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Subspaces and Basis of vector spaces

  1. Oct 27, 2007 #1
    I am totally lost on the following questions. What does exhibit mean?

    1) Show that the given set H is a subspace of ℜ^3 by finding a matrix A such
    that N(A) = H (in this case, N(A) represents the null space of A).

    2) Exhibit a basis for the vector space H.

    a
    b {for all R^3: a=b-c and 2a-b=b-c}
    c


    3) Show that the given set W is a subspace of ℜ^4 by finding a matrix B such
    that Col(B) = W (in this case, Col(B) represents the column space of B).

    4) Exhibit a basis for the vector space W.

    a-3b+c
    2b-11c
    a-3b+9c {for all R^4: a,b,c for all R }
    c+a-b


    Any help is greatly appreciated
     
    Last edited: Oct 27, 2007
  2. jcsd
  3. Oct 27, 2007 #2

    Hurkyl

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    It means "find one, and prove that it is one".
     
  4. Oct 27, 2007 #3
    I am totally lost on how to find a basis for a vector space.

    On #3 I found matrix B, but am confused on how to find a basis for the vector space?

    On #1 and #2 I am totally confused.:confused:
     
  5. Oct 27, 2007 #4
    Basis for Vector Space and Subspaces

    Can someone please explain how to find a basis for a vector space?

    Thanks
     
  6. Oct 27, 2007 #5

    radou

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    Example: if you have a set W = {(a, b, c) in R^3 : a + b - c = 0}, simply write (a, b, c) = (- b + c, b, c) = b(-1, 1, 0) + c(1, 0, 1), where b and c are in R. Obviously, the set {(-1, 1, 0), (1, 0, 1)} spans your subspace W. What condition needs to be satisfied for this set to form a basis for W?
     
  7. Oct 27, 2007 #6
    That,

    a+b-c=0
     
  8. Oct 27, 2007 #7

    radou

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    You better start thinking what you're talking about.
     
  9. Oct 27, 2007 #8
    What are you talking about? If you want to make fun of people that are confused then that is not cool. I told you I am lost and need your help.

    Can you please explain what you are trying to say?
     
  10. Oct 27, 2007 #9

    radou

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    I am not making fun of you.

    What is the definition of a basis?
     
  11. Oct 27, 2007 #10
    This is the textbook definition:
    a basis is a set of vectors that, in a linear combination, can represent every vector in a given vector space, and such that no element of the set can be represented as a linear combination of the others.

    But what that exactly means, I'm not sure.
     
  12. Oct 27, 2007 #11

    HallsofIvy

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    Strictly speaking "exhibit" means "show". Of course, to "show" a basis, you will need to find it as Hurkyl said!

    Is the "given set H" the same as in exercise 2?

    You have two equations in 3 variables. Unless they happen to be dependent, you can solve for two of the variables in terms of the third. That means the subspace will be one dimensional. (Geometrically, each of those equations is the equation of a plane containing the origin. Their intersection is a line through the orgin- a one-dimensional subspace of R3. In particular, you can always take that "third" variable to be equal to 1. If you take c= 1 and solve for a and b, what do you get? Now put those 3 values into your vector.
    Oops! They are dependent. If you set c= 1, you get a= b-1 and 2a= 2b-1 which cannot be solved for a and b! The subspace is actually 2 dimnsional. Take b= 1, c= 0. What is a? Now take b= 0, c= 1. What is a? Show that those two vectors for a basis for the space. (You don't really need to use 0 and 1 but those numbers are simplest.)

    Again, is W the same as in 4?

    Since you can choose a, b, c to be any 3 numbers, that will be three dimensional. A good way to handle this is to let each of a, b, c be equal to 1 while the others are 0. In other words, if a= 1, b= c= 0, what vector do you get? If b= 1, a= c= 0, what vector do you get? If c= 1, a= b= 0, what vector do you get. Can you PROVE that those three vectors for a basis for the subspace. (You will need to use the definition of "basis" to do that.)
     
    Last edited: Oct 27, 2007
  13. Oct 27, 2007 #12
    Yes, for Question 1, H is the same H in question 2.

    And in Question 3, W is the same W in question 4.
     
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