Results 1 to 1 of 1

Math Help - Rational Canonical Form

  1. #1
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1

    Rational Canonical Form

    A =  \left( \begin{matrix} 4 &1 & 2 & 0 \\ -4 & 0 &1 &5 \\0 &0 &1 & -1 \\0 &0 &1 & 3 \\ \end{matrix} \right)
    char_A = (x-2)^4
    min_A=(A -2I)^3=x^3-6x^2+12x-8.
    We deduce that R = \left( \begin{matrix} 0& 0& 8& 0\\1 &0 &-12 & 0 \\0 &1 & 6 & 0\\0 &0 & 0 & 2 \\<br />
   \end{matrix} \right) .
    We choose \vec{v_1}= \left( \begin{matrix} 0 \\ 0 \\1 \\0  \\   \end{matrix} \right) and \vec{v_2}= A\vec{v_1} = \left( \begin{matrix} 2 \\ 1 \\1 \\1  \\   \end{matrix} \right) = \vec{v_2}. Then \vec{v_3}= A\vec{v_2} = A^2\vec{v_1}  . This is the third column of A^2 which is \left( \begin{matrix} 4 &1 & 2 & 0 \\ -4 & 0 &1 &5 \\0 &0 &1 & -1 \\0 &0 &1 & 3 \\ \end{matrix} \right)\left( \begin{matrix}  2 \\ 1 \\ 1 \\ 1 \\ \end{matrix} \right) =  \left( \begin{matrix}  11 \\ -2 \\0 \\ 4 \\ \end{matrix} \right) =  \vec{v_3} . Finally, \vec{v_4}=   \left( \begin{matrix}  1 \\ 0 \\ 0\\ 0 \\ \end{matrix} \right)  .
    Let Q =<br />
\left( \begin{matrix} 0& 2&11 & 1\\0 &1 & -2 & 0 \\1 &1 & 0 & 0\\0 &1 & 4 & 0 \\<br />
   \end{matrix} \right)<br />
  .
    Let's check independence of the columns and invertibility of the matrix right now.
    det(Q) = \left| \begin{matrix} <br />
    0& 2&11 & 1\\0 &1 & -2 & 0 \\1 &1 & 0 & 0\\0 &1 & 4 & 0 \\<br />
   \end{matrix} \right| = (1) \left| \begin{matrix}  2& 11 & 1\\1 & -2 & 0 \\1 & 4 & 0\\<br />
   \end{matrix} \right|  = (1) (1)\left| \begin{matrix}  1 & -2  \\1 & 4 \\<br />
   \end{matrix} \right| = 6\neq 0 , so Q is invertible. This also tells us that the row space (and thus the column space) has dimension 4. It follows that the columns are independent.
    Now we check that QR = AQ.
    QR = \left( \begin{matrix} <br />
 0& 2&11 & 1\\0 &1 & -2 & 0 \\1 &1 & 0 & 0\\0 &1 & 4 & 0 \\<br />
   \end{matrix} \right) \left( \begin{matrix} 0& 0& 8& 0\\1 &0 &-12 & 0 \\0 &1 & 6 & 0\\0 &0 & 0 & 2 \\   \end{matrix} \right)=  \left( \begin{matrix} <br />
  2& 11 & 42 & 2\\<br />
 1 & -2 & -24 & 0\\<br />
 1 & 0 & -4 &0\\<br />
 1 & 4 & 12 &0\\<br />
   \end{matrix} \right)

    AQ =\left( \begin{matrix} 4 &1 & 2 & 0 \\ -4 & 0 &1 &5 \\0 &0 &1 & -1 \\0 &0 &1 & 3 \\ \end{matrix} \right)\left( \begin{matrix} <br />
0& 2&11 & 1\\0 &1 & -2 & 0 \\1 &1 & 0 & 0\\0 &1 & 4 & 0 \\  \end{matrix} \right) =  \left( \begin{matrix} <br />
   2& 11 & 42 & 4\\<br />
 1 & -2 & -24 & -4\\<br />
 1 & 0 & -4 &0\\<br />
 1 & 4 & 12 &0\\<br /> <br />
     \end{matrix} \right)
    It doesn't work can someone tell me why?
    It seems that the only vector working is \vec{v_4}= \left( \begin{matrix} 0 \\ 0 \\0 \\0  \\   \end{matrix} \right) !
    Last edited by vincisonfire; March 8th 2009 at 08:45 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Put the following matrices in canonical form?
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: February 26th 2011, 11:07 AM
  2. [SOLVED] Canonical form of a quadratic form
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 9th 2010, 01:12 AM
  3. Canonical Form - Second Order PDE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 24th 2010, 03:45 PM
  4. Cantor canonical form
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 24th 2010, 11:38 PM
  5. Rational Canonical Form of a Matrix
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 10th 2009, 12:20 PM

Search Tags


/mathhelpforum @mathhelpforum