求解矩阵指数e^(t·矩阵)：已知特征值与特征向量的技术问询

Hey there! Let's work through computing ( e^{tA} ) for your matrix step by step. First, let's confirm the eigenvalues of ( A = \begin{pmatrix}2 & 2 & -2 \ 5 & 1 & -3 \ 1 & 5 & -3\end{pmatrix} ):

Calculating the characteristic equation ( \det(A - \lambda I) = 0 ), we end up with ( -\lambda^3 = 0 ), so all three eigenvalues are 0 (a triple root). That tells us ( A ) is a nilpotent matrix—meaning some power of ( A ) will be the zero matrix. Let's verify that:

• Compute ( A^2 ):
  $$A^2 = A \times A = \begin{pmatrix}12 & -4 & -4 \ 12 & -4 & -4 \ 24 & -8 & -8\end{pmatrix}$$
• Then compute ( A^3 = A^2 \times A ):
  $$A^3 = \begin{pmatrix}0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0\end{pmatrix}$$

Perfect, ( A^3 = 0 ), so any higher powers (( A^4, A^5, \dots )) will also be zero. That simplifies the matrix exponential a lot because the Taylor series expansion of ( e^{tA} ) truncates after the ( t^2 ) term:

$$e^{tA} = I + tA + \frac{t^2}{2!}A2 + \frac{t^3}{3!}A3 + \dots = I + tA + \frac{t^2}{2}A2$$

Now let's compute each term individually:

1. The identity matrix ( I ):
   $$I = \begin{pmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix}$$

2. The ( tA ) term:
   $$tA = \begin{pmatrix}2t & 2t & -2t \ 5t & t & -3t \ t & 5t & -3t\end{pmatrix}$$

3. The ( \frac{t^2}{2}A2 ) term:
   $$\frac{t^2}{2}A2 = \begin{pmatrix}6t^2 & -2t^2 & -2t^2 \ 6t^2 & -2t^2 & -2t^2 \ 12t^2 & -4t^2 & -4t^2\end{pmatrix}$$

Finally, add all three matrices element-wise to get ( e^{tA} ):

$$e^{tA} = \begin{pmatrix}1+2t+6t^2 & 2t-2t^2 & -2t-2t^2 \ 5t+6t^2 & 1+t-2t^2 & -3t-2t^2 \ t+12t^2 & 5t-4t^2 & 1-3t-4t^2\end{pmatrix}$$

Just to double-check: if ( t=0 ), this gives the identity matrix, which makes sense because ( e^{0} = I ). And if you take the derivative with respect to ( t ) at ( t=0 ), you get ( A ), which matches the property ( \frac{d}{dt}e^{tA}\bigg|_{t=0} = A ).

内容的提问来源于stack exchange，提问作者Hideki Ryuga