# How to find the exponential of a nilpotent matrix?

I want to find the exponential $e^{tA}$, where

$A=\left( \begin{array}{ c c } 0 & 1 & 2 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right).$

I know that its eigenvalues are $0,0,0$, its determinant is $0$, and the first vector is $[1,0,0]$. How do I compute the exponential?

• What is the question? – Kevin Carlson Dec 17 '14 at 3:36

I'll assume your question is something like "Given $A = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & 2 \\0 & 0 & 0\end{pmatrix}$, what is $e^{At}$?"
The definition of a matrix exponential is $e^{tA} = \displaystyle\sum_{n = 0}^{\infty}\dfrac{t^n}{n!}A^n = I + tA + \dfrac{t^2}{2!}A^2 + \dfrac{t^3}{3!}A^3 + \cdots$.
In your title, you said that $A$ is nilpotent. Specifically, $A^3 = 0$. So, $A^n = 0$ for all $n \ge 3$.
Hence, all but the first $3$ terms in the series $I + tA + \dfrac{t^2}{2!}A^2 + \dfrac{t^3}{3!}A^3 + \cdots$ are zero.
Can you compute the first $3$ terms and add them up?