Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was asked to show that the exponential map $\exp: \mathfrak{g} \mapsto G$ is not surjective by proving that the matrix $\left(\matrix{-1 & 0 \\ 0 & -2}\right)\in \text{GL}(2,\mathbb{R})$ can't be the exponential of any matrix $A \in \mathfrak{gl}(2,\mathbb{R})$.

My proof (edited)

Lemma: A diagonal matrix $M \in \rm{GL}(2,\mathbb{R})$ is the exponential of a matrix $A\in\mathfrak{gl}(2,\mathbb{R})$ if it has positive eigenvalues or a unique (double) negative eigenvalue.

For any diagonal matrix $A = \left(\matrix{a & 0 \\ 0 & d}\right)$, we have $\displaystyle\text{e}^A=\sum_{k=0}^\infty\dfrac{A^k}{k!} = \left(\matrix{\text{e}^a & 0 \\ 0 & \text{e}^d}\right)$. Then, $$\forall\; \lambda,\,\mu > 0, \qquad\left(\matrix{\lambda & 0 \\ 0 & \mu}\right) = \exp \left(\matrix{\ln\lambda & 0 \\ 0 & \ln\mu}\right)$$

By the Cayley-Hamilton theorem, every square matrix is a root of its characteristic polynomial. Thus I can express any power of $A$ as a linear combination of $A$ itself and the identity matrix. For instance, $A^2= tA-d\mathbb{I}\;$ and $A^3=(t^2-d)A-td\mathbb{I}\;$, where $t=\text{tr} A, \; d=\det A$.
It follows that $M=\text{e}^A=\alpha A+\beta\mathbb{I}\;$ (even if I don't know how to calculate those coefficients).

  • If $M$ has distinct negative eigenvalues, $\alpha \neq 0$: otherwise we'd have $M = \beta \mathbb{I}$, a contradiction. This implies that $A$ must be diagonal too, but then it can't exponentiate to something with negative eigenvalues (see below the conclusion of my exercise).

  • If $M$ has a double negative eigenvalue, it is a negative multiple of $\mathbb{I}$ and $\alpha$ must be $0$. Observing that $\exp\left(\matrix{0 & \pi \\ -\pi & 0}\right)=\left(\matrix{-1 & 0 \\ 0 & -1}\right)=-\mathbb{I}$ and experimenting a bit, it is straightforward to prove that $$\forall\; \lambda>0, \qquad -\lambda\mathbb{I}=\left(\matrix{-\lambda & 0 \\ 0 & -\lambda}\right)=\exp\left(\matrix{\ln\lambda & \pm\, m\pi \\ \mp\, m\pi & \ln\lambda}\right), \qquad \forall m \; \text{odd integer}$$

The conclusion is immediate: the diagonal matrix I'm given must be the exponential of a diagonal matrix (if it exists) $A = \left(\matrix{a & 0 \\ 0 & d}\right)$ such that $\text{e}^a=-1\;\text{and}\;\text{e}^d=-2$, which is impossible.

Alternative approach, problems and questions

My efforts were aimed at using that $\exp(PXP^{-1})=P\exp XP^{-1},\; \forall X\in \mathfrak{gl}(n,\mathbb{R}),\forall P\in\text{GL}(n,\mathbb{R})$. The problem is that if $\exp X$ is a diagonal non-scalar matrix, then it doesn't lie in the center of the algebra and thus doesn't commute with $P$.
That's a pity, because I can (almost) always diagonalize $A=\left(\matrix{a & b \\ c & d}\right)$, writing $A=PXP^{-1}$, where $X$ is diagonal:

  1. If $\det A = 0$, since $\text{tr}A \neq 0$, $A$ has two distinct eigenvalues ($0$ and $\text{tr}A$) and is diagonalizable
  2. If $\det A \neq 0$, the characteristic equation is $\lambda^2-\text{tr}A\lambda +\det A = 0$ and

    • if its discriminant is $\Delta \neq 0$, then again I get two (possibly complex) distinct eigenvalues and $A$ is diagonalizable.
    • if $\Delta = 0$, the only eigenvalue is $\lambda=\dfrac12 \text{tr}A$, whose geometric multiplicity is only $1$, making $A$ not diagonalizable.

This line of thought seems quite promising to me. I was wondering if it exists some way to conclude the proof of the "diagonalizable $A$" case from $P\left(\matrix{\text{e}^{\lambda_1} & 0 \\ 0 & \text{e}^{\lambda_2}}\right)P^{-1} = \left(\matrix{-1 & 0 \\ 0 & -2}\right)$ and then to handle the "non-diagonalizable $A$" case.

