Let $dX/dt=AX$. Find $e^{At}$ where $A=\left[\begin{smallmatrix} 1&1\\0&1\end{smallmatrix}\right]$

Question

I have solved a couple of matrix exponential problems using $PDP^{-1}$ method. However, I am stuck on this problem as it is impossible to find the inverse of a singular matrix.

The problem is :cLet $dX/dt=AX$. Find $e^{At}$ where $A=\begin{bmatrix} 1&1\\0&1\end{bmatrix}$.

My attempt: After $\begin{bmatrix} 1-\lambda&1\\0&1-\lambda\end{bmatrix}$, I find that $\lambda=1,1$. But this would lead to the eigenvector $\begin{bmatrix} 1\\0\end{bmatrix}$. It follows that $\begin{bmatrix} 1&1\\0&0\end{bmatrix}$ does not have an inverse. I am stuck here.

Answer 1

$A$ can be decomposed as $$A=\begin{bmatrix} 1&0\\0&1\end{bmatrix}+\begin{bmatrix} 0&1\\0&0\end{bmatrix}=I+B,$$ notice that $B^2=0,$ hence $e^{tB}=I+tB.$

Answer 2

One can prove by induction (or using Newton's formula) that $$ \forall n\in\mathbb{N},A^n=\begin{pmatrix}1&n\\0&1\end{pmatrix} $$ Thus $$ \exp(tA)=\begin{pmatrix} e^t&\sum_{n=0}^{+\infty}\frac{t^n}{n!}n\\0&e^t \end{pmatrix}=\begin{pmatrix}e^t&te^t\\0&e^t\end{pmatrix} $$

Answer 3

Use the Jordan–Chevalley decomposition: $$A=D+N$$ $$\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}+\begin{bmatrix}0&1\\0&0\end{bmatrix}$$ The first addend is diagonalisable (and in this case already diagonal), the second is nilpotent and the two commute. Then $$e^A=e^De^N$$ and $e^N$ can be worked out directly from the series definition of the matrix exponential because $N^2=\mathbf0$.

You should get $$e^A=\begin{bmatrix}e&e\\0&e\end{bmatrix}$$