I am looking for an easy proof of the fact that the exponential function is a diffeomorphism between the finite dimensional vector space of symmetric real nxn-matrices and the open subset of positive definite symmetric real matrices.

I know that the exp-map is a bijection and that it is smooth. So, there are two ways to proceed:

One way would be to use the Inverse Function Theorem. But this would require to calculate the derivative of exp at a matrix A in the direction of another direction B and then we would have to show that this linear map (seen as a linear map in terms of B) is ivertible. I have no clue how to do that. I know that it suffices to show this for a real diagonal matrix A since we can always diagonalize real symmetric matrices, but since it is not possible to diagonalize A and B simultaniously, it is no fun to calculate the matrix exponential...

A totally different way would be not to use the Inverse Function Theorem, but to construct the inverse (the "logarithm") directly and show that it is smooth. One could use the series expansion of the real logarithm and then plug in positive definite matrices, but this seems to be complicated again...

Are there any ideas how to show this without introducing too much complicated machinery?

Thanks very much in advance, Tom


2 Answers 2


Define $\log(A)$ as follows. Since $A$ is positive definite, $A = U^{-1} D U$ for some orthogonal $U$. Then $\log(A) = U^{-1} \log(D) U$, where $\log(D)$ simply applies $\log$ to the diagonal entries. Clearly this is a two sided inverse to $\exp$. To show it is smooth, use $$ \log(A) = \frac1{2\pi i}\oint \log z \, (z I-A)^{-1} \, dz$$ Use a counterclockwise contour that stays in the right half plane, and includes all the eigenvalues of $A$. Use the standard branch of $\log$ that has a cut along the negative real axis. To show the formula, first diagonalize $A$, and then use the Cauchy integral formula on each diagonal entry.

  • $\begingroup$ That's the best I can think of right now. $\endgroup$ Nov 28, 2013 at 14:30
  • $\begingroup$ I think I used an over-complicated way to show it is smooth. Simply use the fact that the local inverse of a smooth, invertible function is itself smooth. $\endgroup$ Jan 24, 2015 at 23:32
  • $\begingroup$ The inverse of a smooth function is smooth? What about f(x)=x^3 ? $\endgroup$
    – Tom
    Jan 28, 2015 at 15:30
  • $\begingroup$ @Tom Oops. I meant: the local inverse of a smooth function whose derivative is invertible is itself smooth. $\endgroup$ Jan 29, 2015 at 4:18
  • 1
    $\begingroup$ @AsafShachar Think of the integrand as a function of the entries in $A$ and $z$. Just by writing out the formula for the inverse, it is clearly smooth in all the entries of $A$. And integrating out $z$ won't stop any of the smoothness, because you can pass the derivatives under the integral sign. $\endgroup$ Jun 26, 2016 at 14:11

Thanks for your answer!

In the meantime, I found another possibility and for completeness sake, I want to show it here:

Let X be a real symmetric matrix. Without loss of generality, we may assume that X is diagonal with entries (lambda_1, ... lambda_n).

We define the following function h of two real variables as: h(x,y):=(exp(x)-exp(y))/(x-y) for x noteq y and h(x,x):=exp(x).

This function is symmetric in x and y and is always >0 .

Now, let e_{i,j} denote the elementary matrix with one 1 at position (i,j) and zeroes otherwise.

Then we may form the directional derivative of the matrix exponential function at point X in the direction e_{i,j} and obtain:

h(lambda_i,lambda_j) * e_{i,j} .

Now, we fix the following basis of the real vector space of symmetric matrices:

e_{i,j}+e_{j,i} for all (i,j) with 1\leq i\leq j \leq n.

This yields:

The directional derivative of exp at point X in direction e_{i,j}+e_{j,i} is h(lambda_i,lambda_j) * (e_{i,j}+e_{j,i}) and therefore each basis vector is mapped to a positive multiple of itself.

So, the linearisation of exp at point X is a diagonalisable with eigenvalues h(lambda_i,lambda_j)>0.

Hence, the linearisation at point X is invertible and hence exp a local diffeo.

Together with the bijectivity, we get a global diffeo.

The only technical thing at this proof is the calculation of the directional derivative at point X in the direction e_{i,j}, which can be reduced to a calculation of an upper triangular 2x2-matrix.

Do you think, this proof works?

Thank you, Tom

  • $\begingroup$ Can you please elaborate on how to show the directional derivative in the direction $e_{ij}$ is what you claim it is? I am stuck with the case $i \neq j$: Take $X=\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \,, \,i=1 \, \,,j=2$. How do you show that $\frac{d}{dt} exp(X+te_{12})=\frac{d}{dt} exp(\begin{pmatrix} \lambda_1 & t \\ 0 & \lambda_2 \end{pmatrix})=h(\lambda_1,\lambda_2) \cdot e_{1,2}$ I understand terms involving $t^2$ and higher powers of $t$ will vanish after taking derivative at $t=0$, but I still do not know how to finish the calculation. $\endgroup$ Jun 26, 2016 at 11:15
  • $\begingroup$ I tried to understand what I did back in 2013 and I think what I meant was the following: For $\lambda_1=\lambda_2$ there is no problem since $X$ and $e_{1,2}$ commute in that case. To evaluate $exp(X + t e_{12})$ for $\lambda_1\neq\lambda_2$ you have to diagonalize $X + t e_{12}$ first. This is quite ugly and is what I called "the only technical thing" in my answer. For the diagonalisation: It is clear what the eigenvallues are and the eigenvector $v_1$ is also obvious. But some nasty calculations remain... I would like to know if there is a more elegant way, though $\endgroup$
    – Tom
    Jun 27, 2016 at 15:00

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