Matrix Proof

I want show that e^{(At)} = Se^{(\Lambda t)} S^{-1} , where S = (v_{1} v_{2}) ,  A is a 2 \times 2 matrix, and \Lambda = \begin{pmatrix} \lambda_{1} & 0 \\0 & \lambda_{2} \end{pmatrix}

I start by writing the middle sum in summation notation which gives me Se^{(\Lambda t)}S^{-1} = S( \displaystyle\sum_{k = 0}^{\infty} \frac{1}{k!}(\Lambda t)^{k})S^{-1} . Now I can use the identity S^{-1}AS = \Lambda which will then give me S(\displaystyle\sum_{k = 0}^{\infty} \frac{1}{k!}(S^{-1}AS)^{k}t^{k})(S^{-1}) . After pulling terms out of the sum, I will get SS^{-1} (\displaystyle \sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!})SS^{-1} . The SS^{-1} terms create an identity matrix and the middle sum is equivalent to e^{At} as shown below.


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