Results 1 to 3 of 3

Math Help - Basis question

  1. #1
    Senior Member
    Joined
    Jul 2006
    Posts
    364

    Basis question (Answer check please)

    Two subspaces S and T of R^4 are spanned by {(15,1,0,1), (-1,15,0,1) } and { (2,0,1,0), (12, 46, 0, 4), (14, 46, 1, 4)} respectively. Find a basis {X} for S\cap T.

    ---------------

    I am suspicious that the question isn't correct, because I found a solution set for S\cap T that is spanned by 2 vectors, not one. Can the dimension of S\cap T really be 1 as the question implies?

    BTW, my sol'n set is: (if that's correct)

    S=\lbrace \left( \begin{array}{c}a_1\\a_2\\a_3\\a_4\\a_5 \end{array} \right) = <br />
\left( \begin{array}{c}1\\3\\0\\1\\0 \end{array} \right) a_4 +<br />
\left( \begin{array}{c}1\\3\\-1\\0\\1 \end{array} \right) a_5 \mid a_4,a_5 \in R \rbrace

    (if someone can suggest a better way to typeset column vectors I'd like to know)
    Last edited by scorpion007; April 23rd 2009 at 11:23 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Can someone help with this? It is somewhat urgent. If the question is incorrect, I'd like to notify the instructor and let him know.

    Otherwise, could you show where I went wrong?

    I looked at it again, and can see no way for S intersect T to be spanned by 1 vector.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Can someone check if my answer is correct?

    The only vector in S \cap T is the zero vector (0,0,0,0), and hence the basis for S \cap T is {(0,0,0,0)}.

    I am quite unsure about this.

    My approach:

    Let v = (a,b,c,d) be a vector in S \cap T.

    To be in S, v must equal (15x - y, x + 15y, 0, x + y) for arbitrary x, y in R.
    To be in T, v must equal (2x + 12y + 14z, 46y+46z, x+z,4y+4z).

    From this we can see that:

    15x-y=2x+12y+14z
    x+15y=46y+46z
    0=x+z
    x+y=4y+4z

    Solving this homogeneous linear system, it seems like it only has the trivial solution, (x, y, z) = (0, 0, 0).

    Therefore the v must be the vector (0, 0, 0, 0).

    Again, I am very unsure about this.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: August 30th 2011, 04:48 PM
  2. Basis Question
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: May 29th 2010, 09:19 PM
  3. Basis question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 11th 2009, 11:02 AM
  4. another basis question
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 25th 2009, 11:49 PM
  5. Basis Question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 23rd 2008, 11:12 PM

Search Tags


/mathhelpforum @mathhelpforum