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 problem

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Given matrix A= [1,1;0,0] and matrix B = [1,-1;0,0]

    n.b. the semicolon separates the matrix rows

    find exp(A)exp(B)



    2. Relevant equations

    [itex]exp^{A}=\sum\frac{A^{n}}{n!}=I + A + \frac{A^{2}}{2} + \frac{A^{3}}{6} + ...... + \frac{A^{n}}{n!}[/itex]

    for n= 0 to infinity

    3. The attempt at a solution

    The answer the tutor give is the following matrix

    exp(A).exp(B) = [e^2,-(e-1)^2 ; 0,1]

    this would imply that A and B are diagonalisable (since they need to be in order to get the form of the answer)

    However, if I worked out the characteristic ppolynomial of A and B, I would only get one eigenvalue, not two distinct eigenvalues which rules out any chance of diagaonlisation

    Is there a way to diagaonlise A and B ?
     
  2. jcsd
  3. Oct 20, 2011 #2

    Mark44

    Staff: Mentor

    I get two eigenvalues for each matrix, so I don't know what you did that you came up with only one apiece.
     
  4. Oct 20, 2011 #3

    Mark44

    Staff: Mentor

    You can also do this problem without diagonalizing either matrix. Use your relevant equation, which should be slightly amended. It's not a finite sum.
    [tex]exp^{A}=\sum\frac{A^{n}}{n!}=I + A + \frac{A^{2}}{2} + \frac{A^{3}}{6} + ...... + \frac{A^{n}}{n!} + ...[/tex]
     
  5. Oct 20, 2011 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    A standard result for an nxn matrix with all eigenvalues different is: if r_1, r_2,...,r_n are the eigenvalues, there exist matrices E_1,E_2,...,E_n such that for any analytic function f(x) = sum c_n x^n (with radius of convergence containing all the r_j) then f(A) = c_0*I + c_1*A + c_2*A^2 + ... = sum_j f(r_j)* E_j . Here the E_j are the same for any f, so cn be found, for example, by looking at f(x) = 1 = x^0, f(x) = x, f(x) = x^2, etc. Having the E_j, now use f(x) = exp(x) to get exp(A).

    For a 2x2 matrix with eigenvalues 'a' and 'b' we have I = E1 + E2, which comes from using f(x) = 1. Similarly, A = a*E1 + b*E2 (using f(x) = x).

    RGV
     
  6. Oct 20, 2011 #5
    What was your characteristic polynomial for A ? What eigenvalues did you get for A ?


    Also, what char. polynomial and eigenvalues did you get for B ?
     
  7. Oct 20, 2011 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Those are both "Jordan form" matrices so the numbers on the diagonal are the eigenvalues.
     
  8. Oct 20, 2011 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What did you get?
     
  9. Oct 20, 2011 #8
    I got 0 and 1 as the eigenvalues for both matrix A and Matrix B.

    The characteristic polynomial I got for A was [itex]λ^{2}-λ[/itex]

    Similary, my char. polynomial for B is [itex]λ^{2}-λ[/itex]
     
  10. Oct 20, 2011 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    So you got two distinct eigenvalues, and you can diagonalize them both.

    As Mark mentioned, you can calculate eA and eB without diagonalizing though. A matrix satisfies its characteristic polynomial, so A2-A=0. Using that fact, you can easily simplify the series.
     
  11. Oct 21, 2011 #10
    Thanks for that. I was asking in the first placecause I knew the diagonal form of both A and B would be

    [ 0, 1 ; 0, 0]

    which is nowhere close to the tutor's answer
     
  12. Oct 21, 2011 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's not the diagonal form of either A or B.

    Also, your tutor calculated exp(A)exp(B), which isn't the diagonalization of A or B.
     
  13. Oct 21, 2011 #12
    but doesn't the Diagonal matrix contain the eigenvalues down the main diagonal and 0's elsewhere ?

    what don't I understand ?
     
  14. Oct 21, 2011 #13

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The matrix you wrote was
    \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}which isn't diagonal.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




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

  2. Matrix exponential (Replies: 4)

  3. Matrix exponential (Replies: 4)

  4. Matrix Exponentials (Replies: 2)

Loading...