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Let $A,B,C \in \operatorname{Mat}_2(\mathbb{R})$ define the real matrices: $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix},\ B =\begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix},\ C = A+B=\begin{pmatrix} 0 & 1 \\ 0 & 1 \\ \end{pmatrix} $$ Show that $$\exp(A;t) = \begin{pmatrix} 1 & t \\ 0 & 1 \\ \end{pmatrix}\text{ and } \exp(B;t)=\begin{pmatrix} 1 & 0 \\ 0 & e^t \\ \end{pmatrix} \text{ for } t\in \mathbb{R}$$ and show that $$\exp(C;t)=\begin{pmatrix} 1 & e^t-1 \\ 0 & e^t \\ \end{pmatrix}\text{ for } t\in \mathbb{R}$$

I am a bit unsure with the notation used here for $\exp(A;t)=\begin{pmatrix} 1 & t \\ 0 & 1 \\ \end{pmatrix}$ and how am I supposed to show this based on the information for the matrices $A,B,C$ ?

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    $\begingroup$ It looks like they simply mean $e^{tA}$. Do you know how to compute the exponential of a matrix? If not, you should review that. $\endgroup$
    – amd
    May 29, 2018 at 17:56
  • $\begingroup$ It follows from the definition of the exponential mathworld.wolfram.com/MatrixExponential.html $\endgroup$
    – Alan Muniz
    May 29, 2018 at 17:57

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I assume that $e^A = I + A + {1 \over 2!} A^2+ \cdots$.

Note that $A^2 = 0$, hence $e^{At} = I + tA$.

Note that $B^k = B$ for all $k \ge 1$, hence $e^{Bt} = I + B(t + {t^2 \over 2} + \cdots) )= \begin{bmatrix} 1 & 0 \\ 0 & e^t \end{bmatrix}$.

Finally, as Jean-Claude noted, $C^k = C$ for $k \ge 1$.

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  • $\begingroup$ And $C^k=C$ for $k>0$. $\endgroup$ May 29, 2018 at 18:02
  • $\begingroup$ @Jean-ClaudeArbaut: Thanks, was just getting there :-). $\endgroup$
    – copper.hat
    May 29, 2018 at 18:03
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    $\begingroup$ Yes, but I already upvoted :) $\endgroup$ May 29, 2018 at 18:03
  • $\begingroup$ Exactly what I was looking for, thank you! $\endgroup$
    – Simbörg
    May 29, 2018 at 21:14

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