Proving $e^{AT} = e^{\lambda t} \sum_{k=0}^{n-1}\frac{t^{k}}{k!}(A - \lambda I)^{k}$ -final step

Question

Suppose that the Jordan canonical form $J$ of a matrix $A$ is an $n \times n$ Jordan block of the form $$J = \begin{pmatrix} \lambda & 1 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 &\lambda & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \lambda & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \lambda & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \cdots & \lambda & 1 \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & \lambda\end{pmatrix}$$

and I want to show that $\displaystyle e^{AT} = e^{\lambda t} \sum_{k=0}^{n-1}\frac{t^{k}}{k!}(A - \lambda I)^{k}$.

So far, I have shown that

$$e^{At} = e^{(TJT^{-1})}t = e^{\lambda t} \sum_{k=0}^{n-1}\frac{t^{k}N^{k}}{k!} = e^{\lambda t}\sum_{k=0}^{n-k} \frac{t^{k}(J-\lambda I)^{k}}{k!}$$

My question is, how do I finish this? I.e., how do I get $(A-\lambda I)^k$ out of $(J-\lambda I)^{k}$ in my summation?

Thank you for your time and patience.

Answer 1

$$e^{tA} = e^{t(\lambda I + A-\lambda I)} = e^{t\lambda I} e^{t(A-\lambda I)} = e^{\lambda t}\sum_{k=0}^\infty \frac{t^k (A-\lambda I)^k}{k!} = e^{\lambda t} \sum_{k=0}^{n-1} \frac{t^k (A-\lambda I)^k}{k!}$$

as $(A-\lambda I)^n = 0$

Answer 2

By direct calculation, we see that \begin{align} e^{tA} = \sum^\infty_{k=0} \frac{t^kA^k}{k!} = \sum^\infty_{k=0} \frac{t^k (TJT^{-1})^k}{k!} = \sum^\infty_{k=0} \frac{t^k TJ^k T^{-1}}{k!}. \end{align} Like you said \begin{align} J = \lambda I+N \end{align} where $N$ is nilpotent, the we have that \begin{align} J^k = (\lambda I+N)^k = \sum^k_{m=0}\binom{k}{m}N^{k-m}\lambda^m \end{align} which means \begin{align} \sum^\infty_{k=0} \frac{t^kTJ^kT^{-1}}{k!} =&\ T\left(\sum^\infty_{k=0}\frac{1}{k!}\sum^k_{m=0}\binom{k}{m}N^{k-m}\lambda^m\right)T^{-1} = T\left(\sum^\infty_{m=0}\lambda^m\sum^\infty_{k=m} \frac{1}{k!}\binom{k}{m}N^{k-m} \right)T^{-1}\\ =&\ T\left(\sum^\infty_{m=0}\frac{\lambda^m}{m!}\sum^\infty_{k=m}\frac{1}{(k-m)!}N^{k-m} \right)T^{-1} = T\left(\sum^\infty_{m=0}\frac{\lambda^m}{m!}\sum^\infty_{k=0}\frac{N^k}{k!}\right)T^{-1}\\ =&\ T\left(\sum^\infty_{m=0}\frac{\lambda^m}{m!}\sum^{n-1}_{k=0}\frac{N^k}{k!}\right)T^{-1} = T\left(e^{\lambda t}\sum^{n-1}_{k=0}\frac{(J-\lambda I)^k}{k!} \right)T^{-1} = e^{\lambda t}\sum^{n-1}_{k=0}\frac{(A-\lambda I)^k}{k!}. \end{align}