1. Not finding help here? Sign up for a free 30min 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!

Matrix exponential

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    A=[tex]\begin{bmatrix}-1 & 2 & 0\\-2 & -1 & 0\\ 0 & 0 & -3\end{bmatrix}[/tex]
    Use matrix A and compute exp(tA) explicitly.

    2. Relevant equations
    I am having trouble figuring out how to start this. I know how to look at each component of matrix A and to use exp(tA)=I+tA +.....(exponential expansion) but I am not sure how to simplify each component's expansion. I don't know if there is an easier way to break matrix A up? Please help me to get started!

    3.I have noticed that this matrix A is in canonical form with complex eigenvalues and 1 distinct eigenvalue -3. I tried to break this matrix up into 2 matrices (B and C) that commute so that exp(A)=exp(B)exp(C), but matrix C is a nilpotent matrix with all 0's except -3(bottom right corner) and matrix B has an alpha-beta block in the upper right left corner....

    The attempt at a solution
     
  2. jcsd
  3. Nov 23, 2008 #2

    ptr

    User Avatar

    Notice [tex] \begin{bmatrix}-1 & 2 & 0\\-2 & -1 & 0\\ 0 & 0 & -3\end{bmatrix} ^i = \begin{bmatrix}{((-1)^i (2i-1))} & {((-1)^{i-1} 2i)} & 0\\{(-1)^{i} 2i)} & {((-1)^i (2i-1))} & 0\\ 0 & 0 & (-3)^i\end{bmatrix}[/tex]. The series then is easily represented in a closed form.
     
  4. Nov 23, 2008 #3
    Thank you for your help. Is the closed form the exp(A)? How do I incorporate t?
     
  5. Nov 23, 2008 #4

    ptr

    User Avatar

    Yes, sorry, I meant, in a series, not closed form, hahah, do you need the actual sum?
     
  6. Nov 23, 2008 #5
    yes please.
     
  7. Nov 23, 2008 #6

    ptr

    User Avatar

    The actual sum would depend on element of the domain chosen where the image is to be evaluated.
     
  8. Nov 24, 2008 #7
    I checked and noticed that matrix A does not follow component wise the given summation for the series since A[tex]^{3}[/tex]=[tex]\begin{bmatrix} 11&-2&0\\2&11&0\\0&0&-27\end{bmatrix}[/tex] and [tex]A^{4}[/tex]=[tex]\begin{bmatrix} -7&24&0\\-24&-7&0\\0&0&81\end{bmatrix}[/tex]

    I have to be able to use exp(tA) to write the system X'=AX in a general solution form. Please help me to figure out how to write exp(tA).
     
  9. Nov 24, 2008 #8
    u have to diagonalize the mx first
     
  10. Nov 24, 2008 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    mx? There is no "mx" in the problem. If you meant A, the whole point of this problem is that you can't diagonalize A: it is not diagonalizable. Of course, you can separate the -3: obviously e[sups]A[/sup] will have e-3 3rd row 3rd column and 0s elsewhere on the third row and column.

    The whole problem, then, is finding eB where B is the matrix
    [tex]\left[\begin{array}{cc}-1 & 2 \\ -2 & -1\end{array}\right][/itex]
    That has determinant 5 so we can write it as
    [tex]5\left[\begin{array}{cc}-\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{-1}{\sqrt{5}}\end{array}\right][/tex]
    and that matrix can be interpreted as a rotation matrix with [itex]cos(\theta)= -1/\sqrt{5}[/itex] and [itex]sin(\theta)= 2/\sqrt{5}[/itex]. ([itex]\theta[/itex] is about 116 degrees but that isn't important.)
     
    Last edited: Nov 24, 2008
  11. Nov 24, 2008 #10
    I think I have part of the solution. I have [tex]e^{tA}[/tex].

    Now I am asked to write the general solution in the form [tex]X(t)=e^{tA}X_0[/tex],

    where [tex]X_0=X(0)[/tex].

    I have figured the bulk of the work out, I just don't know exactly what form this is going to look like.

    Will it look like this?

    [tex]\begin{bmatrix}X_1(t) \\ X_2(t) \\ X_3(t) \end{bmatrix} = e^{tA} \begin{bmatrix} X_1(0) \\ X_2(0) \\ X_3(0) \end{bmatrix} [/tex].

    EDIT: Also, should I plug that 0 in at some time for anything?
     
    Last edited: Nov 24, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Matrix exponential
  1. Matrix Exponential (Replies: 4)

  2. Matrix exponential (Replies: 4)

  3. Matrix exponential (Replies: 4)

  4. Matrix Exponentials (Replies: 2)

Loading...