Question regarding the exponential of a matrix

[ChrisVer]

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 $be^{a}$)

 

[fresh_42]

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]

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]

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]

$\textbf{A}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix}$

This is valid for $n \ge 1$.

 

[ChrisVer]

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]

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 $0$.

 

[fresh_42]

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 $e^{tA}=e^{atI}\cdot e^{tbN}$.

 

[ChrisVer]

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 $[N,I]=0$ )
$e^{atI}\cdot e^{tbN} = (\sum_n \frac{(at)^n}{n!} ) I^n \cdot (\sum_m \frac{(bt)^m N^m}{m!})$
$e^{atI}\cdot e^{tbN} = e^{at} I \cdot (I + btN)= e^{at} I + bt e^{at} N$