Matrix Proof



Given the matrix equation \dot{u} = Au , where u = \begin{pmatrix} x \\ y \end{pmatrix} and A is a 2 \times 2 matrix, I want to show that \dot{u} = e^{At}u_{0} is the solution where e^{At} = I + At + \frac{1}{2!}A^{2}t^{2} + \frac{1}{3!}A^{3}t^{3} + ... and I is the identity matrix.

I start out by writing the above series in summation notation which gives me \displaystyle\sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!}. I can now take a time derivative of the sum to give me \frac{d}{dt}e^{At} = \displaystyle\sum_{k = 0}^{\infty}\frac{kt^{k - 1}A^{k}}{k!} . Since the first term of the series after differentiation is 0 , I can rewrite and reduce the sum to give me \displaystyle\sum_{k = 1}^{\infty}\frac{t^{k - 1}A^{k}}{(k - 1)!} . Now, I can pull out a single matrix term to give me A \displaystyle\sum_{k = 1}^{\infty}\frac{t^{k - 1}A^{k - 1}}{(k - 1)!} . I can now simplify the sum once again to give me A \displaystyle\sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!} . This is now equivalent to Ae^{At} . Assuming that u = e^{At}u_{0}  is the solution, I can differentiate it with respect to time to give me \dot{u} = \frac{d}{dt}[e^{At}]u_{0} . I just showed that \frac{d}{dt} e^{At} = Ae^{At} , which I can plug in to give me the equation \dot{u} = Ae^{At}u_{0} . Since e^{At}u_{0} = u , I am left with my original matrix equation \dot{u} = Au