Results 1 to 4 of 4

Math Help - Linear Algebra: Linear Independence question

  1. #1
    Junior Member
    Joined
    Dec 2009
    From
    Texas
    Posts
    70
    Awards
    1

    Linear Algebra: Linear Independence question

    Hey guys,

    Can you tell me if this solution looks alright?


    Thanks!
    James
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Dec 2009
    From
    Texas
    Posts
    70
    Awards
    1
    Sorry the picture didn't come up. Here it is again.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    The row reduction looks good to me. I'm not sure I would your matrix equation the way you did. I would write it this way:

    \left[\begin{array}{rrr}\mathbf{v}_{1} &\mathbf{v}_{2} &\mathbf{v}_{3}\end{array}\right]\left[\begin{array}{r}c_{1}\\ c_{2}\\ c_{3}\end{array}\right]=\left[\begin{array}{rrr}1 &2 &0\\ \lambda^{2} &\lambda &0\\ 1 &4 &1\\ 2 &8 &2\end{array}\right]\left[\begin{array}{r}c_{1}\\ c_{2}\\ c_{3}\end{array}\right].

    Your column matrix

    \left[\begin{array}{r}c_{1}\\ c_{2}\\ c_{3}\end{array}\right]

    disappeared on the RHS for some reason. Other than that, as I say, it looks good to me!

    [EDIT] See Deveno's post for a correction.
    Last edited by Ackbeet; May 3rd 2011 at 05:37 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,397
    Thanks
    760
    one must be extremely careful in accounting for each step of the row reduction.

    i found that to eliminate λ from the row-reduced matrix, it was necessary to divide by λ. this suggests that λ = 0 is a special case.

    and indeed, when λ = 0, we have 2(1,0,1,2) + 2(0,0,1,2) = (2,0,4,8), so that {(1,0,1,2), (2,0,4,8), (0,0,1,2)} is a linearly dependent set.

    thus the statement: "regardless of the parameter λ, the only scalar values of c1, c2, c3 such that c1v1 + c2v2 + c3v3 = 0 are c1,c2,c3 all 0" is UNTRUE

    i have exhibited a counter-example.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 1st 2011, 10:00 PM
  2. Linear Independence in linear transformations
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 17th 2009, 04:22 PM
  3. Replies: 7
    Last Post: August 30th 2009, 10:03 AM
  4. Linear Algebra Question - Linear Operators
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 3rd 2009, 04:33 AM
  5. Linear Transformations and Linear Independence
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 6th 2008, 07:36 PM

Search Tags


/mathhelpforum @mathhelpforum