How does the Real Canonical Form help solve the matrix exponential? I am trying to solve the system $\vec{x'}=$$ \left[
  \begin{array}{ c c }
     -13 & -10 \\
     20 & 15
  \end{array} \right]
$ $\vec{x}$
I found that the canonical form for the matrix is 
$$ \left[
  \begin{array}{ c c }
     1 & 2 \\
     -2 & 1
  \end{array} \right]
$$ 
since the eigenvalues were $1\pm 2i$. I am trying to finish the initial problem using this canonical form I found. I am confused what its purpose is to be completely honest..
Thanks
 A: We are given
$$x'= \begin{bmatrix}
     -13 & -10 \\
     20 & 15
  \end{bmatrix}
x$$
Method 1 (M1): Using Eigenvalues / Eigenvectors 
We find the eigenvalues from the characteristic polynomial given by $|A - \lambda I| = 0$ as
$$\lambda_{1,2} = 1~ \pm~ 2 i$$
The corresponding eigenvectors are
$$v_{1, 2} = (-7 ~ \pm ~i,10)$$
We can now write $A = P_{M1} D P_{M1}^{-1}$ where $P_{M1}$ are the column eigenvectors and $D$  the diagonal matrix of eigenvalues. Because we have unique eigenvalues, $D$ is a diagonal matrix (no Jordan block).
This easily allows us to find the matrix exponential as 
$$e^{A t} = P_{M1} e^{D t} P_{M1}^{-1} = e^t \begin{bmatrix}
  \cos (2 t)-7  \sin (2 t) & -5  \sin (2 t) \\
 10  \sin (2 t) & \cos (2 t)+7  \sin (2 t) \\
\end{bmatrix}$$
Method 2 (M2): Real Canonical Form
(Baker–Campbell–Hausdorff Theorem): Suppose we attempt to define a matrix
$C$ by $e^C = e^A~e^B$. If the commutator $[A, B] = AB − BA = 0$, then $C = A + B$.
If a matrix can be written as a sum of two commuting matrices whose exponentials
can be computed, then exponentiation is easy because we can exploit $[aI, b \sigma] = 0$ using the previous theorem. For example,
$$B = \begin{bmatrix}
     a & b \\
     -b & a
  \end{bmatrix} = a I + b \sigma \implies e^{B t} = e^{a ~I ~t} e^{b~ \sigma~ t} = e^{at}\begin{bmatrix}
     \cos(bt) & \sin(b t) \\
     -\sin(bt) & \cos(b t)
  \end{bmatrix}$$
We already found the eigenvalues and eigenvectors above. Using $\lambda_1 = 1 + 2 i$ and $v_1 = (-7 + ~i,10)$, we set $v = u + i w$ as
$$u = \begin{bmatrix}
     -7 \\
     10
  \end{bmatrix}, w = \begin{bmatrix}
     1 \\
     0
  \end{bmatrix} \implies P_{M2} = \begin{bmatrix}
     -7 & 1 \\
    10 & 0
  \end{bmatrix} \implies P_{M2}^{-1} = \begin{bmatrix}
 0 & \dfrac{1}{10} \\
 1 & \dfrac{7}{10} \\
\end{bmatrix}$$
We can now write the Real Canonical Form as 
$$B = P_{M2}^{-1} A P_{M2} = 
\begin{bmatrix}
 1 & 2 \\
 -2 & 1 \\
\end{bmatrix} \implies e^{B t} = e^t\begin{bmatrix}
 \cos(2t) & \sin(2t) \\
 -\sin(2t) & \cos(2t) 
\end{bmatrix}$$
We now compute
$$e^{At} = P_{M2} e^{B t} P_{M2}^{-1} = 
e^t
\begin{bmatrix}
  \cos (2 t)-7 \sin (2 t) & -5  \sin (2 t) \\
 10  \sin (2 t) & \cos (2 t)+7 \sin (2 t) \\
\end{bmatrix}$$
Compare both results.
