<s>[INST] Examples of how to calculate $e^A$ I'm trying to learn the process to discover $e^A$
For example, if $A$ is diagonalizable is easy:
$$A =\begin{pmatrix}
        5 & -6 \\
        3 & -4 \\
        \end{pmatrix}$$
Then we have the canonical form $$J_A =\begin{pmatrix}
        2 & 0 \\
        0 & -1 \\
        \end{pmatrix}$$
because the auto-values are $2$ and $-1$. 
Am I right? so I continue
$$e^A =P\begin{pmatrix}
        e^2 & 0 \\
        0 & e^{-1} \\
        \end{pmatrix}P^{-1}$$
Where $$P=\begin{pmatrix}
        2 & 1 \\
        1 & 1 \\
        \end{pmatrix}$$
Because the auto-vectors are (2,1) and (1,1).
If the auto-values the things become more complicated:
For example:
$$B =\begin{pmatrix}
        0 & 1 \\
       -2 & -2 \\
        \end{pmatrix}$$
The auto-values are $-1+i$ and $-1-i$, then the canonical form is:
$$J_B =\begin{pmatrix}
        -1 & -1 \\
        1 & -1 \\
        \end{pmatrix}$$
I don't know how to discover $e^B$ in the complex case. How do I have to proceed in this case?
I would like to know also if there are some pdfs or books with examples or problems with solutions about this subject.
I really need help
Thanks a lot.
EDIT
I found another example of a matrix whose some auto-values are complexes:
$$
        C=\begin{pmatrix}
        1 & 0 & -2 \\
        -5 & 6 & 11 \\
        5 & -5 & -10 \\
        \end{pmatrix}
$$
Its canonical form
$$
        J_C=\begin{pmatrix}
        1 & 0 & 0 \\
        0 & -2 & 1 \\
        0 & -1 & -2 \\
        \end{pmatrix}
$$
Why $J_A$ is the matrix $A$ in the canonical form? the author doesn't use complexes numbers, why? How do I find $e^A$ in this case? in the same way?
EDIT 2
The book I'm using says that the complex Jordan block related to the auto-value $a+bi$ is
$$
        \begin{pmatrix}
        a & b \\
        -b & a \\
        \end{pmatrix}
$$
The book also says that it's using the real Jordan canonical form in contrast with the complex Jordan canonical form (see answer below).
 [/INST] You might want to have a look at Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
Other references can be found in The matrix exponential: Any good books?
We have:
$$A = \begin{bmatrix}0 & 1 \\-2 & -2\end{bmatrix}$$
We want to find the characteristic polynomial and eigenvalues by solving
$$|A -\lambda I| = 0 \rightarrow \lambda^2+2 \lambda+2 = (\lambda+(1-i)) (\lambda+(1+i))= 0 \rightarrow \lambda_1 = -1+i, ~~\lambda_2 = -1-i $$
If we try and find eigenvectors, we setup and solve: $[A - \lambda I]v_i = 0$.
So, we have:
$$[A - \lambda_1 I]v_1 = 0 = \begin{bmatrix}1-i & 1\\-2 &-1-i\end{bmatrix}v_1 = 0$$
This leads to a RREF of:
$$\begin{bmatrix}1 & \dfrac{1}{2}(1 +1)\\0 & 0\end{bmatrix}v_1 = 0 \rightarrow v_1 = \left(1, -\dfrac{1}{2}(1 + i)\right)$$
Doing the same process for the second eigenvalue yields: $v_2 = \left(1, -\dfrac{1}{2} + \dfrac{i}{2}\right)$
From all of this information, we can write the matrix exponential using the Jordan Normal Form.
We have the diagonal form:
$$A = P \cdot J\cdot P^{-1} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}-1-i & 0 \\ 0 & -1+i\end{bmatrix} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix}$$
So, now we can take advantage of the diagonal form as:
$\displaystyle e^A = P \cdot e^J \cdot P^{-1} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot e^{\begin{bmatrix}-1-i & 0 \\ 0 & -1+i\end{bmatrix}} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}e^{-1-i} & 0 \\ 0 & e^{-1+i}\end{bmatrix} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix} = \begin{bmatrix} (1/2+i/2) e^{-1-i}+(1/2-i/2) e^{-1+i} & (1/2) i e^{-1-i}-(1/2) i e^{-1+i} \\-i e^{-1-i}+i e^{-1+i} & (1/2-i/2) e^{-1-i}+(1/2+i/2) e^{-1+i} \end{bmatrix}$
Other methods are shown in that paper I referenced above.
If we compare this for the real case, the process is the same and we end up with:
$\displaystyle e^A = P \cdot e^J \cdot P^{-1} = \begin{bmatrix}1 & 2\\1 & 1\end{bmatrix} \cdot e^{\begin{bmatrix}-1 & 0\\0 & 2\end{bmatrix}} \cdot \begin{bmatrix}-1 & 2\\1 & -1\end{bmatrix}= \begin{bmatrix}1 & 2\\1 & 1\end{bmatrix} \cdot {\begin{bmatrix}e^{-1} & 0\\0 & e^2\end{bmatrix}} \cdot \begin{bmatrix}-1 & 2\\1 & -1\end{bmatrix} = \begin{bmatrix}\dfrac{-1+2 e^3}{e} & -\dfrac{2 (-1+e^3)}{e}\\ \dfrac{-1+e^3}{e} & \dfrac{2-e^3}{e}\end{bmatrix}$
It is also worth noting that things can also get very ugly, for example, see: 
generalized eigenvector for 3x3 matrix with 1 eigenvalue, 2 eigenvectors
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