Using the Jordan form Complex Let $C$ be a complex $n \times n$ matrix with $\det C \neq 0$. Prove that there is a complex matrix $B$ such that $C = e^B$
Hint: use the Jordan form matrices for comlexas
 A: Suppose $A$ is the Jordan block of size $n$ for some non-zero eigenvalue $\lambda$. That is, 
$$
A = 
\pmatrix{
\lambda&1&&\\
&\lambda&1&\\
&&\ddots\\
&&&\lambda
}
$$
We compute $e^A = e^\lambda K$, where $K$ is the matrix given by
$$
K=\pmatrix{
1&1&\cdots&1\\
&1&\cdots&1\\
&&\ddots&\vdots\\
&&&1
}
$$
Note that $K$ is similar to the Jordan block of size $n$ corresponding to $1$.  
Let $M$ be the matrix given by
$$
M = 
\pmatrix{
1&e^{-\lambda}&\\
&1&e^{-\lambda}&\\
&&\ddots&\\
&&&1
}
$$
Note that $M$ is also similar to the Jordan block of size $n$ corresponding to $1$.  Thus, $M$ and $K$ are similar.  Take $S$ such that $K = SMS^{-1}$.  We have
$$
e^A = e^\lambda K = e^{\lambda}SMS^{-1} = S(e^{\lambda}M)S^{-1}\\
S^{-1}e^A S = e^{S^{-1}AS} = e^\lambda M
$$
That is, we have
$$
e^{S^{-1}AS} = e^{\lambda}M = 
\pmatrix{
e^{\lambda}&1&\\
&e^{\lambda}&1&\\
&&\ddots&\\
&&&e^{\lambda}
}
$$
So, suppose you begin with a matrix of the form
$$
C = \pmatrix{
e^{\lambda}&1&\\
&e^{\lambda}&1&\\
&&\ddots&\\
&&&e^{\lambda}
}
$$
We may take $B = S^{-1}AS$, with $S$ and $A$ as defined above.
Note that any matrix with a non-zero determinant has a Jordan form consisting of blocks of the above form.
A: The fact that $\det C \ne 0$ implies that all the eigenvalues of $C$ are nonzero; thus when $C$ is transformed to Jordan form via a similarity transformation by the appropriate nonsingular matrix $P$, $C \to P^{-1}CP = D$,  we see that $D = \text{diag}(D_1, D_2, \ldots, D_l)$ is block diagonal consisting of Jordan blocks of the form $D_k = \lambda I_m + N_m$, where $\lambda \ne 0$ is an eigenvalue of $C$ and $N$ may be written
$N_m = [n_{ij}] \; \text{with} \; n_{i, i + 1} = 1; \; n_{ij} = 0, j \ne i + 1, 1 \le i, j \le m, \tag{1}$
and $I_m$ is the $m \times m$ identity matrix.  For each such block $D_k$ we may find a matrix $E_k$ such that $e^{E_k} = D_k$ in the following manner:  set
$F_k = -\sum_1^{m - 1} \dfrac{(-N_m)^p}{p \lambda^p}; \tag{2}$
$F_k$ is derived by considering the power series for $\ln(1 + s)$ for small $s$ and substituting $N_m / \lambda$ for $s$; since $N_m^p = 0$ for $p \ge m$ the resulting series becomes the mere polynomial in $N_m$ given by (2); thus it is well-defined.  We put
$E_k = (\ln \lambda) I_m + F_k, \tag{3}$
and we find, since $I_m$ and $F_k$ commute, that is $I_m F_k = F_k I_m$, that
$e^{E_k} = e^{\ln \lambda} e^{F_k} = \lambda e^{F_k}; \tag{4}$
but
$e^{F_k} = I_m + \dfrac{1}{\lambda}N_m, \tag {5}$
which may be readily seen by noting that $e^{\ln (1 + s)} = 1 + s$, since the algebraic maneuvers required in evaluating the coefficients of the power series of $e^{\ln (1 + s)}$ are the same for $s$ and $N_m / \lambda$, given that $N_m^m = 0$.  Thus
$e^{E_k} = \lambda I_m + N_m = D_k, \tag{6}$
which holds for any Jordan block of $D$; taking 
$E = \text{diag}(E_1, E_2, \ldots, E_l) \tag{7}$
thus yields a matrix such that
$D = e^E; \tag{8}$
then
$C = PDP^{-1} = Pe^EP^{-1} = e^{PEP^{-1}}; \tag{9}$
QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
