$$$e^{\left[\begin{array}{cc}t & - t\\0 & t\end{array}\right]}$$$

求$$$e^{\left[\begin{array}{cc}t & - t\\0 & t\end{array}\right]}$$$。

首先，对矩阵进行对角化（步骤参见矩阵对角化计算器）。

由于该矩阵不可对角化，将其写为对角矩阵 $$$D = \left[\begin{array}{cc}t & 0\\0 & t\end{array}\right]$$$ 与幂零矩阵 $$$N = \left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]$$$ 之和。

注意到$$$N^{2} = \left[\begin{array}{cc}0 & 0\\0 & 0\end{array}\right]$$$。

这意味着$$$e^{N} = I + N$$$，即$$$e^{\left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]} = \left[\begin{array}{cc}1 & 0\\0 & 1\end{array}\right] + \left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right] = \left[\begin{array}{cc}1 & - t\\0 & 1\end{array}\right]$$$。

对角矩阵的矩阵指数是一个对角元素逐一取指数得到的矩阵：$$$e^{\left[\begin{array}{cc}t & 0\\0 & t\end{array}\right]} = \left[\begin{array}{cc}e^{t} & 0\\0 & e^{t}\end{array}\right]$$$。

现在，$$$e^{\left[\begin{array}{cc}t & - t\\0 & t\end{array}\right]} = e^{\left[\begin{array}{cc}t & 0\\0 & t\end{array}\right] + \left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]} = e^{\left[\begin{array}{cc}t & 0\\0 & t\end{array}\right]}\cdot e^{\left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]} = \left[\begin{array}{cc}e^{t} & 0\\0 & e^{t}\end{array}\right]\cdot \left[\begin{array}{cc}1 & - t\\0 & 1\end{array}\right]$$$。

最后，将矩阵相乘：

$$$\left[\begin{array}{cc}e^{t} & 0\\0 & e^{t}\end{array}\right]\cdot \left[\begin{array}{cc}1 & - t\\0 & 1\end{array}\right] = \left[\begin{array}{cc}e^{t} & - t e^{t}\\0 & e^{t}\end{array}\right]$$$（步骤请参见矩阵乘法计算器）。

$$$e^{\left[\begin{array}{cc}t & - t\\0 & t\end{array}\right]} = \left[\begin{array}{cc}e^{t} & - t e^{t}\\0 & e^{t}\end{array}\right]$$$A