Results 1 to 3 of 3

Thread: Linearly independent polynomials; answer not checking out.

  1. #1
    Member Jskid's Avatar
    Joined
    Jul 2010
    Posts
    160

    Linearly independent polynomials; answer not checking out.

    Are the following vectors in $\displaystyle P_2$ linearly dependent? If yes, express one vector as a linear combination of the rest.
    $\displaystyle {3t+1,3t^2+1,2t^2+t+1}$

    This forms the matrix $\displaystyle
    \[

    \left[ {\begin{array}{ccc}
    0 & 3 & 2 \\
    3 & 0 & 1 \\
    1 & 1 & 1 \\
    \end{array} } \right]
    \]$ which is row equivalent to $\displaystyle \[

    \left[ {\begin{array}{ccc}
    1 & 0 & \frac{1}{3} \\
    0 & 1 & \frac{2}{3} \\
    0 & 0 & 0 \\
    \end{array} } \right]
    \]$
    So $\displaystyle c_1=\frac{r}{3},c_2=\frac{2r}{3}, c_3=-r$ is a solution for the homogeneous system, where r is any real number.
    Let r = 1, then $\displaystyle \frac{1}{3}P_1 + \frac{2}{3}P_2-1=0$
    Solving for $\displaystyle P_1=\frac{1}{3}-\frac{2}{3}P_2$
    But when I check it doesn't work: $\displaystyle 3t+1 \not= \frac{1}{3}-\frac{2}{3}(3t^2+1)$
    Last edited by Ackbeet; Mar 1st 2011 at 02:26 AM. Reason: Fixed spelling and grammar in title.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641
    $\displaystyle P_1=2t^2+t+1,P_2=3t+1,P_3=3t^2+1$.
    $\displaystyle P_1=\frac{1}{3}P_2+\frac{2}{3}P_3$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641
    $\displaystyle 2t^2+t+1=a(3t+1)+b(3t^2+1)\Rightarrow \left\{\begin{matrix}
    3b=2 \\
    3a=1 \\
    a+b=1
    \end{matrix}\right.$ $\displaystyle \Rightarrow a=\frac{1}{3},b=\frac{2}{3}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. linearly dependent in Q, linearly independent in R
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Apr 12th 2011, 01:44 PM
  2. linearly independent
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 17th 2010, 11:55 AM
  3. linearly independent?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Aug 27th 2010, 09:53 AM
  4. Linearly Independent
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Nov 2nd 2009, 03:31 AM
  5. Linearly independent set.
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Apr 23rd 2009, 05:12 PM

Search Tags


/mathhelpforum @mathhelpforum