Exponential of matrix if $A^2=A$

Question

So the question is: if $A^2=A$ where A is a matrix find $e^A$. I know I should start with series expansion: $$e^A=1+A+ \frac{A^2}{2!} + \frac{A^3}{3!} +... $$ but I don't know how to proceed. Thanks!

Answer 1

Since $A^2=A$, then $A^3=AA^2=AA=A^2=A$, and in general it follows that $A^n=A$. Thus $$e^A=I+A+\frac{A}{2!}+\frac{A}{3!}+\cdots=I+A(1+\tfrac{1}{2!}+\tfrac1{3!}+\cdots)=\boxed{I+(e-1)A}.$$

Answer 2

By induction, we know $A^n = A$ for all $n \geq 1$. Thus \begin{align*} e^A &= I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots \\ &=I+(\sum_{n=0}^{\infty}\frac{1}{n!}-1)A \\ &=I+(e-1)A\end{align*}