Show that any invertible matrix has a logarithm. I was trying to remember how to show that any invertible matrix has a (possibly complex) logarithm. I thought what I came up with was kind of cool, so I thought I'd post my answer here.
 A: It suffices to show that each $\lambda$-Jordan block has a logarithm if $\lambda \neq 0$. 
First note that the exponential of a Jordan block is 
$$
\left[\begin{matrix}
\lambda     & 1         & 0          &  \dotsb     & 0      \\
            & \ddots    & \ddots     &  \ddots     & \vdots \\          
            &           & \ddots     &  \ddots     & 0      \\
            &           &            &  \ddots     & 1      \\
            &           &            &             & \lambda\\
\end{matrix}\right]
\overset{\exp}{\mapsto} 
\left[\begin{matrix}
e^{\lambda} & e^{\lambda} & \frac{e^{\lambda}}{2!} & \dotsb      & \frac{e^{\lambda}}{(k-1)!}  \\
       & e^{\lambda}      & \ddots                 & \ddots      & \vdots                      \\
       &             & \ddots                 & e^{\lambda} & \frac{e^{\lambda}}{2!}      \\
       &             &                        & e^{\lambda} & e^{\lambda}                 \\[7pt]
       &             &                        &             & e^{\lambda}                 \\
\end{matrix}\right].
$$
So, reversing the process, given a Jordan block 
$$J=\left[\begin{matrix} 
a & 1 & \dotsb & 0       \\
  & a & \ddots & \vdots  \\
  &   & \ddots & 1       \\
  &   &        & a
\end{matrix}\right]
$$
with $a\neq 0$, we want to show that $J$ is similar to 
$$
\left[\begin{matrix} 
a\quad & a           & \frac{a}{2!}           & \dotsb      & \frac{a}{(k-1)!}            \\
       & a           & \ddots                 & \ddots      & \vdots                      \\
       &             & \ddots                 & a           & \frac{a}{2!}                \\
       &             &                        & a           & a                           \\[7pt]
       &             &                        &             & a                           \\
\end{matrix}\right],
$$
since we know how to find the logarithm of the above matrix, and since $\log(UMU^{-1})=U\log(M)U^{-1}$.
Now, since the scalar matrix $aI$ has the same form with respect to any basis, we can neglect this part, and just ask if we can conjugate
$$\left[\begin{matrix} 
0 & 1 & \dotsb & 0       \\
  & 0 & \ddots & \vdots  \\
  &   & \ddots & 1       \\
  &   &        & 0
\end{matrix}\right] \tag{$M_1$}
$$
into
$$
\left[\begin{matrix} 
0\quad & a           & \frac{a}{2!}           & \dotsb      & \frac{a}{(k-1)!}            \\
       & 0           & \ddots                 & \ddots      & \vdots                      \\
       &             & \ddots                 & a           & \frac{a}{2!}                \\
       &             &                        & 0           & a                           \\[7pt]
       &             &                        &             & 0                           \\
\end{matrix}\right]. \tag{$M_2$}
$$
First of all, we can see algebraically that these two matrices should be similar, since $M_2= aN + aN^2 + \dots + \dfrac{a}{(k-1)!}N^{k-1}$, where $N$ is the elementary nilpotent matrix of size $k$ (here $N=M_1$, incidentally). Taking powers of $M_2$ shows that $M_2^{k-1}\neq 0 \,\, M_2^k=0$. Therefore $M_1$ and $M_2$ have the same minimal polynomial $x^k$. I claim they also have the same characteristic polynomial, also $x^k$. The only possibility for the invariant factors of  $M_1$ and $M_2$ is $1, 1, \dotsc, 1, x^k$. Therefore they are similar. 
That said, I thought I would tackle the more general problem: Suppose we want to conjugate
$$\left[\begin{matrix} 
0 & 1     & \dotsb      & 0          \\
  & 0     & \ddots      & \vdots     \\
  &       & \ddots      & 1          \\
  &   &        & 0
\end{matrix}\right] \tag{$M_3$}$$ 
into
$$\left[\begin{matrix}
0 & a_{12} & a_{13} & \dotsb   & a_{1k}   \\
  & 0      & a_{23} & \dotsb   & a_{2k}   \\
  &        & \ddots & \ddots   & \vdots   \\
  &        &        & \ddots   & a_{k-1,k}\\
  &        &        &          & 0        
\end{matrix}\right]\tag{$M_4$}$$
Given that $\mathcal{B}=\{v_1, \dotsc, v_k\}$ is an ordered basis such that $M_3 = [T]_{\mathcal{B}\mathcal{B}}$, conjugating $M_3$ into $M_4$ is equivalent to finding an ordered basis $\mathcal{C} = \{w_1, \dotsc, w_k\}$ such that $[T]_{\mathcal{C}\mathcal{C}}=M_4$.
Let $v_1, \dotsc, v_k$ be the basis in $M_1$. Define $w_1, \dotsc, w_k$ as follows: let $w_1=v_1$. Now suppose 
$$w_{i-1}=\sum_{1\leq j\leq i-1}c_jv_{j}$$
Then let 
$$w_i = \sum_{1\leq j\leq i-1}a_{n-j}c_jv_{j+1}.$$
We need to make sure $w_i$ are linearly independent. I claim they will be iff $M_3$ is similar to $M_4$ iff $M_4^{k-1}\neq 0$. One condition which will guarantee this is: 
$$\text{All elements on the superdiagonal of $M_4$ are nonzero.}$$
Perhaps someone can come up with some better conditions for it.
