The matrix $A=\pmatrix{2&-2&-4\\ -1&3&4\\ 1&-2&-3}$ is idempotent i.e. $A^2=A$. And I have a system of differential equations $\dot{v}(t)=Av$ with initial conditions $v(0)=(1,1,1)^T$.
I know the solution to this is given by $v = e^{tA}v(0)$. I can diagonalize A and solve it like that. The question asks me to solve it using the idempotence property. How?
If $A^2=A$ then $$ e^{At} = \sum_{k=0}^\infty \frac 1{k!} t^kA^k = I + \left(\sum_{k=1}^\infty \frac1{k!} t^k\right)A= I + (e^t-1)A. $$