My lecturer introduced two methods in computing exponential matrices... with the second one being remarkably faster in calculation time. I would appreciate any explanation on the logic behind in especially in step $(\star)$.
The matrix is
$$A = \begin{pmatrix} 6 & -2 & 3\\2 & 1 & 6\\1 & -2 & 8\end{pmatrix}$$
First Method
The first method is by just taking the exponential of a direct sum of the Jordan. The eigenvalues are $5,5,5$ with
$$\ker (A-5I)=span\left\{\begin{pmatrix}2\\1\\0 \end{pmatrix},\begin{pmatrix}-3\\0\\1\end{pmatrix}\right\}\\\ker(A-5I) = \mathbb{R}^3\\ J(5) = J_2(5)\oplus J_1(5) = \begin{pmatrix} 5 & 1 & 0\\0 & 5 & 0\\ 0 & 0 & 5\end{pmatrix}\\P = \begin{pmatrix}1 & 1 & -3\\2 & 0 & 0\\1 & 0 & 1\end{pmatrix}. $$
So $$A = PJP^{-1}\\tA = P(tJ_2(5))\oplus (tJ_1(5))P^{-1}\\e^{tA} = Pe^{5t}\begin{pmatrix}1 & t & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}P^{-1} \\ = \begin{pmatrix}1+t & -2t & 3t\\2t & 1-4t & 6t\\t & -2t & 3t+1\end{pmatrix}.$$
Second Method
Notice that the eigenvalues are of multiplicity $3$, and $\ker(A-5I)^3 = \mathbb{R}^3$.
Take the basis vectors to form a Jordan Chain:
$$\begin{pmatrix}1\\0\\0 \end{pmatrix}\longrightarrow \begin{pmatrix}1\\2\\1\end{pmatrix}\longrightarrow \begin{pmatrix}0\\0\\0\end{pmatrix}\\\begin{pmatrix}0\\1\\0 \end{pmatrix}\longrightarrow \begin{pmatrix}-2\\-4\\-2\end{pmatrix}\longrightarrow \begin{pmatrix}0\\0\\0\end{pmatrix}\\\begin{pmatrix}1\\0\\0 \end{pmatrix}\longrightarrow \begin{pmatrix}3\\6\\3\end{pmatrix}\longrightarrow \begin{pmatrix}0\\0\\0\end{pmatrix}.$$
So
$$e^{tA}\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}=e^{5t}\begin{pmatrix}1+t \\ 2t \\ t\end{pmatrix} \\ e^{tA}\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}=e^{5t}\begin{pmatrix}-2t \\ 1-4t \\ -2t\end{pmatrix} \\e^{tA}\begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}=e^{5t}\begin{pmatrix}3t \\ 6t \\ 1+3t\end{pmatrix} $$
Hence,
$$e^{tA} = e^{5t} \begin{pmatrix}1+t & -2t & 3t\\2t & 1-4t & 6t\\t & -2t & 3t+1\end{pmatrix} \qquad (\star)$$
Basically, I'm not sure how the last matrix is formed from those above equations. It looks like we're takign a sum?