TITLE: Image of Matrix Exponential Map
QUESTION [7 upvotes]: It is known that every $A$ belongs to $GL(n,\mathbb C)$ equals to $\exp(B)$ for some $n \times n$ matrix $B$. How to show the following is true? Show that a matrix $M$ belonging to $GL_n(\mathbb R)$ is the exponential of a real matrix if, and only if, it is the square of another real matrix.

REPLY [6 votes]: Let $A = B^2$ be an invertible
   real matrix, where $B$ is also real.  We must show that $A$ is an exponential of a real matrix. (The converse is clear, as in the comments above.)
The first step is to conjugate
   $B$ by a real matrix into real Jordan form.  In real Jordan
   form, a matrix is made up of real Jordan blocks, which have the form
$$\left(\begin{matrix} C & 1 & \\ &C & 1 & \\ & & \ldots \\ & & & C &
   1 \\ & & & & C  \end{matrix}\right)$$
Here $C$ can either be a 1 by 1 real scalar, or a 2 by 2 block of the
   form $$\left(\begin{matrix} a & b\\-b& a \end{matrix}\right).$$ (In
   the latter case, the 1s in the matrix must be interpreted as 2 by 2
   identity matrices.  The latter case corresponds to eigenvalues $a \pm
   bi$.)
Since both exponential and squaring commute with both conjugation and
   "blocking", it is sufficient to prove the result for real Jordan
   blocks, i.e., if $B$ is a real Jordan block, then $B^2$ is an
   exponential of a real matrix.
Now it is a matter of checking the various cases.
If the Jordan block $B$ is a 1 by 1 scalar matrix, then its square is
   a positive real number, and the result just says that positive real
   numbers are exponentials, which is true.
If the Jordan block $B$ is $n$ by $n$ consisting of a single nonzero real scalar on the   diagonal (i.e. has $n$ equal nonzero real eigenvalues), then $A = B^2$
   is an upper triangular matrix with a single positive scalar $c$
   repeated on the diagonal.  We must check that any such matrix is an
   exponential.  
A candidate matrix should have form $(\log c)I + J$ where $J$ is
   strictly upper triangular (0's on the main diagonal).  Since  $\exp((\log c)I + J) = c \exp(J)$, it is sufficient to consider the case $c=1$.  
Thus, we are reduced to showing that $\exp$ is surjective from the
   set of strictly upper triangular matrices (which I will denote $T_0$)
   , to upper triangular matrices with 1 on the diagonal (denoted
   $T_1$).  This isn't too hard to do by brute force (by calculating
   explicitly the exponential of an element of $T_0$ ; remember that all
   terms of the exponential series starting with the $n$th vanish on $T_0$).  Alternatively,
   one can use the Baker-Campbell-Hausdorff formula applied to the
   Lie algebra $T_0$.  Because this Lie algebra is nilpotent, repeated
   Lie brackets eventually vanish, so the right hand side of the BCH
   formula becomes finite and therefore converges for all pairs of
   elements in $T_0$.  Hence the image of $T_0$ under
   $\exp$ is closed under multiplication and inversion, and is hence a
   subgroup of $T_1$.  Because $T_1$ is connected, and the image of
   $\exp$ contains a neighborhood of the identity (true for any Lie
   group), it follows that the image must be all of $T_1$. 
The remaining cases are where the Jordan blocks of $B$ consist of 2
   by 2 blocks, with eigenvalues $a \pm bi$.  If it's a single 2 by 2
   block, the result follows from the surjectivity of $\exp$ in
   $\mathbb{C}$ (essentially, because $(a \pm bi)^2$ has a complex
   logarithm).  If the Jordan block is bigger, then we use a similar argument as above: first, reduce to the case where the eigenvalue is 1, and then use the Lie algebra argument as above.