Question regarding the exponential of a matrix

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ChrisVer
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Homework Statement:

Show that you can write [itex]e^{\textbf{A}t} = f_1(t) \textbf{1} + f_2(t) \textbf{A} [/itex] and determine [itex]f_{1},~f_{2}[/itex] if [itex]\textbf{A}= \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}[/itex]

Relevant Equations:

[itex]e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!} [/itex]
Hi, I think this is a nitpicking question, but oh well let me hear your inputs.
Actually I tried to solve this question straightforwardly, by Taylor expanding the exponential and showing that:
[itex] \textbf{A}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix} [/itex]
i.e.
[itex]e^{\textbf{A}t} = \sum_{n=0}^{\infty} \frac{\textbf{A}^n t^n}{n!}= \begin{pmatrix} \sum_{n=0}^{\infty} \frac{a^nt^n}{n!} & bt \sum_{n=0}^{\infty} \frac{na^{n-1} t^{n-1}}{n!} \\ 0 & \sum_{n=0}^{\infty} \frac{a^nt^n}{n!} \end{pmatrix} [/itex]
(forget [itex]t[/itex] for now)

The problem I faced was related to the top-right element of the matrix when summed. As the exponential gets expanded, it gives me a sum running from n=0 to infinity. However that term results to negative values of n when n=0:
[itex]\sum_{n=0}^{\infty} b n \frac{a^{n-1}}{n!} = b \sum_{n=0}^{\infty} \frac{a^{n-1}}{(n-1)!} [/itex]
Which doesn't make much sense. However, I don't know how I could emit the term with [itex]n=0[/itex]. I could try to say that [itex]m = n-1[/itex] meaning it would change to [itex] \sum_{m=-1}^{\infty} \frac{a^{m}}{m!} [/itex], but I have the exact same problem.

(I already know how the matrix is written and that term should be a [itex]be^{a}[/itex])
 
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Answers and Replies

  • #2
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Why don't you write ##A=a\cdot I + b\cdot N##? Since ##N^2=0## and ##[I,N]=0## this should be the easiest way to write down ##(aI+bN)^n##.
 
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  • #3
ChrisVer
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Thanks. I think that what can actually help me get rid of the [itex]n=0[/itex] term is the original multiplication with [itex]n[/itex], before I cancel it out with the denominator...
 
  • #4
PeroK
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Thanks. I think that what can actually help me get rid of the [itex]n=0[/itex] term is the original multiplication with [itex]n[/itex], before I cancel it out with the denominator...
That term should never be in there, s ##A^0 = I##, so the zeroth term is ##0## and has no terms in ##a## or ##b##.
 
  • #5
PeroK
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[itex] \textbf{A}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix} [/itex]
This is valid for ##n \ge 1##.
 
  • #6
ChrisVer
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This is valid for ##n \ge 1##.
Hm, yes. But I think it is expandable to ##n=0## too, no? (Or it happens to be the case because of this "type" of matrix)
I say that because ##nba^{n-1} =0##, and so we obtain the unit matrix as expected. I agree that I was too fast to cancel out ##n## with ##n!##, and I missed that I could omit the sum for ##n=0## right away.
 
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PeroK
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Hm, yes. But I think it is expandable to ##n=0## too, no? (Or it happens to be the case because of this "type" of matrix)
I say that because ##nba^{n-1} =0##, and so we obtain the unit matrix as expected. I agree that I was too fast to cancel out ##n## with ##n!##, and I missed that I could omit the sum for ##n=0## right away.
Okay, but you can drop the ##n = 0## term, otherwise, when you cancel ##n##, you are cancelling ##0##.
 
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  • #8
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You should really calculate ## (aI+bN)^n ## by the binomial formula. With ##(...)^0=I## and ##N^2=0## you can almost write the answer without any calculation. Since ##[I,N]=0## you can also calculate ##e^{tA}=e^{atI}\cdot e^{tbN}##.
 
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  • #9
ChrisVer
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You should really calculate ## (aI+bN)^n ## by the binomial formula. With ##(...)^0=I## and ##N^2=0## you can almost write the answer without any calculation. Since ##[I,N]=0## you can also calculate ##e^{tA}=e^{atI}\cdot e^{tbN}##.
Yup I agree about the commutator part and splitting in 2 exponentials. Using only the [itex](Ia+Nb)^n[/itex] however, only helps you to calculate [itex]A^n[/itex].
[itex]e^{aI + bN} = e^{atI}\cdot e^{tbN} [/itex] (from [itex][N,I]=0[/itex] )
[itex]e^{atI}\cdot e^{tbN} = (\sum_n \frac{(at)^n}{n!} ) I^n \cdot (\sum_m \frac{(bt)^m N^m}{m!})[/itex]
[itex]e^{atI}\cdot e^{tbN} = e^{at} I \cdot (I + btN)= e^{at} I + bt e^{at} N[/itex]
 
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