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I'm studying for an exam and I can't find this in my notes or in the book, but it's on a past exam...

Given $A = \begin{bmatrix}-1 & 1\\0 & -1\end{bmatrix}$,

$e^{tA} = \begin{bmatrix}e^{-t} & te^{-t}\\0 & e^{-t}\end{bmatrix}$

True/False? The answer is true according to the answer key, but I don't see why. What theorem/concept would tell me that this is true? It's a true false question so I'm sure it should be something I can see at a glance or with very little work ...

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2 Answers 2

up vote 6 down vote accepted

Let $I = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$.

Since $A = -I+B$ and the matrices $I$ and $B$ commute, we have $e^{tA} = e^{t(-I+B)} = e^{-tI}e^{tB}$.

Trivially, $e^{-tI} = \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-t}\end{bmatrix}$. Also, since $B^2 = 0$, we have $e^{tB} = I+tB = \begin{bmatrix} 1 & t \\ 0 & 1\end{bmatrix}$.

Therefore, $e^{tA} = e^{-tI}e^{tB} = \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-t}\end{bmatrix}\begin{bmatrix} 1 & t \\ 0 & 1\end{bmatrix} = \begin{bmatrix} e^{-t} & te^{-t} \\ 0 & e^{-t}\end{bmatrix}$.


Better answer: Suppose $\begin{bmatrix} e^{-t} & te^{-t} \\ 0 & e^{-t}\end{bmatrix} = e^{tA}$ for some matrix $A$.

Differentiate to get $\begin{bmatrix} -e^{-t} & (1-t)e^{-t} \\ 0 & -e^{-t}\end{bmatrix} = Ae^{tA}$. Then plug in $t = 0$ to get $\begin{bmatrix} -1 & 1 \\ 0 & -1\end{bmatrix} = A$.

This assumes that the given matrix is the matrix exponential of some matrix, but if it is a true/false question and you don't have a lot of time, making that assumption might be the way to go.

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Ooh, I do remember the bit about the commutative matrices. The book (and my notes) just weren't really clear on what to DO with it. You are a godsend, thank you :D –  Daniel Ball Jul 25 at 2:52
    
Actually, no. I'm still stuck on the last bit. Why does e^tB = I + tB because B^2 = 0? –  Daniel Ball Jul 25 at 4:35
1  
got it. Exponential of a nilpotent matrix. It's not in those words, but it's mentioned as the infinite series becoming finite since the B^n becomes the zero vector in the exponentiation series. –  Daniel Ball Jul 25 at 4:54

See the general formula for functions of Jordan blocks, which this is.

You'd have to figure out how matrix exponents were defined in your course. There is no general formula for triangular matrices. The hard way is to notice that $A=-I+J$, where $J^2=0$, and then to write $e^{tA}$ as a power series, powers of $-I+J$ can be computed explicitly by induction.

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While that may be applicable, that phrase has never come up in my differential equations course, so I don't think that's what they're after; the exam is tomorrow too, so I'm not sure I want to spend my time learning something not actually on the exam ... Can you think of anything more direct? –  Daniel Ball Jul 25 at 2:44

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