# 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.