Wikipedia says that
"If $A = UDU^{−1}$ and D is diagonal, then $e^{A} = Ue^{D}U^{−1}$"
Why is this the case?
I understand that $e^D$ yields a matrix where $M_{i,j} = e^{D_{i,j}}$, but how is it possible that we are able to factor $U$ and $U^{-1}$ out? I can not seem to find the logic anywhere.
On
It is know that $t \mapsto e^{tA}$ is the only solution of the differential equation $$X' =AX, \quad X(0) =1.$$ If you prove that $Ue^{tD}U^{-1}$ is a solution of $X'=UDU^{-1}X$, then by uniqueness you will have $$e^{tUDU^{-1}}=Ue^{tD}U^{-1}$$ and you will get the conclusion by evalutating at $t=1$.
On
The expression $e^A$ does not refer to the matrix such that $(e_A)_{i,j} = e^{A_{i,j}}$, just as the expression $A^2$ doesn't mean the matrix with $i,j$-entry equal to $(A_{i,j})^2$.
What is meant by $e^A$ is the matrix $$ e^A = \sum^\infty_{n=0} \frac{A^n}{n!}. $$
This is the limit of the matrices $B_n := \sum^n_{i=1} \frac{A^i}{i!}$. You do know that each $U D^i U^{-1} = (U D U^{-1})^i = A^i$, and you know that the diagonal entry $D^i_{j,j} = (D_{j,j})^i$. So $$ U^{-1} (B_n)_{j,j} U = \sum^n_{i=1} \frac{(D_{j,j})^i}{i!}, $$ and hence $\lim_{n \to \infty} U^{-1} (B_n)_{j,j} U = e^{D_{j,j}}$. That is, $\lim_{n \to \infty} U^{-1} B_n U = e^{D}$, and since multiplication of matrices is continuous, we obtain $$ U^{-1} (\lim_{n \to \infty} B_n) U = e^{D} $$ Multiplying on the left by $U$ and on the right by $U^{-1}$, we get $$ e^A = U e^D U^{-1}. $$
$$e^A=I+A+\frac{1}{2}A^2+\cdots$$ $$=UU^{-1}+UDU^{-1}+\frac{1}{2}UDU^{-1}UDU^{-1}+\cdots$$ $$=UU^{-1}+UDU^{-1}+\frac{1}{2}UD^2U^{-1}+\cdots$$ $$=U\bigg(I+D+\frac{1}{2}D^2+\cdots\bigg)U^{-1}$$ $$Ue^DU^{-1}$$