$ e^{At}$ for $A = B^{-1} \lvert \cdots \rvert B $ For a homework problem, I have to compute $ e^{At}$ for
$$ A = B^{-1}  \begin{pmatrix}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \end{pmatrix} B$$
I know how to compute the result for $2 \times 2$ matrices where I can calculate the eigenvalues, but this is $3 \times 3$, and I cannot compute eigenvalues, so is there any identity or something which allows computing such exponentials?
Thanks!
 A: Joseph is right and  $ D^n=\begin{pmatrix}(-1)^n&0&0\\0&2^n&0\\0&0&3^n \end{pmatrix}so (Dt)^n=\begin{pmatrix}(-t)^n&0&0\\0&(2t)^n &0\\0&0&(3t)^n\end{pmatrix}$ so $\left(I+Dt+\frac{1}{2!}(Dt)^2+\frac{1}{3!}(Dt)^3+\cdots\right)=\begin{pmatrix}e^{-t}&0&0\\0&e^{2t}&0\\0&0&e^{3t}\end{pmatrix}$
A: Note that $A$ has been diagonalized, and that: $$e^{At}=I+At+\frac{1}{2!}(At)^2+\frac{1}{3!}(At)^3+\cdots  $$ $$ \Longrightarrow e^{BDB^{-1}t}=I+BDB^{-1}t+\frac{1}{2!}(BDB^{-1}t)^2+\frac{1}{3!}(BDB^{-1}t)^3+\cdots $$ $$ =BB^{-1}+B(Dt)B^{-1}+\frac{1}{2!}B(Dt)^2B^{-1}+\frac{1}{3!}B(Dt)^3B^{-1}+\cdots $$ $$ =B\left(I+Dt+\frac{1}{2!}(Dt)^2+\frac{1}{3!}(Dt)^3+\cdots\right)B^{-1}$$ What is the expression in the middle?
Edit: Forgot the $t$.
A: Joseph G. has it right.  In addition, you say, "... I cannot compute eigenvalues" but the eigenvalues are simply the solutions to
$\left|\begin{array}{ccc}
-1-\lambda& 0& 0 \\
0& 2-\lambda& 0 \\
0& 0& 3-\lambda \end{array}\right|=0$
which boils down to $(-1-\lambda)(2-\lambda)(3-\lambda)$, so $\lambda=-1,2,3$.
