Results 1 to 4 of 4

Math Help - Vector Spaces and linear transformation

  1. #1
    Newbie
    Joined
    May 2012
    From
    hull
    Posts
    1

    Vector Spaces and linear transformation

    (a) Suppose that
    V is vector spaces over a field F and that U and W are subspaces of V .
    Show that
    U \W is also a subspace of V

    (b) Define a real linear transformation
    L1 : R4 -->R2 by

    L1(x1, x2, x3, x4) = (3x1 + x2 + 2x3 x4, 2x1 + 4x2 + 5x3 x4)
    and let
    U1 denote the kernel of L1. Define a real linear transformation L2 : R4 --> R2 by
    L
    L2=(x1, x2, x3, x4) = (5x1 + 7x2 + 11x3 + 3x4, 2x1 + 6x2 + 9x3 + 4x4)
    and let
    U2 denote the kernel of L2. Construct bases for U1, U2, U1 nU2 and U1 + U2.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,312
    Thanks
    693

    Re: Vector Spaces and linear transformation

    (a) what do you mean by U\W? if you mean the set difference, this cannot possibly be a subspace of V, since since {0} is in the intersection, and thus NOT in U\W.

    on the other hand, if you mean the quotient of U by W, this isn't necessarily even defined, since W may not be a subspace of U. something is missing, here...

    (b) well first we need to determine the two kernels (solve two homogeneous sets of linear equations). for example:

    ker(L1) = {(x1,x2,x3,x4) in R4 : L1(x1,x2,x3,x4) = (0,0)}

    this is equivalent to the homogeneous system of linear equations:

    3x1 + x2 + 2x3 - x4 = 0
    2x1 + 4x2 + 5x3 - x4 = 0

    or the matrix equation:

    \begin{bmatrix}3&1&2&-1\\2&4&5&-1 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix}0\\0 \end{bmatrix}

    it should be clear that this matrix has rank 2, so the kernel has dimension 2.

    row-reduction gives us:

    \begin{bmatrix}1&0&3/10&-3/10\\0&1&11/10&-1/10 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix}0\\0 \end{bmatrix}

    so if we set x3 = s, x4 = t, we get:

    10x1 = -3s + 3t
    10x2 = -11s + t

    so the kernel consists of vectors of the form (-3s+3t,-11s+t,10s,10t) = s(-3,-11,10,0) + t(3,1,0,10), so {(-3,-11,10,0),(3,1,0,10)} is a basis for U1.

    i leave it to you to construct a basis for U2, and of U1∩U2, U1+U2.

    you may find it helpful to know that dim(U1∩U2) ≤ min(dim(U1),dim(U2)) and that:

    dim(U1+U2) = dim(U1) + dim(U2) - dim(U1∩U2).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2013
    From
    essex
    Posts
    3
    Thanks
    1

    Re: Vector Spaces and linear transformation

    i think should be UnW is also a subspace of V . So how should we solve this question.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,389
    Thanks
    1325

    Re: Vector Spaces and linear transformation

    Let x and y be vectors in U\cap W. Show that x+ y is also in U\cap W and ax, for any number a, is in U\cap W.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. vector spaces and linear maps
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 26th 2011, 08:30 AM
  2. vector spaces, linear independence and functions
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: March 12th 2011, 08:38 PM
  3. Linear mappings and vector spaces proof
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: February 17th 2011, 05:08 AM
  4. linear algebra abstract vector spaces
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: August 8th 2010, 10:49 PM
  5. Linear transformation between function spaces
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 26th 2009, 01:26 PM

Search Tags


/mathhelpforum @mathhelpforum