3
$\begingroup$

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 $A\in \exp(M_n(\mathbb{R}))$

$\endgroup$
  • 1
    $\begingroup$ But $A_1 = \pmatrix{0&-1\\1&0}^2$ $\endgroup$ – Omnomnomnom Feb 15 '17 at 16:23
  • 2
    $\begingroup$ Do you know about Jordan canonical form? $\endgroup$ – Omnomnomnom Feb 15 '17 at 16:24
  • $\begingroup$ I think you have much of the conceptual framework for this problem, but your reasoning above is hurried, and you may be rushing past solutions without seeing them. It is greatly preferred not to put so many separate problem parts into one Question as this discourages Readers from starting with the misconceptions encountered at the beginning. $\endgroup$ – hardmath Feb 15 '17 at 16:28
  • $\begingroup$ Oh yes indeed I made a mistake with $A_1$. Yes I know the Jordan canonical form. $\endgroup$ – Jennifer Feb 15 '17 at 16:30
  • $\begingroup$ I'm confused by your reasoning for $A_1$: why does $A_1 = B^2$ (for some real $B$) imply that $A_1$ is the exponential of a real matrix? Of course, the converse holds. $\endgroup$ – Omnomnomnom Feb 15 '17 at 17:21
2
$\begingroup$

$A$ is the exponential of a real matrix iff, in the Jordan form of $A$:

  • When $\lambda$ is complex (non-real), the Jordan blocks of $\lambda$ are exactly the same as those associated with $\overline{\lambda}$
  • When $\lambda$ is negative, any Jordan blocks associated with $\lambda$ appear twice

$A_5$ is already in Jordan form, so that's easy.

$A_6$ satisfies $$ rk(A_6 + I) = 2,\\ rk(A_6 + I)^2 = 1,\\ (A_6 + I)^3 = 0 $$ So, $A_6$ has Jordan form $$ A_6 = \pmatrix{-1&1\\&-1&1\\&&-1\\&&&-1} $$ so, it is not the exponential of a real matrix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.