I need to show if the following matrices are the exponential of a real matrix. $$A_1=\begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}\\ A_2=\begin{bmatrix}-1 & 1 \\ 0 & -1\end{bmatrix}\\ A_3=\begin{bmatrix}-1 & 0 \\ 0 & -4\end{bmatrix}\\ A_4=\begin{bmatrix}-1 & -1\\ 0 & -1\end{bmatrix}\\ A_5=\begin{bmatrix}A_2& 0 \\ 0 & A_2\end{bmatrix}\\ A_6=\begin{bmatrix}A_2& I_2 \\ 0 & A_2\end{bmatrix}$$
I know the following results about the exponential of a matrix :
- $\exp(M_n(\mathbb{R}))=\{A\in Gl_n(\mathbb{R}),\exists B\in M_n(\mathbb{R}),A=B^2\}$
- $M$ is diagonalisable if and only if $\exp(M)$ is diagonalisable
- if $M$ is the exponential of a real matrix then the negative eigen values of $M$ have an even multiplicity
For the matrix $A_1$ I managed to show that the equation $A_1=B^2$ has a solution in $M_n(\mathbb{R})$, so $A_1$ is the exponential of a real matrix.
For the matrix $A_2$, I assume that the matrix is the exponential of a real matrix, $A_2=\exp(B)$, let $\lambda_1,\lambda_2$ the eigen values of $B$, then $e^{\lambda_1}=e^{\lambda_2}=-1$, then $\lambda_1,\lambda_2 \in i\pi+2i\pi\mathbb{Z}$. Two cases, if $\lambda_1=\lambda_2$, these two values are complex non real, so they are conjugated which is impossible, second case, then $B$ has two different eigen values, so $B$ is diagonalisable, which is impossible because $A$ is not diagonalisable. So $A$ is not the exponential of a real matrix.
For the matrix $A_3$, there are two negative eigen values of odd multiplicity so it can't be the exponential of a real matrix.
For the matrix $A_4$, I can apply the same reasoning than for $A_2$.
Apparently I should be able to prove that $A_5$ and $A_6$ are the exponential of real matrices. Is there an easier method to prove that than trying to solve the equation $A=B^2$. Also is there an universal method to prove this kind of result ?
Proof of the first fact.
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.
If we have $A=B^2$ with $B$ real, $\det(B)^2=\det(A)\neq0$, so $B$ is invertible. So with the lemma $\exists P \in \mathbb{C}[X]$ such that $B=\exp(P(B))$. $B$ is real so : $$A=\overline BB=\overline{\exp(P (B))}\exp(P(B))=\exp(\overline{P}(B))\exp(P(B))=\exp((\overline{P}+P)(B))$$ Because $\overline{P}(B)$ and $P(B)$ commute. The polynomial $\overline{P}+P$ is real, so $A\in \exp(M_n(\mathbb{R}))$