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Homework Help: A = λI + N, N is a nilpotent matrix

  1. Apr 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A is a quadratic matrix

    A = λI + N

    where λ is a real number, I is the identity matrix and N is a nilpotent matrix

    Compute exp(tA) as a function of t, λ and N^i for i ? 1,2..,n-1

    My attempt

    exp(At) = exp((λI + N)t) = exp(λIt)exp(Nt) =
    [tex] \sum_{n=0}^{\infty} \frac{(Iλt)^n}{n!} \sum_{n=0}^{\infty} \frac{(Nt)^n}{n!} [/tex]

    Its supposed to be Iλt and not I955;t
     
    Last edited: Apr 14, 2010
  2. jcsd
  3. Apr 14, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Then try making a proper lambda in LaTeX: [itex]I \lambda t[/itex] ;)

    What is exp(c I), where c is a constant and I is the identity matrix?

    What happens when you start writing out the sum
    [tex]
    \sum_{n=0}^{\infty} \frac{(Nt)^n}{n!} = \sum_{n = 0}^\infty \frac{t^n}{n!} N^n
    [/tex]?
     
  4. Apr 14, 2010 #3
    Thanks

    exp(cI) = ones + exp(c)I - I
    where one is a sqare matrix with only ones,
    and exp(ctI) = ones + exp(ct)I - I

    and e^N=
    [tex] I + N \frac{1}{2}N^2 + \frac{1}{6}N^3 + \cdots + \frac{1}{(n-1)!}N^{n-1}. [/tex]

    so
    [tex] e^{At} = ( ones + e^{ \lambda t} I - I )(I + N \frac{1}{2}N^2 + \frac{1}{6}N^3 + \cdots + \frac{1}{(n-1)!}N^{n-1} ) [/tex]

    correct?
     
    Last edited: Apr 14, 2010
  5. Apr 14, 2010 #4

    Mark44

    Staff: Mentor

    No, this doesn't make sense at all. exp(A) = I + A + A2/2! + ... + An/n! + ... Replace A by cI and see what you get.
     
  6. Apr 14, 2010 #5
    Thanks, it should be exp(c)*I
     
  7. Apr 14, 2010 #6

    Mark44

    Staff: Mentor

    exp(c)*I is a diagonal matrix with exp(c) in each position along the diagonal.
     
  8. Apr 14, 2010 #7
    Am I wrong again?
     
  9. Apr 14, 2010 #8

    Mark44

    Staff: Mentor

    I think that what you need to do now is to replace each factor on the right by its corresponding series (not in closed form), and use the fact that N is nilpotent, which will simplify exp(Nt) considerably.
     
  10. Apr 15, 2010 #9
    Thanks
     
  11. Apr 15, 2010 #10

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    So what you got is
    [tex]e^c \left( I + \sum_{i = 1}^{n - 1} \frac{t^i}{i!} N^i \right) [/tex]
    right?
     
  12. Apr 15, 2010 #11
    Yes, thanks to both of you for all the help
    [tex]
    e^{At} = (e^{\lambda t})(I + \sum_{i = 1}^{n-1} \frac{t^i}{i!}N^i)
    [/tex]
     
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