Given a matrix $A$ which has the form of a Jordan normal matrix:
$$ A = \left(\begin{array}{cccc}
\log(\phi_1) & 1& …& 0\\
& \log( \phi_2)& & \\
& & \ddots & 1 \\
0 & & & \log (\phi_n) \\
\end{array}\right)$$
how can we prove that the Jordan normal form of $ \exp(A)= VJV^{-1} $ is $J$:
$$ J = \left(\begin{array}{cccc}
\phi_1 & 1& …& 0\\
& \phi_2& & \\
& & \ddots & 1 \\
0 & & & \phi_n \\
\end{array}\right)$$
that has the exact same block form as $A$, but only with the exponential of the eigenvalues?
EDIT:
I tried the $2 \times 2$ case to see if it worked and this is what I got:
$$ A = \left(\begin{array}{cc}
\log(\phi_1) & 1\\
0 & \log( \phi_2) \\
\end{array}\right)$$
We can separate this into to $2$ matricies:
$$ A = \left(\begin{array}{cc}
\log(\phi_1) & 0\\
0 & \log( \phi_2) \\
\end{array}\right)
+
\left(\begin{array}{cc}
0 & 1\\
0 & 0 \\
\end{array}\right)
$$
Exponentiating:
$$ e^A = \left(\begin{array}{cc}
\phi_1 & 0\\
0 & \phi_2 \\
\end{array}\right)
\times
\left[I + \left(\begin{array}{cc}
0 & 1\\
0 & 0 \\
\end{array}\right) \right]
= \left(\begin{array}{cc}
\phi_1 & \phi_1\\
0 & \phi_2 \\
\end{array}\right)$$
Now, its eigenvalues have all algebraic multiplicity of 1 so the Jordan normal form of $e^A$ is:
$$ J = \left(\begin{array}{cc}
\phi_1 & 0\\
0 & \phi_2 \\
\end{array}\right)$$
which is not quite what we wanted, I dont know what to do.
Best Answer
Hint: If $J$ is the $n\times n$ Jordan block with zero eigenvalue, compute the kernel of $exp(J)$.