How to find exponential of triangular matrix 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 ...
 A: 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. 
A: 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.
