Results 1 to 4 of 4

Math Help - linear dependency

  1. #1
    Junior Member
    Joined
    Nov 2008
    From
    Vermont, New England, USA, North America, Earth, Sol Solar System, Orion Arm, Milky Way Galaxy
    Posts
    61

    linear dependency

    For the first part, I got that they are dependent...I'm just double checking that that's right. For the second part, how does that prove or disprove that they span R4, and can I get a hint in the right direction for solving it, please?

    Decide whether or not the following vectors are linearly independent, by solving c1v1+c2v2+c3v3+c4v4=0:
    v1=(1,1,0,0)
    v2=(1,0,1,0)
    v3=(0,0,1,1)
    v4=(0,1,0,1)

    Decide also if they span R4 by trying to solve c1v1+c2v2+c3v3+c4v4=(0,0,0,1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,397
    Thanks
    760

    Re: linear dependency

    what the problem is asking you to solve, is the system:

    \begin{bmatrix}1&1&0&0\\1&0&0&1\\0&1&1&0\\0&0&1&1 \end{bmatrix} \begin{bmatrix}c_1\\c_2\\c_3\\c_4 \end{bmatrix} = \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}

    which hopefully you know how to do by now.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2008
    From
    Vermont, New England, USA, North America, Earth, Sol Solar System, Orion Arm, Milky Way Galaxy
    Posts
    61

    Re: linear dependency

    Quote Originally Posted by Deveno View Post
    what the problem is asking you to solve, is the system:

    \begin{bmatrix}1&1&0&0\\1&0&0&1\\0&1&1&0\\0&0&1&1 \end{bmatrix} \begin{bmatrix}c_1\\c_2\\c_3\\c_4 \end{bmatrix} = \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}

    which hopefully you know how to do by now.
    so that matrix reduces to
    \begin{bmatrix}1&1&0&0\\0&-1&0&1\\0&0&1&1\\0&0&0&0 \end{bmatrix} \begin{bmatrix}c_1\\c_2\\c_3\\c_4 \end{bmatrix} = \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}

    so it doesn't span R4, right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,397
    Thanks
    760

    Re: linear dependency

    from the fourth row, 0 = 1, which is impossible.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Dependency between orbits
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: October 20th 2011, 09:20 AM
  2. Linear Dependency
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 4th 2010, 10:35 PM
  3. geomtric and algebraic arguements for linear dependency
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: October 19th 2009, 02:28 PM
  4. Expressing a non-trivial linear dependency relationship
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 10th 2009, 01:13 AM
  5. Proving Linear Dependency
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: November 8th 2007, 12:56 PM

/mathhelpforum @mathhelpforum