Find $\exp\bigg[t\left ( \begin{matrix} -1& 1 \\ 0 & -1 \\ \end{matrix} \right )\bigg]$ by definition

Question

Find $\exp\bigg[t\left ( \begin{matrix} -1& 1 \\ 0 & -1 \\ \end{matrix} \right )\bigg]$ by definition

Denote $A=\left ( \begin{matrix} -1& 1 \\ 0 & -1 \\ \end{matrix} \right )$

I find out that $$A^n=\left ( \begin{matrix} -1& n \\ 0 & -1 \\ \end{matrix} \right ) \forall n\in\mathbb{N} \text{ such that } n\mod 2 =1$$

$$A^n=\left ( \begin{matrix} 1& -n \\ 0 & 1 \\ \end{matrix} \right ) \forall n\in\mathbb{N} \text{ such that } n\mod 2 =0$$

Denote $k=2n,j=2n+1$.

$e^{tA}=\sum_{k=1}^\infty \frac{t^k}{k!}A^k + \sum_{j=1}^\infty \frac{t^j}{j!}A^j$

I got a bit confused how to find $e^{tA}$ by definition

Thanks !

Answer 1

From your insight, it may be more helpful to rewrite the powers of $A$ (you made a slight error) as

$$A^n = (-1)^n \begin{pmatrix} 1 & -n \\ 0 & 1 \end{pmatrix}$$

So,

$$\sum_{n=0}^{\infty} \frac{t^n}{n!}A^n = \sum_{n=0}^\infty \frac{(-t)^n}{n!} \begin{pmatrix} 1 & -n \\ 0 & 1 \end{pmatrix} = \sum_{n=0}^\infty \begin{pmatrix} \frac{(-t)^n}{n!} & -\frac{(-t)^n}{(n-1)!} \\ 0 & \frac{(-t)^n}{n!} \end{pmatrix}$$

Then you can compute these sums component-wise using the Taylor expansion for $e$.

Answer 2

Actually,$$A^n=\begin{bmatrix}(-1)^n&(-1)^{n-1}n\\0&(-1)^n\end{bmatrix}.$$So,$$(tA)^n=\begin{bmatrix}(-1)^nt^n&(-1)^{n-1}nt^n\\0&(-1)^nt^n\end{bmatrix},$$and therefore\begin{align}\exp(tA)&=\sum_{n=0}^\infty\frac1{n!}\begin{bmatrix}(-1)^nt^n&(-1)^{n-1}nt^n\\0&(-1)^nt^n\end{bmatrix}\\&=\begin{bmatrix}e^{-t}&te^{-t}\\0&e^{-t}\end{bmatrix}.\end{align}