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

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In summary: Can you please elaborate?Thanks!Vector spaces are a set of vectors that are closed under addition and scalar multiplication. This means that you can add any two vectors in a vector space and get another vector, and you can multiply any two vectors in a vector space and get another vector. The zero vector is also in a vector space.
  • #1
kesun
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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]?
 
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  • #2
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
Mark44 said:
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.

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
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
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
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
kesun said:
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)].
Yes.
kesun said:
Vector space is a set of vectors that are closed under addition and scalar multiplication, and also contains the zero vector.
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)}.
kesun said:
I don't think I am grasping this definition well either, so I am having some difficult time putting them together..
 
  • #8
Mark44 said:
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.

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
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
Mark44 said:
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.

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

1. What is a vector?

A vector is a mathematical object that has both magnitude and direction. It is often represented graphically as an arrow, with the length of the arrow indicating the magnitude and the direction of the arrow indicating the direction of the vector.

2. What is Sp(V)?

Sp(V) stands for "span of V" and is the set of all possible linear combinations of the vectors in V. In other words, it is the set of all vectors that can be created by multiplying each vector in V by a scalar and adding them together.

3. How do you determine if a vector belongs to Sp(V)?

To determine if a vector belongs to Sp(V), you can check if it can be written as a linear combination of the vectors in V. This can be done by setting up a system of equations and solving for the scalar values. If there exists a solution, then the vector belongs to Sp(V).

4. Why is it important to determine if a vector belongs to Sp(V)?

Determining if a vector belongs to Sp(V) is important because it allows us to understand the properties and relationships between vectors. It also helps us to solve problems involving linear algebra, such as finding solutions to systems of linear equations.

5. Can a vector belong to more than one Sp(V)?

Yes, a vector can belong to more than one Sp(V) if it is a linear combination of different sets of vectors. However, if the same set of vectors is used, then a vector can only belong to one Sp(V).

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