If A is a real non-singular square matrix, then there exists a real matrix $B$ such that $e^B$ $=$ $A^2$ If we consider the matrix exponential map on $M_n(\mathbb R)$, then what will be the image set of the exponential map?
I have seen this. From there I can say that $exp(M_n(\mathbb C))=GL_n(\mathbb C)$. But what about the set $exp(M_n(\mathbb R))$?  
In the exercise 66 of this link I have $exp(M_n(\mathbb R))\neq GL_n(\mathbb R)$. Even it says that :
If A is a real non-singular square matrix, then there exists a real matrix B such that $e^B=A^2$
How can I prove this statement? 
Any help please.
Thanks in advance.
 A: Let me prove a lemma first.

Lemma : Let $M\in GL_n(\mathbb{C}), \exists P\in \mathbb{C}[X]$ such that $M=\exp(P(M))$.
With Dunford decomposition $M=D+N$, where $D,N$ are polynomials in $M$ and commute. Thus it exist $P$ a complex polynom such that : $$D=Pdiag(\lambda_1,\dots,\lambda_n)P^{-1}$$
The complex exponential is surjective thus $\forall i,\exists \nu_i$ such that $e^{\nu_i}=\lambda_i$. Let $Q$ the Lagrange polynomial such that $Q(\lambda_i)=\nu_i$.
$$Q(D)=PQ(diag(\lambda_1,\dots,\lambda_n))P^{-1}=Pdiag(\nu_1,\dots,\nu_n)P^{-1}$$
$$\exp(Q(D))=Pdiag(e^{\nu_1},\dots,e^{\nu_n})P^{-1}=D$$. 
$Q(D)$ is a polynomial in $M$, so the lemma is true if the matrix is diagonalisable.
Also $M$ is invertible and $D$ has the same eigen values so $D$ is invertible. So $(D^{-1}N)^k=D^{-k}N^k$, $N$ is nilpotent so $D^{-1}N$ is also nilpotent. Thus : 
$$I_n+D^{-1}N=\exp(\sum_{k=1}^n \frac{(-1)^{k-1}}{k}(D^{-1}N)^k)$$
Also we have the characteristic polynomial $$\chi_D=(-1)^nX^n+\sum_{k=0}^{n-1}a_kX^k$$
$a_0\neq0$, because $D$ is invertible, so with Cayley-Hamilton theorem :
$$D^{-1}=-\frac{1}{a_0}((-1)^nD^{n-1}+\sum_{k=1}^{n-1}a_kD^{k-1})$$
$D$ and $N$ are polynomials in $M$ so $\exists B\in \mathbb{C}[X]$ such that $I_n+D^{-1}N=\exp(B(M))$.
Finally :
$$M=D(I_n+D^{-1}N)=\exp(A(M))\exp(B(M))=\exp((A+B)(M))$$
because $A(M)$ and $B(M)$ commute. The lemma is proved.

Now we will prove that $\exp(M_n(\mathbb R))=\{A\in GL_n(\mathbb{R}) | \exists B \in M_n(\mathbb R),A=B^2\}$.
Let $A \in \exp (Mn(\mathbb{R}))$, then $\exists B  \in Mn(\mathbb{R})$ such that $A=\exp(B)$, so $A = (\exp(\frac 1 2 B))^2$ is the square of an invertible matrix so it is invertible.
Let show the other inclusion. If $A\in GLn(\mathbb R))$ such that $\exists B \in M_n(\mathbb R)$ and $A=B^2$, so we have $\det(B)^2=\det(A)\neq 0$ so $B$ is invertible. By the lemma $\exists P \in \mathbb C [X]$ such that $B=\exp(P(B))$. Since $B$ is real we have $A=B\overline B=\exp(P(B))\overline{\exp(P(B))}=\exp(P(B))\exp(\overline P(B))=\exp((P + \overline P)(B))$ because $P(B)$ and $\overline P (B)$ commute. And since $P+\overline P$ is a real polynomial we have $A\in \exp(Mn(\mathbb R))$. 
We have proved the equality of the two sets.

Finaly we can prove that $\{A \in GL_n(\mathbb{R}) | \exists B \in M_n(\mathbb R),A=B^2\} \neq GLn(\mathbb R)$ by taking the matrix $\begin{pmatrix} 
1 & 0 \\
0  & -1 
\end{pmatrix}$. Indeed it is invertible because its determinant is not zero but it can't be the square of a real matrix beacause its determinant is strictly negative.
