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

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

首先，將矩陣對角化（步驟請參見 matrix diagonalization calculator）。

由於該矩陣不可對角化，將其寫成對角矩陣 $$$D = \left[\begin{array}{cc}t & 0\\0 & t\end{array}\right]$$$ 與冪零矩陣 $$$N = \left[\begin{array}{cc}0 & - t\\0 & 0\end{array}\right]$$$ 之和。

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

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

對角矩陣的矩陣指數是將其對角元素分別取指數所得的矩陣：$$$e^{\left[\begin{array}{cc}t & 0\\0 & t\end{array}\right]} = \left[\begin{array}{cc}e^{t} & 0\\0 & e^{t}\end{array}\right]$$$

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

最後，將矩陣相乘：

$$$\left[\begin{array}{cc}e^{t} & 0\\0 & e^{t}\end{array}\right]\cdot \left[\begin{array}{cc}1 & - t\\0 & 1\end{array}\right] = \left[\begin{array}{cc}e^{t} & - t e^{t}\\0 & e^{t}\end{array}\right]$$$（步驟請參見 矩陣乘法計算器）。

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