, if your augmented matrix had a fourth column of 0s, and you row-reduced it, you would use exactly the same operations as you did earlier, and you would have found that there is a unique solution to the equation c1*v1 + c2*v2 + c3*v3 = 0, which is the definition of linear independence; namely that this equation has a single solution.

What you're doing is solving Ac = w, where the columns of A are your three vectors in V, c is a column vector (c1, c2, c3), and w is the vector you're given.

For a unique solution, your row reduction has to end up with a row of zeroes on the bottom of your augmented matrix, and three rows above it with leading entries of 1. If you ended up with a bottom row in the augmented matrix that looks like 0 0 0 | 7 (for example), that means that there is no solution, and w is not in the span(V). If you ended up with two rows of zeroes in the augmented matrix, that means that there are an infinite number of solutions to the equation c1*v1 + c2*v2 + c3*v3 = w, so it must be the case that one of the three vectors in V is a linear combination of the other two, which means that the set is linearly dependent, and therefore can't be a basis for span(V).

6. Mar 3, 2009

### kesun

I'm finally getting it...That was a great explanation! Thanks a lot! :D

I have one side question though, which still involves span: how to determine the vector spaces of polynomials spanned by S, for S equals, say, {(1+x+x^2, x+x^2+x^3, 1+x^2+x^3}?

By definition, something that is spanned by S is probably presented like [a (1+x+x^2) + b (x+x^2+x^3) + c (1+x^2+x^3)]. Vector space is a set of vectors that are closed under addition and scalar multiplication, and also contains the zero vector. I don't think I am grasping this definition well either, so I am having some difficult time putting them together..

7. Mar 4, 2009

### Staff: Mentor

Yes.
This vector space is a function space, a kind of vector space in which functions play the same role that vectors do in a vector space. Just as vector spaces can have different dimensions, with some examples being R, R^2, R^3, and so on, there are function spaces of different dimensions. The functions in S belong to P3, the space of polynomial functions of degree <= 3.

P3 is very similar to R^4, and in fact the two spaces are isomorphic to each other, meaning that there is a one-to-one, onto function that maps vectors in R^4 to functions in P3. That the two spaces are isomorphic says that they have essentially the same "shape." For example, the function 1 + x + 0x^2 + 2x^3 maps to the vector (1, 1, 0, 2) in R^4. The standard basis for P3 is {1, x, x^2, x^3}, and the standard basis for R^4 is {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)}.

8. Mar 4, 2009

### kesun

So is it correct to say that the vector space of polynomials spanned by S has a dimension of 3, since it's in R^4?

9. Mar 4, 2009

### Staff: Mentor

More accurately, the subspace of P3 spanned by S has dimension 3. That's assuming that the three vectors in your set S are linearly independent, which I didn't check.

10. Mar 4, 2009

### kesun

Ah, I see. Thanks a lot for your help! That clarified everything for me. :D