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

Every invertible square matrix with complex entries can be written as the exponential of a complex matrix. I wish to ask if it is true that

Every invertible real matrix with positive determinant can be written as the exponential of a real matrix. (We need +ve determinant condition because if $A=e^X$ then $\det A=e^{\operatorname{tr}(X)} > 0$.) If not is there a simple characterization of such real matrices (with +ve determinant) which are exponentials of other matrices ?

share|cite|improve this question
As far as I know there exists an open neighborhood $U$ in $T_IGL_n({\bf R})$ such that ${\rm exp}|_U$ is a diffeomorphism. That is, any matrix around $I$ having small pertubation can be written by exponential. – HK Lee Oct 11 '13 at 9:25
up vote 7 down vote accepted

No, a real matrix has a real logarithm if and only if it is nonsingular and in its (complex) Jordan normal form, every Jordan block corresponding to a negative eigenvalue occurs an even number of times. So, it is possible that a matrix with positive determinant is not the matrix exponential of a real matrix. Here are two counterexamples: $\pmatrix{-1&1\\ 0&-1}$ and $\operatorname{diag}(-2,-\frac12,1,\ldots,1)$. For more details, see

Walter J. Culver, On the existence and uniqueness of the real logarithm of a matrix, Proceedings of the American Mathematical Society, 17(5): 1146-1151, 1966.

share|cite|improve this answer
Thanks! This is very useful as the criterion is not hard to check for a given matrix. This is exactly what I was looking for. – user90041 Oct 11 '13 at 16:16

Another characterization is as follows: $A$ is the exponential of a real matrix iff $A$ is the square of a real invertible matrix. In particular, remark that if $A=e^X$, then $A=(e^{X/2})^2$.

Concerning the Lee's post, if $A$ is in a neighborhood of $I$, then $A=I+B$ with $||B||<1$ and we can take $X=B-B^2/2+B^3/3+\cdots$.

EDIT: An outline of the proof. Let $A=B^2$ where $B$ is invertible real; we may assume that $B$ is in Jordan form. For each Jordan block $B_k=\lambda I_k+J_k$ of $B$ s.t. $\lambda<0$, change $B_k$ with $-B_k$. Finally you obtain a matrix $C$ s.t. $C^2=A$. It is not difficult to show that such a matrix $C$ which has no $<0$ eigenvalues is the exp of a real matrix.

share|cite|improve this answer
Thanks! But could you please give a reference where this is proved ? I am unable to see immediately why A should be exponential of a real matrix if it is a square of another real matrix. – user90041 Oct 11 '13 at 15:02

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.