I'm trying to show that the exponential map from the symmetric matrices to the positive definite symmetric matrices is a smooth diffeomorphism. To show that the map is smooth, I'm just relying on the fact that it's the sum of smooth maps.

I'm trying to use the inverse function theorem to show that the differential of the exponential map at every $s \in S_n(\mathbb{R})$ is invertible, and therefore that the map has a smooth inverse at every $s \in S_n(\mathbb{R})$, and therefore, that its inverse is smooth globally.

I'm not quite sure how to show this. One idea was to use:

$$ de^{A} = d\sum_{p=0}^{\infty} \frac{1}{p!}A^p = 0 $$ $$ \sum_{p=0}^{\infty} d(\frac{1}{p!}A^p) = 0$$

Because $A$ is a symmetric matrix, we can diagonalize. $$ \sum_{p=0}^{\infty} d(\frac{1}{p!}qA^pq^{-1}) = 0 $$ Then the entries on the diagonal look like $D^p = qAqq^{-1}...qq^{-1}Aq^{-1}$

$$ \sum_{p=0}^{\infty} d(\frac{1}{p!}D^p) = 0 $$ Where $D_{ij} = 0 $ unless $i = j$, and $(\frac{1}{p!}D^p)_{ii} = \frac{1}{p!}(D_{ii})^p $. I'm not sure how best to show that the differential is injective though.


Note that $(*)$ $x\in\mathbb{R}\rightarrow e^x\in (0,+\infty)$ is a $C^{\infty}$ diffeomorphism.

Let $f:X\in S\rightarrow e^X\in S^+$. From the normal convergence of the series associated to $\exp()$, we deduce that $f$ is smooth.

  1. $f$ is one to one. Assume that $A,B\in S$ and $f(A)=f(B)$. We may assume that $A=diag(\lambda_1 I_{\alpha_1},\cdots,\lambda_kI_{\alpha_k})$ where the sequence $\lambda_i$ is real increasing. From $(*)$ and $spectrum(e^U)=e^{spectrum(U)}$, we deduce that $spectrum(A)=spectrum(B)$ (with multiplicities); since $A,B$ are diagonalizable, they are similar: $B=PAP^T$ where $P\in O(n)$ satisfies $Pdiag(e^{\lambda_i})=diag(e^{\lambda_i})P$. Thus $P=diag(P_1,\cdots,P_k)$ where $P_i\in O(\alpha_i)$ and $Pdiag(\lambda_i)=diag(\lambda_i)P$, that is $A=B$.

  2. $f$ is onto. Let $A\in S^+$ and let $B=\log(A)$ ( $\log(A)$, the principal logarithm, is well defined because the eigenvalues of $A$ are $>0$). If $A=Pdiag(\lambda_i)P^T$, then $B=Pdiag(log(\lambda_i))P^T\in S_n$ and, clearly, $e^B=A$.

  3. $f$ is a diffeomorphism. $Df_X:H\in S\rightarrow e^X\dfrac{1-e^{-ad_X}}{ad_X}(H)\in S$ where $ad_X:Y\rightarrow [X,Y]$; note that if $spectrum(X)=(\lambda_i)$, then $spectrum(ad_X)=\{\lambda_i-\lambda_j|i,j\}$ is a subset of $\mathbb{R}$. Remark that $Df_X$ is bijective iff $0\notin spectrum(\dfrac{1-e^{-ad_X}}{ad_X})$ iff $ad_X$ does not admit any eigenvalue in the form $2ki\pi$ where $k$ is a non-zero integer. Clearly, this last condition is fulfilled and we are done.

  • $\begingroup$ This is off topic, but may I ask where you learned linear algebra from. For instance which text or notes could you suggest. $\endgroup$ Sep 29 '16 at 9:47

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.