Bonus question

I was also trying to work with the other property $\exp(X+Y)=\exp X \exp Y \quad \text{iff}\quad [X,Y]=0$, exploiting some decomposition of $A$, such as $$A = \left(\matrix{a & b \\ c & d}\right) = A_{\text{sym}}+A_{\text{antisym}}\;, \quad\text{where}\quad A_{\text{sym}}=\frac12(A+A^T) \;,\;A_{\text{antisym}}=\frac12(A-A^T)$$ $$\text{or} \quad A = \left(\matrix{a & b \\ c & d}\right) = A_{\text{diag}}+A_{\text{antidiag}}\;, \quad\text{where}\quad A_{\text{diag}}=\left(\matrix{a & 0 \\ 0 & d}\right) \;,\;A_{\text{antisym}}=\left(\matrix{0 & b \\ c & 0}\right)\;. $$

Unfortunately, both decomposition don't help, because $[A_{\text{sym}},A_{\text{antisym}}] \neq 0$, $[A_{\text{diag}},A_{\text{antidiag}}] \neq 0$.

Is there a chance that something like this could lead to a solution?

Update: Thanks to Pink Elephants, who pointed out a flaw in my proof, I modified the statement of the lemma and added some detail. I think now it works well for my exercise.

share|cite|improve this question
Try $A = \left(\matrix{1 & 0 \\ 0 & -2}\right)$. And show that this cannot be an exponential, since exponential cannot be negative in $\mathbb{R}$. – Jayesh Badwaik Sep 22 '12 at 15:35
That easily follows from $\det(\rm{e}^A)=\rm{e}^{\rm{tr}A}$, because your matrix has negative determinant. But I'm asking - more specifically - about the non surjectivity onto $\rm{GL}^+(2,\mathbb{R})$, i.e. the component of $\rm{GL}(2,\mathbb{R})$ connected to the identity, which is subtler. ;) – Andrea Orta Sep 22 '12 at 15:41
I believe there is a problem with your Lemma. When you write $e^A=\alpha A+\beta I$, if $\alpha$ is 0, then we can't conclude $A$ is diagonal just from the fact that $e^A$ is. I believe $A=\left(\matrix{0 & \pi \\ -\pi & 0}\right)$ is a counterexample. $A$ is similar (over $\mathbb{C}$) to $\left(\matrix{\pi i & 0 \\ 0 & -\pi i}\right)$, so $e^A=\left(\matrix{-1 & 0 \\ 0 & -1}\right)$ – Julian Rosen Sep 22 '12 at 15:49
See this question and the answers given there. – Marc van Leeuwen Sep 22 '12 at 16:22
@Pink Elephants: thanks for your useful comment! The statement of the lemma was clearly wrong, I've weakened it. Now it provides a classification of the matrices in $\text{GL}(2,\mathbb{R})$ which can be the exponential of matrices of $\mathfrak{gl}(2,\mathbb{R})$. Marc: thanks for your link! – Andrea Orta Sep 23 '12 at 16:08
up vote 7 down vote accepted

I looked at the question linked by Marc van Leeuwen, but thought I would outline a different (though fairly standard) approach. It works for complex matrices. I'll write $\exp(A)$ for $e^{A}$, when $A$ is an $n \times n$ complex matrix. Note that $\exp(MAM^{-1}) = M\exp(A)M^{-1}$ for any invertible $n \times n$ matrix $M.$ Since $MAM^{-1}$ is in Jordan normal form for a suitable matrix $M,$ we first try to understand $\exp(A)$ when $A$ is in Jordan normal form. In that case, on considering the Jordan blocks of $A$ one at a time, we may write $A = D+N$ where $D$ is diagonal, $N$ is upper triangular with $0$s on the main diagonal, and $DN = ND.$ Then it is clear that $\exp(A) = \exp(D)\exp(N).$ Now $\exp(A)$ is diagonal, and $\exp(N)$ is upper triangular with $1$s on the main diagonal. It follows that if $R$ is the set of roots of the characteristic polyomial of $A,$ then $\exp(R)$ is the set of roots of the characteristic polynomial of $\exp(A),$ with algebraic multiplicities matching as expected. Since $MAM^{-1}$ and $Mexp(A)M^{-1}$ have the same characteristic polynomials as $A$ and $\exp(A)$ respectively, the same statement holds for any complex $n \times n$ matrix. This is all general theory.

Now we can ask if there is a real $2 \times 2$ matrix $A$ such that $\exp(A) = \left( \begin{array}{clcr}-1 & 0\\0 & -2 \end{array} \right)$. If $A$ has eigenvalues $\alpha$ and $\beta$ (possibly equal), then $\exp(A)$ has eigenvalues $e^{\alpha}$ and $e^{\beta}.$ Hence we may suppose that $e^{\alpha} = -1$ and $e^{\beta} = -2.$ Hence $\alpha$ and $\beta$ are certainly non-real complex numbers. However, $A$ is a real matrix, so $\alpha$ and $\beta$ are the roots of the same real quadratic equation. Hence they must be complex conjugates. But then $e^{\alpha}$ and $e^{\beta}$ have the same absolute value, which is a contradiction.

share|cite|improve this answer
Nice answer, +1. For this particular case you don't need to consider Jordan normal forms though: if $v$ is an eigenvector of $A$ for $\lambda$ then it is also an eigenvector of $\exp A$ for $\exp\lambda$, so with the given matrix having two distinct negative eigenvalues, $A$ would have to have two distinct non-real and non-conjugate eigenvalues; your last paragraph applies directly. – Marc van Leeuwen Sep 24 '12 at 13:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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