1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the third vector diff equation

  1. Aug 11, 2009 #1
    solve y'=Ay+b system of equation where
    [tex]
    \vec{A}=\begin{pmatrix}
    3 & 1 & 0\\
    -1 & 1 & 0\\
    1& 1 &2
    \end{pmatrix},\vec{B}=\begin{pmatrix}
    e^{2x}\\0
    \\ e^{2x}

    \end{pmatrix}
    [/tex]

    i cant see where is my mistake in my method
    1st page
    http://i26.tinypic.com/24ou8u8.gif
    2nd page
    http://i29.tinypic.com/1j40ue.gif
     
  2. jcsd
  3. Aug 11, 2009 #2

    Mark44

    Staff: Mentor

    Your problem is that you have three repeated eigenvalues (lambda = 2), but only two eigenvectors. To be able to diagonalize your matrix (which is what the technique you are using requires), you need to find a basis for the three-dimensional eigenspace, but you have found only two of these basis vectors. Taking a linear combination your two vectors doesn't get you anywhere. No matter what vector you end up with by doing this, your three vectors are linearly dependent, and hence don't form a basis for your eigenspace.

    I don't recall what, if anything, you can do in this case, as I'm not anywhere close to my reference books.
     
  4. Aug 11, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I haven't looked at your attachment but since Mark44 says you have 2 as a triple eigenvalue, the characteristic equation must be [itex](\lambda- 2)^3= 0[/itex]. Now, it is true that a matrix always satisfies its own characteristic equation so, writing "A" for the matrix, we must have [itex](A- 2I)^3 v= 0[/itex] for all vectors v. Mark44 says you have two independent eigenvectors, say, [itex]v_1[/itex] and [itex]v_2[/itex] so you have [itex](A- 2I)v_1= 0[/itex] and [itex](A- 2I)v_2= 0[/itex]. From that it follows that [itex](A- 2I)^2v_1= 0[/itex] and [itex](A- 2I)^2v_2= 0[/itex]. You want to find a third, independent, vector, w, such that neither [itex](A- 2I)w= 0[/itex] nor [itex](A- 2I)^2w= 0[/itex] but [itex](A- 2I)^2w= 0[/itex]. Finding a vector, w, such that [itex](A- 2I)w= v_1[/itex] or [itex](A- 2I)w= v_2[/itex] will do that.
     
  5. Aug 12, 2009 #4
    i cant understand how and why you transformed
    [tex]
    (A- 2I)v_1= 0
    [/tex]
    into
    [tex]
    (A- 2I)^2v_1= 0
    [/tex]

    and i cant practically understand how to find the third vector from this:"
    such that neither [tex] (A- 2I)w= 0[/tex] nor [tex] (A- 2I)^2w= 0
    [/tex] but [tex] (A- 2I)^2w= 0 .[/tex] Finding a vector, w, such that
    [tex] (A- 2I)w= v_1[/tex] or [tex] (A- 2I)w= v_2 will do
    that.[/tex]"

    there is a conflicting conditions in the first sentence.
    and i need to do 3 matrix multiplications some with power 2
    which is another multiplication.
    and i need to see that it differs 0.

    ???
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding the third vector diff equation
  1. Find DIff equations (Replies: 3)

Loading...