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Don't have a damn idea how to do this

  1. Nov 19, 2009 #1
    Suppose that A is a real 2 x 2 matrix with one eigenvalue [tex]\lambda[/tex] of multiplicity two. Show that the solution to the initial value problem y' = Ay with y(0) = v is given by

    y(t) = e^[tex]\lambda[/tex]t [v + t(A - [tex]\lambda[/tex]I)v]

    Hint: Verify the result by direct substitution. Remember that (A - [tex]\lambda[/tex]I)^2 = 0I, so A(A - [tex]\lambda[/tex] I) = [tex]\lambda[/tex] (A - [tex]\lambda[/tex] I).

    Obviously, y(0) = v, but I couldn't figure what else to do.
     
  2. jcsd
  3. Nov 19, 2009 #2

    Hurkyl

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    Have you tried direct substitution? :tongue:
     
  4. Nov 19, 2009 #3
    I don't see how it does any good.

    y' = A [ e^[tex]\lambda[/tex]t [v + t(A - [tex]\lambda[/tex]I)v] ]

    I can play with that, but it still leaves the y' on the other side.

    Maybe A^2 y = A [ e^[tex]\lambda[/tex]t [v + t(A - [tex]\lambda[/tex]I)v] ]
     
  5. Nov 20, 2009 #4

    Hurkyl

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    Why not substitute that y as well?
     
  6. Nov 20, 2009 #5
    Okay. Let's see. If I do that, then I think I can then bring the terms to the left and equate everything to the zero vector, right?!
     
  7. Nov 20, 2009 #6

    Mark44

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    Shackleford, can you check to make sure that you have given us the problem exactly as it was given to you? I worked on this for maybe a couple of hours last night, and didn't get very far.

    In the hint it says to remember that (A - [itex]\lambda[/itex]I)^2 = 0, and I wasn't able to convince myself why this is true. I came up with a 2x2 matrix that has repeated eigenvalues, and verified this equation using that matrix. If x is an eigenvector, then clearly (A - [itex]\lambda[/itex]I)x = 0, so (A - [itex]\lambda[/itex]I)^2*x is 0 as well.

    If the technique is to just substitute the expession for y(t) into the differential equation, it should be more straightforward that it seems to be, so I'd like to make sure that we're working on the right problem.

    Another strategy that I thought of was diagonalization, but with the repeated eigenvalue there's no guarantee that there are two independent eigenvectors, so that's probably a dead end.
     
  8. Nov 20, 2009 #7
    Yes. I will when I get home from work.

    Your reasoning for (A - [itex]\lambda[/itex]I)^2 = 0 is mostly how the book explains it. However, the vector x is any vector in R2, if I remember correctly. It then goes on to explain how to find an additional eigenvector for the general solution.
     
  9. Nov 20, 2009 #8
    It is correct.
     
  10. Nov 21, 2009 #9

    Mark44

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    OK, I think I have it.

    Suppose y(t) = e[itex]\lambda[/itex]t(v + t(A - [itex]\lambda[/itex]I)v) is a solution of the differential equation y' = Ay, and that [itex]\lambda[/itex] is the eigenvalue of A as described in your problem statement.

    From the diff. equation, we have y' - Ay = 0, which must be true for all t.

    This implies that:
    [tex]\lambda e^{\lambda t}[\bold{v} + t(A - \lambda I)\bold{v}] + e^{\lambda t}[(A - \lambda I)\bold{v}] - e^{\lambda t}[A\bold{v} + At(A - \lambda I)\bold{v}~=~\bold{0}[/tex]
    [tex]\Rightarrow -e^{\lambda t}[(A - \lambda I)^2\bold{v}] - e^{\lambda t}[(A - \lambda I)\bold{v} -(A - \lambda I)\bold{v}]~=~\bold{0}[/tex]
    [tex]\Rightarrow -e^{\lambda t}[(A - \lambda I)^2\bold{v}] ~=~\bold{0}[/tex]
    The equation above has to be true for all t, so what can you conclude about
    [tex](A - \lambda I)^2 \bold{v}?[/tex]

    The hint is somewhat incomplete. I believe it should say:
    Remember that [itex](A - \lambda I)^2\bold{v} = \bold{0}[/itex], so [itex]A(A - \lambda I)\bold{v} = \lambda (A - \lambda I)\bold{v}[/itex].

    What does this say about [itex](A - \lambda I)\bold{v}[/itex] relative to the matrix A?

    All of the above was laboriously transcribed into LaTeX from my notes, so it's possible I have typed something incorrect somewhere. I've checked it over and don't see anything wrong, but it's very easy to miss something when half of what I type is script.
     
  11. Nov 21, 2009 #10

    HallsofIvy

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    Of course it's true. A has [itex]\lambda[/itex] as a double eigenvalue so this is just saying that A satisfies its own characteristic equation.

     
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