Results 1 to 8 of 8

Math Help - Basis & Dimension Question

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    28

    Arrow Basis & Dimension Question

    Let u and v distinct vectors of a vector space V. Show that if {u,v} is a basis for V and a and b are nonzero scalars, then both {u + v}, au} and {au,bv} are also bases for V.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,395
    Thanks
    1481
    Awards
    1
    Quote Originally Posted by leungsta View Post
    Let u and v distinct vectors of a vector space V. Show that if {u,v} is a basis for V and a and b are nonzero scalars, then both {u + v, au} and {au,bv} are also bases for V.
    Suppose that w = \alpha u + \beta v then w = \beta \left( {u + v} \right) + \frac{{\alpha  - \beta }}{a}\left( {au} \right).
    In other words, if w is in span{u,v} then w is in the span{(u+v),au}.
    Now do the converse.

    Do the same two proofs for {au,bv}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    28
    Hmmm...I'm not understanding your solution....where is that equation from?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,395
    Thanks
    1481
    Awards
    1
    Quote Originally Posted by leungsta View Post
    Hmmm...I'm not understanding your solution
    OK, How much do understand about what it means to be a basis of a space?
    Do you know about spanning sets? Do you understand about dimension?
    If {u,v} is a basis for a v-space then the dimension is 2.
    What does that mean for any other spanning set of two vectors?

    Quote Originally Posted by leungsta View Post
    Where is that equation from?
    If you understand about a basis then think about that question.

    If you have no clue as to what is going on here, then it is time for you have a one-on-one with your course instructor. These ideas are fundamental to your study.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    Posts
    28
    I understand that a subset B of vector space V over R is a basis o V if B is linearly independant and Span B = V... I also understand dimensions. I'm just not very clear on proofs. And as for your question of "If {u,v} is a basis for a v-space then the dimension is 2. What does that mean for any other spanning set of two vectors?"

    This means that for any other spanning set of two vectors, the dimension is also 2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,395
    Thanks
    1481
    Awards
    1
    Quote Originally Posted by leungsta View Post
    "If {u,v} is a basis for a v-space then the dimension is 2. What does that mean for any other spanning set of two vectors?"
    This means that for any other spanning set of two vectors, the dimension is also 2
    Good. It does mean that any spanning set has at least two vectors.
    It also means that any two independent vectors will be a basis.

    So we have two approaches to doing this problem. I chose the ‘spanning’ option.
    But can you show that {(u+v),au} are independent?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2008
    Posts
    28
    Quote Originally Posted by Plato View Post
    Good. It does mean that any spanning set has at least two vectors.
    It also means that any two independent vectors will be a basis.

    So we have two approaches to doing this problem. I chose the ‘spanning’ option.
    But can you show that {(u+v),au} are independent?
    So if w is in span {u,v} then w is in span {u + v, au}. And if {u,v} are linear independant then {u + v, au} must also be independant?

    Sorry i'm really not good with proofs...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,395
    Thanks
    1481
    Awards
    1
    Quote Originally Posted by leungsta View Post
    And if {u,v} are linear independant then {u + v, au} must also be independant?
    NO! That is completely wrong.
    The fact that you are given that {u,v} is a basis means they are independent.
    Now I asked you if you understood the definition of basis.
    I don’t think that you do.

    But here goes if \lambda \left( {u + v} \right) + \eta \left( {au} \right) = 0 then because u & v are independent we get \left( {\lambda  + a\eta } \right)u + \lambda v = 0\; \Rightarrow \;\left( {\lambda  + a\eta } \right) = 0\;\& \;\lambda  = 0.
    But that means a\eta  = 0\;\& \;a \ne 0\; \Rightarrow \;\eta  = 0.
    Now if you do understand about independence then you see that u+v & au are independent.

    Now, I cannot go beyond this in helping you.
    If this still leaves you clueless, please go to your instructor. Look, simply using the old excuse “I am not good at proofs” does not cut it. The simple fact is: People who cannot do proofs have no business in mathematics courses. There is a reason why people who do well in mathematics course are rewarded.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Dimension and basis
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: November 12th 2011, 01:37 PM
  2. Basis and dimension
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: October 4th 2010, 12:02 AM
  3. Another basis and dimension question
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: October 3rd 2010, 06:05 PM
  4. another basis and dimension question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 22nd 2009, 06:46 AM
  5. Basis and dimension question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 24th 2008, 05:25 PM

Search Tags


/mathhelpforum @mathhelpforum