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Let $S$ be the set of positive definite matrices, and $\mathfrak s$ the set of symmetric matrices. The exponential map $\exp: \mathfrak s \rightarrow S$ is a smooth bijection. To argue that it is a diffeomorphism, I would like to define a smooth inverse $\log: S \rightarrow \mathfrak s$. For $B \in S$, the series

$$\log(B) \;\; =\;\; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} (I - B)^k$$ only converges when the eigenvalues of $B$ are between $0$ and $2$. Perhaps a way we can define $\log$ in general is by setting $\log(B) = P^{-1} \log(PBP^{-1})P$, where $P$ is any orthogonal matrix such that $PBP^{-1}$ is diagonal, and for diagonal positive definite matrices we define the logarithm of $\operatorname{diag}(\alpha_1, ... , \alpha_n)$ to be $\operatorname{diag}(\log(\alpha_1), ... , \log(\alpha_n))$.

How can we show that this map is well defined and smooth?

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  • $\begingroup$ By the way, I guess you can also use the series. Just take $\log(B)=\log\left(\frac{1}{\|B\|}B\right)+\log(\|B\|)I$, where $I$ is the identity matrix and $\|B\|$ is the norm of $B$, that is, the largest eigenvalue. $\endgroup$
    – Daniel
    Apr 15 '20 at 19:00
  • $\begingroup$ Thanks. That seems like a pretty good way to do it. Is it obvious that $B \mapsto ||B||$ is smooth on $\operatorname{GL}_n(\mathbb R)$? $\endgroup$
    – D_S
    Apr 15 '20 at 20:33
  • $\begingroup$ Good point, I guess it is not smooth. But anyway, you can always move a neighborhood of any symmetric matrix to the desired interval $(0,2)$. It's probably better to use some constant, so $\log(B)=\log\left(\frac{1}{\alpha}B\right)+\log\alpha\,I$. Then you have to show that this is well defined. $\endgroup$
    – Daniel
    Apr 16 '20 at 6:10
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What you are describing is called functional calculus. You can try to google this term, although you will probably find mostly stuff regarding operators on Hilbert spaces rather then just finite-dimensional matrices.

On finite dimension, this process is very simple. Take any normal matrix $A$ and any continuous function $f\colon\sigma(A)\to\mathbb{C}$, where $\sigma(A)$ is the spectrum of $A$. (In case of positive operators, you may take functions $f\colon \mathbb{R}^+\to\mathbb{C}$.)

Now, what you are suggesting is essentially to choose a basis of eigenvectors $x_i$ corresponding to eigenvalues $\lambda_i$ (so $Ax_i=\lambda_i x_i$) and define a new operator $B$ by $Bx_i=f(\lambda_i)x_i$. This is surely well defined and we can denote $B=:f(A)$

Slightly "more standard" formulation is the following. You can always do so-called spectral decomposition of the operator $A$. This means to write it in the form $A=\sum_i \lambda_i P_i$, where $\lambda_i$ are the eigenvalues of $A$ and $P_i$ are orthogonal projections to the corresponding eigenspaces. Then you can define $$f(A):=\sum f(\lambda_i)P_i.$$

Now you can try to prove all the stuff you need. Namely that the mapping of matrices induced by $f$ is again continuous, how does the derivative look like and so on.

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  • $\begingroup$ You comment should be the answer. The above doesn't address the question. $\endgroup$
    – copper.hat
    Apr 15 '20 at 19:20
  • $\begingroup$ What? He did not ask about the series.He asked about the idea of diagonalizing the matrix and then applying the function on the diagonal entries, which is precisely functional calculus as I described here. $\endgroup$
    – Daniel
    Apr 16 '20 at 6:03

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