ジョルダン標準形
S = P [ 3 0 0 3 ] P − 1 = [ 3 0 0 3 ] , N = P [ 0 1 0 0 ] P − 1 = [ − 2 2 − 2 2 ] , A = P ( [ 3 0 0 3 ] + [ 0 1 0 0 ] ) P − 1 = S + N {\displaystyle S=P{\begin{bmatrix}3&0\\0&3\end{bmatrix}}P^{-1}={\begin{bmatrix}3&0\\0&3\end{bmatrix}},~N=P{\begin{bmatrix}0&1\\0&0\end{bmatrix}}P^{-1}={\begin{bmatrix}-2&2\\-2&2\end{bmatrix}},~A=P\left({\begin{bmatrix}3&0\\0&3\end{bmatrix}}+{\begin{bmatrix}0&1\\0&0\end{bmatrix}}\right)P^{-1}=S+N}
この分解は N = 0 や SN = NS が成り立つので、 行列の指数関数や冪乗の計算に役立つ。
e A t = e S t + N t = e S t e N t = e S t ( I + N t ) = e 3 t [ 1 − 2 t 2 t − 2 t 1 + 2 t ] {\displaystyle e^{At}=e^{St+Nt}=e^{St}e^{Nt}=e^{St}(I+Nt)=e^{3t}{\begin{bmatrix}1-2t&2t\\-2t&1+2t\end{bmatrix}}}
A n = ( S + N ) n = S n + n S n − 1 N = 3 n − 1 [ 3 − 2 n 2 n − 2 n 3 + 2 n ] {\displaystyle A^{n}=(S+N)^{n}=S^{n}+nS^{n-1}N=3^{n-1}{\begin{bmatrix}3-2n&2n\\-2n&3+2n\end{bmatrix}}}