In order to find $e^{AT}$ in order to find $e^{AT}$ We can't just take the exponential of A as we would do in its diagonalized form. We need to diagonalize $A=S^{-1}e^{\delta(t)}S$ in order to find $e^{AT}$ why is this the case? I know we can't take the exponential of the matrix right away, do we need to take the exponential of the diagonal and multiply by $S$ in order to reach the answer all the time?
If yes, why?
 A: It's not that you can't write down the series
$$
\exp(tA) = \sum_{k=0}^{\infty} \frac{t^{k} A^{k}}{k!},
\tag{1}
$$
it's that the entries of $A^{k}$ usually aren't easy to express in terms of the entries of $A$ (try it yourself for a $2 \times 2$ matrix!), so (1) isn't an explicit description of the entries of $\exp(tA)$.
By contrast, diagonalizing $A = S^{-1}DS$ (or generally, putting $A$ into Jordan canonical form) permits the entries of $A^{k} = S^{-1}D^{k}S$ to be calculated, so that
$$
\exp(tA) = S^{-1} \exp(tD) S
\tag{2}
$$
is a useful, relatively explicit description.
A: In general,
$$
e^M = I + M + \frac{M^2}{2!} + \frac{M^3}{3!} + \ldots
$$
but computing this sum involves multiplying $M$ by itself many many times and summing up the results and is generally computationally horrible, even if we only go out 10 terms or so. 
But observe that
\begin{align}
Q^{-1}M^2Q 
&= Q^{-1}M(QQ^{-1})M Q \\
&= (Q^{-1}MQ)(Q^{-1}M Q)
\end{align}
and a similar trick works for $M^3$, or $M^4$, or any power. So for any invertible matrix $Q$, we can write
\begin{align}
Q^{-1}e^MQ 
&= Q^{-1}(I + M + \frac{M^2}{2!} + \frac{M^3}{3!} + \ldots) Q\\
&= Q^{-1}IQ + Q^{-1}MQ + Q^{-1}\frac{M^2}{2!}Q + Q^{-1}\frac{M^3}{3!}Q + \ldots\\
&= I + Q^{-1}MQ + \frac{(Q^{-1}MQ)^2}{2!} + \frac{(Q^{-1}MQ)^3}{3!} + \ldots
&= e^ {Q^{-1}MQ}.
\end{align}
So suppose you're trying to exponentiate $A$, and it's a pain. But you know a matrix $Q$ with $Q^{-1}AQ$ diagonal, say $D$. Then you can compute
$$
e^{Q^{-1}AQ} = e^D
$$
But you also know that this is the same as $Q^{-1} e^A Q$. So you can write
$$
Q^{-1} e^A Q = e^D\\
e^A = Q e^D Q^{-1}.
$$
Now computing $e^A$ is easy: you exponentiate the diagonal matrix (easy! just exponentiate each diagonal entry!) and then conjugate by $Q$. 
So the reason we do this is that exponentiating diagonals is easy (computationally and theoretically), and reducing general exponentiation to this simple case makes the general case easier. 
Of course, to have this work, you need to know such a diagonalizing matrix $Q$, which isn't trivial (and in general may not exist...but does if $A$ is symmetric, for instance). But it may well be easier than computing the exponential without this diagonalization trick. 
A: Definitions with matrices are not that obvious as you might expect. Consider for example the matrix inverse. If $x$ is a number, its inverse $x^{-1}$ should satisfy $x x^{-1} = 1$. A simple computation gives of course $x^{-1} = \frac{1}{ x}$. With a matrix $A$, this is not as simple, We require $A A^{-1} = I$, where $I$ is the identity matrix. In this case, what should $\frac 1 A$ mean? It is not the inverse of each of its entries (in general).  However, with diagonal matrices, things are asier, because the inverse of a diagonal matrix is indeed the inverse of its diagonal entries,
$$\begin{pmatrix}a & 0 & 0\\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix} ^{-1} = \begin{pmatrix}a^{-1} & 0 & 0\\ 0 & b^{-1} & 0 \\ 0 & 0 & c^{-1} \end{pmatrix}$$
A similar thing holds with exponentials. The exponential of a diagonal matrix is the exponent of its diagonal entries,
$$\exp\begin{pmatrix}a & 0 & 0\\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix} = \begin{pmatrix}e^a & 0 & 0\\ 0 & e^b & 0 \\ 0 & 0 & e^c \end{pmatrix}.$$
If you do a diagonalization,  $A^k = U^{-1}D^kU$, we can follow the definition of the matrix exponential, which is given by the infinite summation,
$$\exp(tA) = \sum_{k=0}^{\infty} \frac{t^{k} A^{k}}{k!} = U^{-1}\left(\sum_{k=0}^{\infty} \frac{t^{k} D^{k}}{k!} \right)U = U^{-1}e^{tD}U,$$
And Yes! We simplified the infinite summation to something we know: the exponential of a diagonal matrix. 
