# Question regarding the exponential of a matrix

Gold Member

## Homework Statement:

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

## Relevant Equations:

$e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}$
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:
$\textbf{A}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix}$
i.e.
$e^{\textbf{A}t} = \sum_{n=0}^{\infty} \frac{\textbf{A}^n t^n}{n!}= \begin{pmatrix} \sum_{n=0}^{\infty} \frac{a^nt^n}{n!} & bt \sum_{n=0}^{\infty} \frac{na^{n-1} t^{n-1}}{n!} \\ 0 & \sum_{n=0}^{\infty} \frac{a^nt^n}{n!} \end{pmatrix}$
(forget $t$ 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:
$\sum_{n=0}^{\infty} b n \frac{a^{n-1}}{n!} = b \sum_{n=0}^{\infty} \frac{a^{n-1}}{(n-1)!}$
Which doesn't make much sense. However, I don't know how I could emit the term with $n=0$. I could try to say that $m = n-1$ meaning it would change to $\sum_{m=-1}^{\infty} \frac{a^{m}}{m!}$, but I have the exact same problem.

(I already know how the matrix is written and that term should be a $be^{a}$)

• etotheipi

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fresh_42
Mentor
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##.

• ChrisVer
Gold Member
Thanks. I think that what can actually help me get rid of the $n=0$ term is the original multiplication with $n$, before I cancel it out with the denominator...

PeroK
Homework Helper
Gold Member
Thanks. I think that what can actually help me get rid of the $n=0$ term is the original multiplication with $n$, 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##.

PeroK
Homework Helper
Gold Member
$\textbf{A}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix}$
This is valid for ##n \ge 1##.

Gold Member
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.

PeroK
Homework Helper
Gold Member
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##.

• ChrisVer
fresh_42
Mentor
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}##.

• etotheipi
Gold Member
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 $(Ia+Nb)^n$ however, only helps you to calculate $A^n$.
$e^{aI + bN} = e^{atI}\cdot e^{tbN}$ (from $[N,I]=0$ )
$e^{atI}\cdot e^{tbN} = (\sum_n \frac{(at)^n}{n!} ) I^n \cdot (\sum_m \frac{(bt)^m N^m}{m!})$
$e^{atI}\cdot e^{tbN} = e^{at} I \cdot (I + btN)= e^{at} I + bt e^{at} N$

• fresh_42